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Replced factor 2 by 4 in Hypergeom. function
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Johannes Trost
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To wrap things up here is the result for the OP's integral: $$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{2\ a} \right) $$$$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{4\ a} \right) $$

To wrap things up here is the result for the OP's integral: $$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{2\ a} \right) $$

To wrap up here is the result for the OP's integral: $$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{4\ a} \right) $$

Source Link
Johannes Trost
  • 1.3k
  • 11
  • 17

To wrap things up here is the result for the OP's integral: $$ \int_{0}^{\infty} dr \ e^{-a r^2}\ L_{k}^{1}(b \ r^2) \ J_{1}( c\ r)\ r^d = \\ \frac{(k+1)! \ c}{4 \ a^{\frac{d}{2}+1}}\sum_{n=0}^{k}\left(-\frac{b}{a}\right)^n \ \frac{\Gamma(n+\frac{d}{2}+1)}{(k-n)!\ n!\ (n+1)!} \ _{1}F_{1}\left(n+\frac{d}{2}+1,2;-\frac{c^2}{2\ a} \right) $$