Integrand size = 23, antiderivative size = 300 \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\frac {3 e^{-d+\frac {e^2}{4 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {e+2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{16 \sqrt {f-c \log (f)}}-\frac {e^{-3 d+\frac {9 e^2}{12 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {3 e+2 x (3 f-c \log (f))}{2 \sqrt {3 f-c \log (f)}}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 e^{d-\frac {e^2}{4 (f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+2 x (f+c \log (f))}{2 \sqrt {f+c \log (f)}}\right )}{16 \sqrt {f+c \log (f)}}+\frac {e^{3 d-\frac {9 e^2}{4 (3 f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 e+2 x (3 f+c \log (f))}{2 \sqrt {3 f+c \log (f)}}\right )}{16 \sqrt {3 f+c \log (f)}} \] Output:
3/16*exp(-d+e^2/(4*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(e+2*x*(f-c*ln(f)))/ (f-c*ln(f))^(1/2))/(f-c*ln(f))^(1/2)-1/16*exp(-3*d+9*e^2/(12*f-4*c*ln(f))) *f^a*Pi^(1/2)*erf(1/2*(3*e+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-e^2/(4*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(e+2*x* (f+c*ln(f)))/(f+c*ln(f))^(1/2))/(f+c*ln(f))^(1/2)+1/16*exp(3*d-9*e^2/(12*f +4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(3*e+2*x*(3*f+c*ln(f)))/(3*f+c*ln(f))^( 1/2))/(3*f+c*ln(f))^(1/2)
Time = 4.13 (sec) , antiderivative size = 480, normalized size of antiderivative = 1.60 \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\frac {e^{-\frac {1}{4} e^2 \left (\frac {1}{f+c \log (f)}+\frac {9}{3 f+c \log (f)}\right )} f^a \sqrt {\pi } \left (3 e^{\frac {1}{4} e^2 \left (\frac {1}{f-c \log (f)}+\frac {1}{f+c \log (f)}+\frac {9}{3 f+c \log (f)}\right )} \text {erf}\left (\frac {e+2 f x-2 c x \log (f)}{2 \sqrt {f-c \log (f)}}\right ) \sqrt {f-c \log (f)} \left (9 f^3+9 c f^2 \log (f)-c^2 f \log ^2(f)-c^3 \log ^3(f)\right ) (\cosh (d)-\sinh (d))-(f-c \log (f)) \left (e^{\frac {1}{4} e^2 \left (\frac {9}{3 f-c \log (f)}+\frac {1}{f+c \log (f)}+\frac {9}{3 f+c \log (f)}\right )} \text {erf}\left (\frac {3 e+6 f x-2 c x \log (f)}{2 \sqrt {3 f-c \log (f)}}\right ) \sqrt {3 f-c \log (f)} \left (3 f^2+4 c f \log (f)+c^2 \log ^2(f)\right ) (\cosh (3 d)-\sinh (3 d))+(3 f-c \log (f)) \left (3 e^{\frac {9 e^2}{4 (3 f+c \log (f))}} \text {erfi}\left (\frac {e+2 f x+2 c x \log (f)}{2 \sqrt {f+c \log (f)}}\right ) \sqrt {f+c \log (f)} (3 f+c \log (f)) (\cosh (d)+\sinh (d))-e^{\frac {e^2}{4 (f+c \log (f))}} \text {erfi}\left (\frac {3 e+6 f x+2 c x \log (f)}{2 \sqrt {3 f+c \log (f)}}\right ) (f+c \log (f)) \sqrt {3 f+c \log (f)} (\cosh (3 d)+\sinh (3 d))\right )\right )\right )}{16 \left (9 f^4-10 c^2 f^2 \log ^2(f)+c^4 \log ^4(f)\right )} \] Input:
Integrate[f^(a + c*x^2)*Sinh[d + e*x + f*x^2]^3,x]
Output:
(f^a*Sqrt[Pi]*(3*E^((e^2*((f - c*Log[f])^(-1) + (f + c*Log[f])^(-1) + 9/(3 *f + c*Log[f])))/4)*Erf[(e + 2*f*x - 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^((e^2*(9/(3*f - c*Log[f]) + (f + c*Log[f])^(-1) + 9/(3*f + c*Log[f])))/4)*Erf[(3*e + 6*f*x - 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^((9*e^2)/(4*( 3*f + c*Log[f])))*Erfi[(e + 2*f*x + 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^(e^2/(4*(f + c *Log[f])))*Erfi[(3*e + 6*f*x + 2*c*x*Log[f])/(2*Sqrt[3*f + c*Log[f]])]*(f + c*Log[f])*Sqrt[3*f + c*Log[f]]*(Cosh[3*d] + Sinh[3*d])))))/(16*E^((e^2*( (f + c*Log[f])^(-1) + 9/(3*f + c*Log[f])))/4)*(9*f^4 - 10*c^2*f^2*Log[f]^2 + c^4*Log[f]^4))
Time = 0.90 (sec) , antiderivative size = 300, 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.
