Integrand size = 23, antiderivative size = 323 \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=-\frac {3 e^{-d+\frac {b^2 \log ^2(f)}{4 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{16 \sqrt {f-c \log (f)}}+\frac {e^{-3 d+\frac {b^2 \log ^2(f)}{12 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (3 f-c \log (f))}{2 \sqrt {3 f-c \log (f)}}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 e^{d-\frac {b^2 \log ^2(f)}{4 (f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (f+c \log (f))}{2 \sqrt {f+c \log (f)}}\right )}{16 \sqrt {f+c \log (f)}}+\frac {e^{3 d-\frac {b^2 \log ^2(f)}{4 (3 f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (3 f+c \log (f))}{2 \sqrt {3 f+c \log (f)}}\right )}{16 \sqrt {3 f+c \log (f)}} \] Output:
-3/16*exp(-d+b^2*ln(f)^2/(4*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(b*ln(f)-2* x*(f-c*ln(f)))/(f-c*ln(f))^(1/2))/(f-c*ln(f))^(1/2)+1/16*exp(-3*d+b^2*ln(f )^2/(12*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(b*ln(f)-2*x*(3*f-c*ln(f)))/(3* f-c*ln(f))^(1/2))/(3*f-c*ln(f))^(1/2)-3/16*exp(d-b^2*ln(f)^2/(4*f+4*c*ln(f )))*f^a*Pi^(1/2)*erfi(1/2*(b*ln(f)+2*x*(f+c*ln(f)))/(f+c*ln(f))^(1/2))/(f+ c*ln(f))^(1/2)+1/16*exp(3*d-b^2*ln(f)^2/(12*f+4*c*ln(f)))*f^a*Pi^(1/2)*erf i(1/2*(b*ln(f)+2*x*(3*f+c*ln(f)))/(3*f+c*ln(f))^(1/2))/(3*f+c*ln(f))^(1/2)
Time = 4.54 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.56 \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\frac {e^{-\frac {b^2 \log ^2(f) (2 f+c \log (f))}{2 (f+c \log (f)) (3 f+c \log (f))}} f^a \sqrt {\pi } \left (3 e^{\frac {1}{4} b^2 \log ^2(f) \left (\frac {1}{f-c \log (f)}+\frac {1}{f+c \log (f)}+\frac {1}{3 f+c \log (f)}\right )} \text {erf}\left (\frac {2 f x-(b+2 c x) \log (f)}{2 \sqrt {f-c \log (f)}}\right ) \sqrt {f-c \log (f)} \left (9 f^3+9 c f^2 \log (f)-c^2 f \log ^2(f)-c^3 \log ^3(f)\right ) (\cosh (d)-\sinh (d))-(f-c \log (f)) \left (e^{\frac {1}{4} b^2 \log ^2(f) \left (\frac {1}{3 f-c \log (f)}+\frac {1}{f+c \log (f)}+\frac {1}{3 f+c \log (f)}\right )} \text {erf}\left (\frac {6 f x-(b+2 c x) \log (f)}{2 \sqrt {3 f-c \log (f)}}\right ) \sqrt {3 f-c \log (f)} \left (3 f^2+4 c f \log (f)+c^2 \log ^2(f)\right ) (\cosh (3 d)-\sinh (3 d))+(3 f-c \log (f)) \left (3 e^{\frac {b^2 \log ^2(f)}{12 f+4 c \log (f)}} \text {erfi}\left (\frac {2 f x+(b+2 c x) \log (f)}{2 \sqrt {f+c \log (f)}}\right ) \sqrt {f+c \log (f)} (3 f+c \log (f)) (\cosh (d)+\sinh (d))-e^{\frac {b^2 \log ^2(f)}{4 (f+c \log (f))}} \text {erfi}\left (\frac {6 f x+(b+2 c x) \log (f)}{2 \sqrt {3 f+c \log (f)}}\right ) (f+c \log (f)) \sqrt {3 f+c \log (f)} (\cosh (3 d)+\sinh (3 d))\right )\right )\right )}{16 \left (9 f^4-10 c^2 f^2 \log ^2(f)+c^4 \log ^4(f)\right )} \] Input:
Integrate[f^(a + b*x + c*x^2)*Sinh[d + f*x^2]^3,x]
Output:
