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In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

In Maple, once you have the integral you might use holexprtodiffeq in the gfun package to find a linear differential equation it satisfies.

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty. It's probably best to leave these in implicit form.

Unfortunately, it looks like holexprtodiffeq will not handle expressions involving EllipticF. In principle, I guess the differential operator should be determined by the partial differential equations satisfied by EllipticF. $$\eqalign{ {\frac {\partial ^{2}}{\partial {k}^{2}}}f \left( x,k \right) &+{\frac { \left( 3\,{k}^{2}-1 \right) }{{k}^{3}-k}} {\frac {\partial }{\partial k}}f \left( x,k \right) +{\frac { \left( -{x}^{3}+x \right) }{ (k^2-1)(k^2 x^2-1)}} {\frac {\partial }{\partial x}}f \left( x,k \right)+{\frac {f \left( x ,k \right) }{{k}^{2}-1}}=0 \cr {\frac {\partial ^{2}}{\partial x \partial k}}f \left( x,k \right) & +{\frac { \left( {\frac {\partial }{ \partial x}}f \left( x,k \right) \right) k{x}^{2}}{{k}^{2}{x}^{2}-1}} =0\cr {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,k \right) &-{ \frac { \left( -2\,{k}^{2}{x}^{3}+ \left( {k}^{2}+1 \right) x \right) {\frac {\partial }{\partial x}}f \left( x,k \right) }{1+{k}^{2}{x}^{4} + \left( -{k}^{2}-1 \right) {x}^{2}}}=0} $$

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

In Maple, once you have the integral you might use holexprtodiffeq in the gfun package to find a linear differential equation it satisfies.

For example, this tells me that $$\int_0^1 \frac{dt}{\sqrt{(z-t)^2 + t + 1}} = \ln(3+2\sqrt{z^2-2z+3}-2z) - \ln(1 + 2 \sqrt{z^2+1}-2z)$$ satisfies the differential equation $$ \left( 16\,{z}^{7}+4\,{z}^{6}+4\,{z}^{5}-592\,{z}^{4}+352\,{z}^{3}- 254\,{z}^{2}-128\,z+138 \right) {\frac {\rm d}{{\rm d}z}}f \left( z \right) + \left( 32\,{z}^{8}-8\,{z}^{7}-17\,{z}^{6}-501\,{z}^{5}+400 \,{z}^{4}-790\,{z}^{3}-265\,{z}^{2}-177\,z+126 \right) {\frac {{\rm d} ^{2}}{{\rm d}{z}^{2}}}f \left( z \right) + \left( 8\,{z}^{9}-6\,{z}^{8 }+11\,{z}^{7}-53\,{z}^{6}+89\,{z}^{5}-155\,{z}^{4}-103\,{z}^{3}-135\,{ z}^{2}-189\,z-27 \right) {\frac {{\rm d}^{3}}{{\rm d}{z}^{3}}}f \left( z \right) =0$$

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty. It's probably best to leave these in implicit form.

Unfortunately, it looks like holexprtodiffeq will not handle expressions involving EllipticF. In principle, I guess the differential operator should be determined by the partial differential equations satisfied by EllipticF. $$\eqalign{ {\frac {\partial ^{2}}{\partial {k}^{2}}}f \left( x,k \right) &+{\frac { \left( 3\,{k}^{2}-1 \right) }{{k}^{3}-k}} {\frac {\partial }{\partial k}}f \left( x,k \right) +{\frac { \left( -{x}^{3}+x \right) }{ (k^2-1)(k^2 x^2-1)}} {\frac {\partial }{\partial x}}f \left( x,k \right)+{\frac {f \left( x ,k \right) }{{k}^{2}-1}}=0 \cr {\frac {\partial ^{2}}{\partial x \partial k}}f \left( x,k \right) & +{\frac { \left( {\frac {\partial }{ \partial x}}f \left( x,k \right) \right) k{x}^{2}}{{k}^{2}{x}^{2}-1}} =0\cr {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,k \right) &-{ \frac { \left( -2\,{k}^{2}{x}^{3}+ \left( {k}^{2}+1 \right) x \right) {\frac {\partial }{\partial x}}f \left( x,k \right) }{1+{k}^{2}{x}^{4} + \left( -{k}^{2}-1 \right) {x}^{2}}}=0} $$

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty.

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

In Maple, once you have the integral you might use holexprtodiffeq in the gfun package to find a linear differential equation it satisfies.

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty. It's probably best to leave these in implicit form.

Unfortunately, it looks like holexprtodiffeq will not handle expressions involving EllipticF. In principle, I guess the differential operator should be determined by the partial differential equations satisfied by EllipticF. $$\eqalign{ {\frac {\partial ^{2}}{\partial {k}^{2}}}f \left( x,k \right) &+{\frac { \left( 3\,{k}^{2}-1 \right) }{{k}^{3}-k}} {\frac {\partial }{\partial k}}f \left( x,k \right) +{\frac { \left( -{x}^{3}+x \right) }{ (k^2-1)(k^2 x^2-1)}} {\frac {\partial }{\partial x}}f \left( x,k \right)+{\frac {f \left( x ,k \right) }{{k}^{2}-1}}=0 \cr {\frac {\partial ^{2}}{\partial x \partial k}}f \left( x,k \right) & +{\frac { \left( {\frac {\partial }{ \partial x}}f \left( x,k \right) \right) k{x}^{2}}{{k}^{2}{x}^{2}-1}} =0\cr {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,k \right) &-{ \frac { \left( -2\,{k}^{2}{x}^{3}+ \left( {k}^{2}+1 \right) x \right) {\frac {\partial }{\partial x}}f \left( x,k \right) }{1+{k}^{2}{x}^{4} + \left( -{k}^{2}-1 \right) {x}^{2}}}=0} $$

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t= \pm 2\,{\frac {1}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) } $$

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty.