Integrand size = 21, antiderivative size = 219 \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=-\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 d-\frac {(2 e-b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{2 d-\frac {(2 e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}} \] Output:
-1/4*f^(a-1/4*b^2/c)*Pi^(1/2)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))/c^(1 /2)/ln(f)^(1/2)-1/8*exp(-2*d-1/4*(2*e-b*ln(f))^2/c/ln(f))*f^a*Pi^(1/2)*erf i(1/2*(2*e-b*ln(f)-2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)+1 /8*exp(2*d-1/4*(2*e+b*ln(f))^2/c/ln(f))*f^a*Pi^(1/2)*erfi(1/2*(2*e+b*ln(f) +2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))/c^(1/2)/ln(f)^(1/2)
Time = 0.39 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.84 \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\frac {e^{-\frac {e (e+b \log (f))}{c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (-2 e^{\frac {e (e+b \log (f))}{c \log (f)}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )+e^{\frac {2 b e}{c}} \text {erfi}\left (\frac {-2 e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (2 d)-\sinh (2 d))+\text {erfi}\left (\frac {2 e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (2 d)+\sinh (2 d))\right )}{8 \sqrt {c} \sqrt {\log (f)}} \] Input:
Integrate[f^(a + b*x + c*x^2)*Sinh[d + e*x]^2,x]
Output:
(f^(a - b^2/(4*c))*Sqrt[Pi]*(-2*E^((e*(e + b*Log[f]))/(c*Log[f]))*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])] + E^((2*b*e)/c)*Erfi[(-2*e + (b + 2*c *x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[2*d] - Sinh[2*d]) + Erfi[(2*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[2*d] + Sinh[2*d])))/ (8*Sqrt[c]*E^((e*(e + b*Log[f]))/(c*Log[f]))*Sqrt[Log[f]])
Time = 0.61 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \sinh ^2(d+e x) f^{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 6038 |
\(\displaystyle \int \left (\frac {1}{4} e^{-2 d-2 e x} f^{a+b x+c x^2}+\frac {1}{4} e^{2 d+2 e x} f^{a+b x+c x^2}-\frac {1}{2} f^{a+b x+c x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{-\frac {(2 e-b \log (f))^2}{4 c \log (f)}-2 d} \text {erfi}\left (\frac {-b \log (f)-2 c x \log (f)+2 e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{2 d-\frac {(b \log (f)+2 e)^2}{4 c \log (f)}} \text {erfi}\left (\frac {b \log (f)+2 c x \log (f)+2 e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{8 \sqrt {c} \sqrt {\log (f)}}\) |
Input:
Int[f^(a + b*x + c*x^2)*Sinh[d + e*x]^2,x]
Output:
-1/4*(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c ])])/(Sqrt[c]*Sqrt[Log[f]]) - (E^(-2*d - (2*e - b*Log[f])^2/(4*c*Log[f]))* f^a*Sqrt[Pi]*Erfi[(2*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]]) ])/(8*Sqrt[c]*Sqrt[Log[f]]) + (E^(2*d - (2*e + b*Log[f])^2/(4*c*Log[f]))*f ^a*Sqrt[Pi]*Erfi[(2*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])] )/(8*Sqrt[c]*Sqrt[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 = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-2 e}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {\ln \left (f \right ) b e -2 d \ln \left (f \right ) c -e^{2}}{\ln \left (f \right ) c}}}{8 \sqrt {-c \ln \left (f \right )}}-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {2 e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {\ln \left (f \right ) b e -2 d \ln \left (f \right ) c +e^{2}}{\ln \left (f \right ) c}}}{8 \sqrt {-c \ln \left (f \right )}}+\frac {f^{a} \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\) | \(210\) |
Input:
int(f^(c*x^2+b*x+a)*sinh(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-1/8*erf(-(-c*ln(f))^(1/2)*x+1/2*(b*ln(f)-2*e)/(-c*ln(f))^(1/2))/(-c*ln(f) )^(1/2)*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp((ln(f)*b*e-2*d*ln(f)*c-e^2)/ln(f)/ c)-1/8*erf(-(-c*ln(f))^(1/2)*x+1/2*(2*e+b*ln(f))/(-c*ln(f))^(1/2))/(-c*ln( f))^(1/2)*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp(-(ln(f)*b*e-2*d*ln(f)*c+e^2)/ln( f)/c)+1/4*f^a*Pi^(1/2)*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/ 2)*x+1/2*ln(f)*b/(-c*ln(f))^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (167) = 334\).