(f^a*Sqrt[Pi]*(3*E^((b^2*Log[f]^2*((f - c*Log[f])^(-1) + (f + c*Log[f])^(- 1) + (3*f + c*Log[f])^(-1)))/4)*Erf[(2*f*x - (b + 2*c*x)*Log[f])/(2*Sqrt[f - c*Log[f]])]*Sqrt[f - c*Log[f]]*(9*f^3 + 9*c*f^2*Log[f] - c^2*f*Log[f]^2 - c^3*Log[f]^3)*(Cosh[d] - Sinh[d]) - (f - c*Log[f])*(E^((b^2*Log[f]^2*(( 3*f - c*Log[f])^(-1) + (f + c*Log[f])^(-1) + (3*f + c*Log[f])^(-1)))/4)*Er f[(6*f*x - (b + 2*c*x)*Log[f])/(2*Sqrt[3*f - c*Log[f]])]*Sqrt[3*f - c*Log[ f]]*(3*f^2 + 4*c*f*Log[f] + c^2*Log[f]^2)*(Cosh[3*d] - Sinh[3*d]) + (3*f - c*Log[f])*(3*E^((b^2*Log[f]^2)/(12*f + 4*c*Log[f]))*Erfi[(2*f*x + (b + 2* c*x)*Log[f])/(2*Sqrt[f + c*Log[f]])]*Sqrt[f + c*Log[f]]*(3*f + c*Log[f])*( Cosh[d] + Sinh[d]) - E^((b^2*Log[f]^2)/(4*(f + c*Log[f])))*Erfi[(6*f*x + ( b + 2*c*x)*Log[f])/(2*Sqrt[3*f + c*Log[f]])]*(f + c*Log[f])*Sqrt[3*f + c*L og[f]]*(Cosh[3*d] + Sinh[3*d])))))/(16*E^((b^2*Log[f]^2*(2*f + c*Log[f]))/ (2*(f + c*Log[f])*(3*f + c*Log[f])))*(9*f^4 - 10*c^2*f^2*Log[f]^2 + c^4*Lo g[f]^4))
Time = 0.93 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6038, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sinh ^3\left (d+f x^2\right ) f^{a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 6038 |
| \(\displaystyle \int \left (-\frac {1}{8} e^{-3 d-3 f x^2} f^{a+b x+c x^2}+\frac {3}{8} e^{-d-f x^2} f^{a+b x+c x^2}-\frac {3}{8} e^{d+f x^2} f^{a+b x+c x^2}+\frac {1}{8} e^{3 d+3 f x^2} f^{a+b x+c x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt {\pi } f^a e^{\frac {b^2 \log ^2(f)}{4 f-4 c \log (f)}-d} \text {erf}\left (\frac {b \log (f)-2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{16 \sqrt {f-c \log (f)}}+\frac {\sqrt {\pi } f^a e^{\frac {b^2 \log ^2(f)}{12 f-4 c \log (f)}-3 d} \text {erf}\left (\frac {b \log (f)-2 x (3 f-c \log (f))}{2 \sqrt {3 f-c \log (f)}}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 \sqrt {\pi } f^a e^{d-\frac {b^2 \log ^2(f)}{4 (c \log (f)+f)}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+f)}{2 \sqrt {c \log (f)+f}}\right )}{16 \sqrt {c \log (f)+f}}+\frac {\sqrt {\pi } f^a e^{3 d-\frac {b^2 \log ^2(f)}{4 (c \log (f)+3 f)}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+3 f)}{2 \sqrt {c \log (f)+3 f}}\right )}{16 \sqrt {c \log (f)+3 f}}\) |
Input:
Int[f^(a + b*x + c*x^2)*Sinh[d + f*x^2]^3,x]
Output:
(-3*E^(-d + (b^2*Log[f]^2)/(4*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(b*Log[f] - 2*x*(f - c*Log[f]))/(2*Sqrt[f - c*Log[f]])])/(16*Sqrt[f - c*Log[f]]) + ( E^(-3*d + (b^2*Log[f]^2)/(12*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(b*Log[f] - 2*x*(3*f - c*Log[f]))/(2*Sqrt[3*f - c*Log[f]])])/(16*Sqrt[3*f - c*Log[f]] ) - (3*E^(d - (b^2*Log[f]^2)/(4*(f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[(b*Log[ f] + 2*x*(f + c*Log[f]))/(2*Sqrt[f + c*Log[f]])])/(16*Sqrt[f + c*Log[f]]) + (E^(3*d - (b^2*Log[f]^2)/(4*(3*f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[(b*Log[ f] + 2*x*(3*f + c*Log[f]))/(2*Sqrt[3*f + c*Log[f]])])/(16*Sqrt[3*f + c*Log [f]])
Int[(F_)^(u_)*Sinh[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sinh[v] ^n, x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[ v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 1.06 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-3 f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-3 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-12 d \ln \left (f \right ) c -36 d f}{4 \left (3 f +c \ln \left (f \right )\right )}}}{16 \sqrt {-c \ln \left (f \right )-3 f}}+\frac {\operatorname {erf}\left (-x \sqrt {3 f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {3 f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+12 