\(\int f^{a+b x+c x^2} \sinh ^2(d+e x+f x^2) \, dx\) [364]

Optimal result

Integrand size = 26, antiderivative size = 239 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=-\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 d+\frac {(2 e-b \log (f))^2}{8 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {2 e-b \log (f)+2 x (2 f-c \log (f))}{2 \sqrt {2 f-c \log (f)}}\right )}{8 \sqrt {2 f-c \log (f)}}+\frac {e^{2 d-\frac {(2 e+b \log (f))^2}{8 f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 e+b \log (f)+2 x (2 f+c \log (f))}{2 \sqrt {2 f+c \log (f)}}\right )}{8 \sqrt {2 f+c \log (f)}} \] 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+(2*e-b*ln(f))^2/(8*f-4*c*ln(f)))*f^a*Pi^(1/2) 
*erf(1/2*(2*e-b*ln(f)+2*x*(2*f-c*ln(f)))/(2*f-c*ln(f))^(1/2))/(2*f-c*ln(f) 
)^(1/2)+1/8*exp(2*d-(2*e+b*ln(f))^2/(8*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2 
*(2*e+b*ln(f)+2*x*(2*f+c*ln(f)))/(2*f+c*ln(f))^(1/2))/(2*f+c*ln(f))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 4.02 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.42 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=-\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-\frac {4 e^2+b^2 \log ^2(f)}{8 f+4 c \log (f)}} f^{a+\frac {4 b e f}{-4 f^2+c^2 \log ^2(f)}} \sqrt {\pi } \left (e^{\frac {f \left (4 e^2+b^2 \log ^2(f)\right )}{4 f^2-c^2 \log ^2(f)}} f^{\frac {b e}{2 f+c \log (f)}} \text {erf}\left (\frac {2 (e+2 f x)-(b+2 c x) \log (f)}{2 \sqrt {2 f-c \log (f)}}\right ) \sqrt {2 f-c \log (f)} (2 f+c \log (f)) (\cosh (2 d)-\sinh (2 d))+f^{\frac {b e}{2 f-c \log (f)}} \text {erfi}\left (\frac {2 (e+2 f x)+(b+2 c x) \log (f)}{2 \sqrt {2 f+c \log (f)}}\right ) (2 f-c \log (f)) \sqrt {2 f+c \log (f)} (\cosh (2 d)+\sinh (2 d))\right )}{8 \left (-4 f^2+c^2 \log ^2(f)\right )} \] Input:

Integrate[f^(a + b*x + c*x^2)*Sinh[d + e*x + f*x^2]^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]]) - (f^(a + (4*b*e*f)/(-4*f^2 + c^2*Log[f]^2))*S 
qrt[Pi]*(E^((f*(4*e^2 + b^2*Log[f]^2))/(4*f^2 - c^2*Log[f]^2))*f^((b*e)/(2 
*f + c*Log[f]))*Erf[(2*(e + 2*f*x) - (b + 2*c*x)*Log[f])/(2*Sqrt[2*f - c*L 
og[f]])]*Sqrt[2*f - c*Log[f]]*(2*f + c*Log[f])*(Cosh[2*d] - Sinh[2*d]) + f 
^((b*e)/(2*f - c*Log[f]))*Erfi[(2*(e + 2*f*x) + (b + 2*c*x)*Log[f])/(2*Sqr 
t[2*f + c*Log[f]])]*(2*f - c*Log[f])*Sqrt[2*f + c*Log[f]]*(Cosh[2*d] + Sin 
h[2*d])))/(8*E^((4*e^2 + b^2*Log[f]^2)/(8*f + 4*c*Log[f]))*(-4*f^2 + c^2*L 
og[f]^2))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

\(\displaystyle \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx\)

\(\Big \downarrow \) 6038

\(\displaystyle \int \left (\frac {1}{4} e^{-2 d-2 e x-2 f x^2} f^{a+b x+c x^2}+\frac {1}{4} e^{2 d+2 e x+2 f x^2} f^{a+b x+c x^2}-\frac {1}{2} f^{a+b x+c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a \exp \left (\frac {(2 e-b \log (f))^2}{8 f-4 c \log (f)}-2 d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (2 f-c \log (f))+2 e}{2 \sqrt {2 f-c \log (f)}}\right )}{8 \sqrt {2 f-c \log (f)}}+\frac {\sqrt {\pi } f^a \exp \left (2 d-\frac {(b \log (f)+2 e)^2}{4 c \log (f)+8 f}\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+2 f)+2 e}{2 \sqrt {c \log (f)+2 f}}\right )}{8 \sqrt {c \log (f)+2 f}}\)