Time = 0.10 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.57 \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\frac {2 \, \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )}{4 \, c}\right ) + \sqrt {\pi } \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right ) - \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, e^{2} - 4 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, e^{2} - 4 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + 2 \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) - \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, e^{2} + 4 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + 4 \, e^{2} + 4 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - 2 \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right )}{8 \, c \log \left (f\right )} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(e*x+d)^2,x, algorithm="fricas")
Output:
1/8*(2*sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*(b^2 - 4*a*c)*log(f)/c) + sqrt( pi)*sinh(-1/4*(b^2 - 4*a*c)*log(f)/c))*erf(1/2*(2*c*x + b)*sqrt(-c*log(f)) /c) - sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 4*e^2 - 4*(2*c*d - b*e)*log(f))/(c*log(f))) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)* log(f)^2 + 4*e^2 - 4*(2*c*d - b*e)*log(f))/(c*log(f))))*erf(1/2*((2*c*x + b)*log(f) + 2*e)*sqrt(-c*log(f))/(c*log(f))) - sqrt(-c*log(f))*(sqrt(pi)*c osh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 4*e^2 + 4*(2*c*d - b*e)*log(f))/(c*log( f))) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 4*e^2 + 4*(2*c*d - b*e )*log(f))/(c*log(f))))*erf(1/2*((2*c*x + b)*log(f) - 2*e)*sqrt(-c*log(f))/ (c*log(f))))/(c*log(f))
\[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\int f^{a + b x + c x^{2}} \sinh ^{2}{\left (d + e x \right )}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)*sinh(e*x+d)**2,x)
Output:
Integral(f**(a + b*x + c*x**2)*sinh(d + e*x)**2, x)
Time = 0.05 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.84 \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right ) + 2 \, e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (2 \, d - \frac {{\left (b \log \left (f\right ) + 2 \, e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right ) - 2 \, e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (-2 \, d - \frac {{\left (b \log \left (f\right ) - 2 \, e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{4 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(e*x+d)^2,x, algorithm="maxima")
Output:
1/8*sqrt(pi)*f^a*erf(sqrt(-c*log(f))*x - 1/2*(b*log(f) + 2*e)/sqrt(-c*log( f)))*e^(2*d - 1/4*(b*log(f) + 2*e)^2/(c*log(f)))/sqrt(-c*log(f)) + 1/8*sqr t(pi)*f^a*erf(sqrt(-c*log(f))*x - 1/2*(b*log(f) - 2*e)/sqrt(-c*log(f)))*e^ (-2*d - 1/4*(b*log(f) - 2*e)^2/(c*log(f)))/sqrt(-c*log(f)) - 1/4*sqrt(pi)* f^a*erf(sqrt(-c*log(f))*x - 1/2*b*log(f)/sqrt(-c*log(f)))/(sqrt(-c*log(f)) *f^(1/4*b^2/c))
Time = 0.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.02 \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b \log \left (f\right ) - 2 \, e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 8 \, c d \log \left (f\right ) - 4 \, b e \log \left (f\right ) + 4 \, e^{2}}{4 \, c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b \log \left (f\right ) + 2 \, e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 8 \, c d \log \left (f\right ) + 4 \, b e \log \left (f\right ) + 4 \, e^{2}}{4 \, c \log \left (f\right )}\right )}}{8 \, \sqrt {-c \log \left (f\right )}} \] Input:
integrate(f^(c*x^2+b*x+a)*sinh(e*x+d)^2,x, algorithm="giac")
Output:
1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*x + b/c))*e^(-1/4*(b^2*log(f) - 4 *a*c*log(f))/c)/sqrt(-c*log(f)) - 1/8*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2 *x + (b*log(f) - 2*e)/(c*log(f))))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 + 8*c*d*log(f) - 4*b*e*log(f) + 4*e^2)/(c*log(f)))/sqrt(-c*log(f)) - 1/8*s qrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*x + (b*log(f) + 2*e)/(c*log(f))))*e^(- 1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 - 8*c*d*log(f) + 4*b*e*log(f) + 4*e^2)/ (c*log(f)))/sqrt(-c*log(f))
Timed out. \[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=\int f^{c\,x^2+b\,x+a}\,{\mathrm {sinh}\left (d+e\,x\right )}^2 \,d x \] Input:
int(f^(a + b*x + c*x^2)*sinh(d + e*x)^2,x)
Output:
int(f^(a + b*x + c*x^2)*sinh(d + e*x)^2, x)
\[ \int f^{a+b x+c x^2} \sinh ^2(d+e x) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \sinh \left (e x +d \right )^{2}d x \right ) \] Input:
int(f^(c*x^2+b*x+a)*sinh(e*x+d)^2,x)
Output:
f**a*int(f**(b*x + c*x**2)*sinh(d + e*x)**2,x)