d \ln \left (f \right ) c -36 d f}{4 \left (c \ln \left (f \right )-3 f \right )}}}{16 \sqrt {3 f -c \ln \left (f \right )}}-\frac {3 \,\operatorname {erf}\left (-x \sqrt {f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+4 d \ln \left (f \right ) c -4 d f}{4 \left (c \ln \left (f \right )-f \right )}}}{16 \sqrt {f -c \ln \left (f \right )}}+\frac {3 \,\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-4 d \ln \left (f \right ) c -4 d f}{4 \left (f +c \ln \left (f \right )\right )}}}{16 \sqrt {-c \ln \left (f \right )-f}}\) | \(326\) |
Input:
int(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/16*erf(-(-c*ln(f)-3*f)^(1/2)*x+1/2*ln(f)*b/(-c*ln(f)-3*f)^(1/2))/(-c*ln (f)-3*f)^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2-12*d*ln(f)*c-36*d*f)/(3* f+c*ln(f)))+1/16*erf(-x*(3*f-c*ln(f))^(1/2)+1/2*ln(f)*b/(3*f-c*ln(f))^(1/2 ))/(3*f-c*ln(f))^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2+12*d*ln(f)*c-36* d*f)/(c*ln(f)-3*f))-3/16*erf(-x*(f-c*ln(f))^(1/2)+1/2*ln(f)*b/(f-c*ln(f))^ (1/2))/(f-c*ln(f))^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2+4*d*ln(f)*c-4* d*f)/(c*ln(f)-f))+3/16*erf(-(-c*ln(f)-f)^(1/2)*x+1/2*ln(f)*b/(-c*ln(f)-f)^ (1/2))/(-c*ln(f)-f)^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2-4*d*ln(f)*c-4 *d*f)/(f+c*ln(f)))
Leaf count of result is larger than twice the leaf count of optimal. 852 vs. \(2 (275) = 550\).
Time = 0.10 (sec) , antiderivative size = 852, normalized size of antiderivative = 2.64 \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^3,x, algorithm="fricas")
Output:
1/16*((sqrt(pi)*(c^3*log(f)^3 + 3*c^2*f*log(f)^2 - c*f^2*log(f) - 3*f^3)*c osh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 36*d*f + 12*(c*d + a*f)*log(f))/(c*log( f) - 3*f)) + sqrt(pi)*(c^3*log(f)^3 + 3*c^2*f*log(f)^2 - c*f^2*log(f) - 3* f^3)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 36*d*f + 12*(c*d + a*f)*log(f))/( c*log(f) - 3*f)))*sqrt(-c*log(f) + 3*f)*erf(-1/2*(6*f*x - (2*c*x + b)*log( f))*sqrt(-c*log(f) + 3*f)/(c*log(f) - 3*f)) - 3*(sqrt(pi)*(c^3*log(f)^3 + c^2*f*log(f)^2 - 9*c*f^2*log(f) - 9*f^3)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) - f)) + sqrt(pi)*(c^3*log(f)^3 + c^2*f*log(f)^2 - 9*c*f^2*log(f) - 9*f^3)*sinh(-1/4*((b^2 - 4*a*c)*log(f) ^2 - 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) - f)))*sqrt(-c*log(f) + f)*er f(-1/2*(2*f*x - (2*c*x + b)*log(f))*sqrt(-c*log(f) + f)/(c*log(f) - f)) + 3*(sqrt(pi)*(c^3*log(f)^3 - c^2*f*log(f)^2 - 9*c*f^2*log(f) + 9*f^3)*cosh( -1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f - 4*(c*d + a*f)*log(f))/(c*log(f) + f )) + sqrt(pi)*(c^3*log(f)^3 - c^2*f*log(f)^2 - 9*c*f^2*log(f) + 9*f^3)*sin h(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f - 4*(c*d + a*f)*log(f))/(c*log(f) + f)))*sqrt(-c*log(f) - f)*erf(1/2*(2*f*x + (2*c*x + b)*log(f))*sqrt(-c*log (f) - f)/(c*log(f) + f)) - (sqrt(pi)*(c^3*log(f)^3 - 3*c^2*f*log(f)^2 - c* f^2*log(f) + 3*f^3)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 36*d*f - 12*(c*d + a*f)*log(f))/(c*log(f) + 3*f)) + sqrt(pi)*(c^3*log(f)^3 - 3*c^2*f*log(f)^ 2 - c*f^2*log(f) + 3*f^3)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 36*d*f - ...