| \(\displaystyle \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 6038 |
| \(\displaystyle \int \left (\frac {3}{8} f^{a+c x^2} \exp \left (-3 \left (d+e x+f x^2\right )+2 d+2 e x+2 f x^2\right )-\frac {3}{8} f^{a+c x^2} \exp \left (-3 \left (d+e x+f x^2\right )+4 d+4 e x+4 f x^2\right )+\frac {1}{8} f^{a+c x^2} \exp \left (-3 \left (d+e x+f x^2\right )+6 d+6 e x+6 f x^2\right )-\frac {1}{8} f^{a+c x^2} e^{-3 \left (d+e x+f x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \sqrt {\pi } f^a e^{\frac {e^2}{4 f-4 c \log (f)}-d} \text {erf}\left (\frac {2 x (f-c \log (f))+e}{2 \sqrt {f-c \log (f)}}\right )}{16 \sqrt {f-c \log (f)}}-\frac {\sqrt {\pi } f^a e^{\frac {9 e^2}{12 f-4 c \log (f)}-3 d} \text {erf}\left (\frac {2 x (3 f-c \log (f))+3 e}{2 \sqrt {3 f-c \log (f)}}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 \sqrt {\pi } f^a e^{d-\frac {e^2}{4 (c \log (f)+f)}} \text {erfi}\left (\frac {2 x (c \log (f)+f)+e}{2 \sqrt {c \log (f)+f}}\right )}{16 \sqrt {c \log (f)+f}}+\frac {\sqrt {\pi } f^a e^{3 d-\frac {9 e^2}{4 (c \log (f)+3 f)}} \text {erfi}\left (\frac {2 x (c \log (f)+3 f)+3 e}{2 \sqrt {c \log (f)+3 f}}\right )}{16 \sqrt {c \log (f)+3 f}}\) |
Input:
Int[f^(a + c*x^2)*Sinh[d + e*x + f*x^2]^3,x]
Output:
(3*E^(-d + e^2/(4*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(e + 2*x*(f - c*Log[f] ))/(2*Sqrt[f - c*Log[f]])])/(16*Sqrt[f - c*Log[f]]) - (E^(-3*d + (9*e^2)/( 12*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(3*e + 2*x*(3*f - c*Log[f]))/(2*Sqrt[ 3*f - c*Log[f]])])/(16*Sqrt[3*f - c*Log[f]]) - (3*E^(d - e^2/(4*(f + c*Log [f])))*f^a*Sqrt[Pi]*Erfi[(e + 2*x*(f + c*Log[f]))/(2*Sqrt[f + c*Log[f]])]) /(16*Sqrt[f + c*Log[f]]) + (E^(3*d - (9*e^2)/(4*(3*f + c*Log[f])))*f^a*Sqr t[Pi]*Erfi[(3*e + 2*x*(3*f + c*Log[f]))/(2*Sqrt[3*f + c*Log[f]])])/(16*Sqr t[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 = 1.44 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-3 f}\, x +\frac {3 e}{2 \sqrt {-c \ln \left (f \right )-3 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {3 d \ln \left (f \right ) c +9 d f -\frac {9 e^{2}}{4}}{3 f +c \ln \left (f \right )}}}{16 \sqrt {-c \ln \left (f \right )-3 f}}-\frac {\operatorname {erf}\left (x \sqrt {3 f -c \ln \left (f \right )}+\frac {3 e}{2 \sqrt {3 f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {3 \left (4 d \ln \left (f \right ) c -12 d f +3 e^{2}\right )}{4 \left (c \ln \left (f \right )-3 f \right )}}}{16 \sqrt {3 f -c \ln \left (f \right )}}+\frac {3 \,\operatorname {erf}\left (x \sqrt {f -c \ln \left (f \right )}+\frac {e}{2 \sqrt {f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 d \ln \left (f \right ) c -4 d f +e^{2}}{4 \left (c \ln \left (f \right )-f \right )}}}{16 \sqrt {f -c \ln \left (f \right )}}+\frac {3 \,\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-f}\, x +\frac {e}{2 \sqrt {-c \ln \left (f \right )-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 d \ln \left (f \right ) c +4 d f -e^{2}}{4 f +4 c \ln \left (f \right )}}}{16 \sqrt {-c \ln \left (f \right )-f}}\) | \(302\) |
Input:
int(f^(c*x^2+a)*sinh(f*x^2+e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/16*erf(-(-c*ln(f)-3*f)^(1/2)*x+3/2*e/(-c*ln(f)-3*f)^(1/2))/(-c*ln(f)-3* f)^(1/2)*Pi^(1/2)*f^a*exp(3/4*(4*d*ln(f)*c+12*d*f-3*e^2)/(3*f+c*ln(f)))-1/ 16*erf(x*(3*f-c*ln(f))^(1/2)+3/2*e/(3*f-c*ln(f))^(1/2))/(3*f-c*ln(f))^(1/2 )*Pi^(1/2)*f^a*exp(-3/4*(4*d*ln(f)*c-12*d*f+3*e^2)/(c*ln(f)-3*f))+3/16*erf (x*(f-c*ln(f))^(1/2)+1/2*e/(f-c*ln(f))^(1/2))/(f-c*ln(f))^(1/2)*Pi^(1/2)*f ^a*exp(-1/4*(4*d*ln(f)*c-4*d*f+e^2)/(c*ln(f)-f))+3/16*erf(-(-c*ln(f)-f)^(1 /2)*x+1/2*e/(-c*ln(f)-f)^(1/2))/(-c*ln(f)-f)^(1/2)*Pi^(1/2)*f^a*exp(1/4*(4 *d*ln(f)*c+4*d*f-e^2)/(f+c*ln(f)))
Leaf count of result is larger than twice the leaf count of optimal. 848 vs. \(2 (253) = 506\).
Time = 0.12 (sec) , antiderivative size = 848, normalized size of antiderivative = 2.83 \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(f^(c*x^2+a)*sinh(f*x^2+e*x+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*(4*a*c*log(f)^2 - 9*e^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*(4*a*c*log(f)^2 - 9*e^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*(2*c*x*log(f) - 6*f*x - 3*e) *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*(4*a*c*log(f)^2 - e^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*(4*a*c*log(f)^2 - e^2 + 4*d* f - 4*(c*d + a*f)*log(f))/(c*log(f) - f)))*sqrt(-c*log(f) + f)*erf(1/2*(2* c*x*log(f) - 2*f*x - e)*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*(4*a*c*l og(f)^2 - e^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*(4*a*c*lo g(f)^2 - e^2 + 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) + f)))*sqrt(-c*log( f) - f)*erf(1/2*(2*c*x*log(f) + 2*f*x + e)*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*(4*a*c*log(f)^2 - 9*e^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*(4*a*c*log(f)^2 - 9*e^2 + 36*d*f + 12*(c*d + a*f)*log(f))...