\[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sinh ^{3}{\left (d + f x^{2} \right )}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)*sinh(f*x**2+d)**3,x)
Output:
Integral(f**(a + b*x + c*x**2)*sinh(d + f*x**2)**3, x)
Time = 0.06 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.89 \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 3 \, f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) - 3 \, f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) + 3 \, f\right )}} + 3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} - \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) - f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) + f\right )}} + d\right )}}{16 \, \sqrt {-c \log \left (f\right ) - f}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) + f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) - f\right )}} - d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 3 \, f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) + 3 \, f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) - 3 \, f\right )}} - 3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^3,x, algorithm="maxima")
Output:
1/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - 3*f)*x - 1/2*b*log(f)/sqrt(-c*log(f ) - 3*f))*e^(-1/4*b^2*log(f)^2/(c*log(f) + 3*f) + 3*d)/sqrt(-c*log(f) - 3* f) - 3/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - f)*x - 1/2*b*log(f)/sqrt(-c*lo g(f) - f))*e^(-1/4*b^2*log(f)^2/(c*log(f) + f) + d)/sqrt(-c*log(f) - f) + 3/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + f)*x - 1/2*b*log(f)/sqrt(-c*log(f) + f))*e^(-1/4*b^2*log(f)^2/(c*log(f) - f) - d)/sqrt(-c*log(f) + f) - 1/16* sqrt(pi)*f^a*erf(sqrt(-c*log(f) + 3*f)*x - 1/2*b*log(f)/sqrt(-c*log(f) + 3 *f))*e^(-1/4*b^2*log(f)^2/(c*log(f) - 3*f) - 3*d)/sqrt(-c*log(f) + 3*f)
Time = 0.15 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.14 \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - 3 \, f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) + 3 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 12 \, c d \log \left (f\right ) - 12 \, a f \log \left (f\right ) - 36 \, d f}{4 \, {\left (c \log \left (f\right ) + 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} + \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) + f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - f}} - \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) - f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + 3 \, f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) - 3 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 12 \, c d \log \left (f\right ) + 12 \, a f \log \left (f\right ) - 36 \, d f}{4 \, {\left (c \log \left (f\right ) - 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^3,x, algorithm="giac")
Output:
-1/16*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) - 3*f)*(2*x + b*log(f)/(c*log(f) + 3*f)))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 - 12*c*d*log(f) - 12*a*f*log (f) - 36*d*f)/(c*log(f) + 3*f))/sqrt(-c*log(f) - 3*f) + 3/16*sqrt(pi)*erf( -1/2*sqrt(-c*log(f) - f)*(2*x + b*log(f)/(c*log(f) + f)))*e^(-1/4*(b^2*log (f)^2 - 4*a*c*log(f)^2 - 4*c*d*log(f) - 4*a*f*log(f) - 4*d*f)/(c*log(f) + f))/sqrt(-c*log(f) - f) - 3/16*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) + f)*(2*x + b*log(f)/(c*log(f) - f)))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 + 4*c*d *log(f) + 4*a*f*log(f) - 4*d*f)/(c*log(f) - f))/sqrt(-c*log(f) + f) + 1/16 *sqrt(pi)*erf(-1/2*sqrt(-c*log(f) + 3*f)*(2*x + b*log(f)/(c*log(f) - 3*f)) )*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 + 12*c*d*log(f) + 12*a*f*log(f) - 36*d*f)/(c*log(f) - 3*f))/sqrt(-c*log(f) + 3*f)
Timed out. \[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\mathrm {sinh}\left (f\,x^2+d\right )}^3 \,d x \] Input:
int(f^(a + b*x + c*x^2)*sinh(d + f*x^2)^3,x)
Output:
int(f^(a + b*x + c*x^2)*sinh(d + f*x^2)^3, x)
\[ \int f^{a+b x+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \sinh \left (f \,x^{2}+d \right )^{3}d x \right ) \] Input:
int(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^3,x)
Output:
f**a*int(f**(b*x + c*x**2)*sinh(d + f*x**2)**3,x)