\[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\int f^{a + c x^{2}} \sinh ^{3}{\left (d + e x + f x^{2} \right )}\, dx \] Input:
integrate(f**(c*x**2+a)*sinh(f*x**2+e*x+d)**3,x)
Output:
Integral(f**(a + c*x**2)*sinh(d + e*x + f*x**2)**3, x)
Time = 0.05 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.88 \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 3 \, f} x - \frac {3 \, e}{2 \, \sqrt {-c \log \left (f\right ) - 3 \, f}}\right ) e^{\left (3 \, d - \frac {9 \, e^{2}}{4 \, {\left (c \log \left (f\right ) + 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} - \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - f} x - \frac {e}{2 \, \sqrt {-c \log \left (f\right ) - f}}\right ) e^{\left (d - \frac {e^{2}}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - f}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + f} x + \frac {e}{2 \, \sqrt {-c \log \left (f\right ) + f}}\right ) e^{\left (-d - \frac {e^{2}}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 3 \, f} x + \frac {3 \, e}{2 \, \sqrt {-c \log \left (f\right ) + 3 \, f}}\right ) e^{\left (-3 \, d - \frac {9 \, e^{2}}{4 \, {\left (c \log \left (f\right ) - 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} \] Input:
integrate(f^(c*x^2+a)*sinh(f*x^2+e*x+d)^3,x, algorithm="maxima")
Output:
1/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - 3*f)*x - 3/2*e/sqrt(-c*log(f) - 3*f ))*e^(3*d - 9/4*e^2/(c*log(f) + 3*f))/sqrt(-c*log(f) - 3*f) - 3/16*sqrt(pi )*f^a*erf(sqrt(-c*log(f) - f)*x - 1/2*e/sqrt(-c*log(f) - f))*e^(d - 1/4*e^ 2/(c*log(f) + f))/sqrt(-c*log(f) - f) + 3/16*sqrt(pi)*f^a*erf(sqrt(-c*log( f) + f)*x + 1/2*e/sqrt(-c*log(f) + f))*e^(-d - 1/4*e^2/(c*log(f) - f))/sqr t(-c*log(f) + f) - 1/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + 3*f)*x + 3/2*e/s qrt(-c*log(f) + 3*f))*e^(-3*d - 9/4*e^2/(c*log(f) - 3*f))/sqrt(-c*log(f) + 3*f)
Time = 0.15 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.17 \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - 3 \, f} {\left (2 \, x + \frac {3 \, e}{c \log \left (f\right ) + 3 \, f}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 12 \, c d \log \left (f\right ) + 12 \, a f \log \left (f\right ) - 9 \, e^{2} + 36 \, d f}{4 \, {\left (c \log \left (f\right ) + 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} + \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - f} {\left (2 \, x + \frac {e}{c \log \left (f\right ) + f}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) + 4 \, a f \log \left (f\right ) - e^{2} + 4 \, d f}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) - f}} - \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + f} {\left (2 \, x - \frac {e}{c \log \left (f\right ) - f}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - 4 \, a f \log \left (f\right ) - e^{2} + 4 \, d f}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + 3 \, f} {\left (2 \, x - \frac {3 \, e}{c \log \left (f\right ) - 3 \, f}\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 12 \, c d \log \left (f\right ) - 12 \, a f \log \left (f\right ) - 9 \, e^{2} + 36 \, d f}{4 \, {\left (c \log \left (f\right ) - 3 \, f\right )}}\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} \] Input:
integrate(f^(c*x^2+a)*sinh(f*x^2+e*x+d)^3,x, algorithm="giac")
Output:
-1/16*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) - 3*f)*(2*x + 3*e/(c*log(f) + 3*f)) )*e^(1/4*(4*a*c*log(f)^2 + 12*c*d*log(f) + 12*a*f*log(f) - 9*e^2 + 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 + e/(c*log(f) + f)))*e^(1/4*(4*a*c*log(f)^2 + 4*c*d*log(f ) + 4*a*f*log(f) - e^2 + 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 - e/(c*log(f) - f)))*e^(1/4*(4 *a*c*log(f)^2 - 4*c*d*log(f) - 4*a*f*log(f) - e^2 + 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 - 3*e/(c*log(f) - 3*f)))*e^(1/4*(4*a*c*log(f)^2 - 12*c*d*log(f) - 12*a*f*lo g(f) - 9*e^2 + 36*d*f)/(c*log(f) - 3*f))/sqrt(-c*log(f) + 3*f)
Timed out. \[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+a}\,{\mathrm {sinh}\left (f\,x^2+e\,x+d\right )}^3 \,d x \] Input:
int(f^(a + c*x^2)*sinh(d + e*x + f*x^2)^3,x)
Output:
int(f^(a + c*x^2)*sinh(d + e*x + f*x^2)^3, x)
\[ \int f^{a+c x^2} \sinh ^3\left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}} \sinh \left (f \,x^{2}+e x +d \right )^{3}d x \right ) \] Input:
int(f^(c*x^2+a)*sinh(f*x^2+e*x+d)^3,x)
Output:
f**a*int(f**(c*x**2)*sinh(d + e*x + f*x**2)**3,x)