Input:

Int[f^(a + b*x + c*x^2)*Sinh[d + e*x + f*x^2]^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/(8*f - 4*c*Log 
[f]))*f^a*Sqrt[Pi]*Erf[(2*e - b*Log[f] + 2*x*(2*f - c*Log[f]))/(2*Sqrt[2*f 
 - c*Log[f]])])/(8*Sqrt[2*f - c*Log[f]]) + (E^(2*d - (2*e + b*Log[f])^2/(8 
*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(2*e + b*Log[f] + 2*x*(2*f + c*Log[f]) 
)/(2*Sqrt[2*f + c*Log[f]])])/(8*Sqrt[2*f + c*Log[f]])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6038

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]
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-x \sqrt {2 f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-2 e}{2 \sqrt {2 f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-4 \ln \left (f \right ) b e +8 d \ln \left (f \right ) c -16 d f +4 e^{2}}{4 \left (c \ln \left (f \right )-2 f \right )}}}{8 \sqrt {2 f -c \ln \left (f \right )}}-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 f}\, x +\frac {2 e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-2 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+4 \ln \left (f \right ) b e -8 d \ln \left (f \right ) c -16 d f +4 e^{2}}{4 \left (2 f +c \ln \left (f \right )\right )}}}{8 \sqrt {-c \ln \left (f \right )-2 f}}+\frac {f^{a} \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(249\)

Input:

int(f^(c*x^2+b*x+a)*sinh(f*x^2+e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/8*erf(-x*(2*f-c*ln(f))^(1/2)+1/2*(b*ln(f)-2*e)/(2*f-c*ln(f))^(1/2))/(2* 
f-c*ln(f))^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2-4*ln(f)*b*e+8*d*ln(f)* 
c-16*d*f+4*e^2)/(c*ln(f)-2*f))-1/8*erf(-(-c*ln(f)-2*f)^(1/2)*x+1/2*(2*e+b* 
ln(f))/(-c*ln(f)-2*f)^(1/2))/(-c*ln(f)-2*f)^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b 
^2*ln(f)^2+4*ln(f)*b*e-8*d*ln(f)*c-16*d*f+4*e^2)/(2*f+c*ln(f)))+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))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (197) = 394\).

Time = 0.09 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.16 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+e*x+d)^2,x, algorithm="fricas")
 

Output:

-1/8*((sqrt(pi)*(c^2*log(f)^2 + 2*c*f*log(f))*cosh(-1/4*((b^2 - 4*a*c)*log 
(f)^2 + 4*e^2 - 16*d*f + 4*(2*c*d - b*e + 2*a*f)*log(f))/(c*log(f) - 2*f)) 
 + sqrt(pi)*(c^2*log(f)^2 + 2*c*f*log(f))*sinh(-1/4*((b^2 - 4*a*c)*log(f)^ 
2 + 4*e^2 - 16*d*f + 4*(2*c*d - b*e + 2*a*f)*log(f))/(c*log(f) - 2*f)))*sq 
rt(-c*log(f) + 2*f)*erf(-1/2*(4*f*x - (2*c*x + b)*log(f) + 2*e)*sqrt(-c*lo 
g(f) + 2*f)/(c*log(f) - 2*f)) + (sqrt(pi)*(c^2*log(f)^2 - 2*c*f*log(f))*co 
sh(-1/4*((b^2 - 4*a*c)*log(f)^2 + 4*e^2 - 16*d*f - 4*(2*c*d - b*e + 2*a*f) 
*log(f))/(c*log(f) + 2*f)) + sqrt(pi)*(c^2*log(f)^2 - 2*c*f*log(f))*sinh(- 
1/4*((b^2 - 4*a*c)*log(f)^2 + 4*e^2 - 16*d*f - 4*(2*c*d - b*e + 2*a*f)*log 
(f))/(c*log(f) + 2*f)))*sqrt(-c*log(f) - 2*f)*erf(1/2*(4*f*x + (2*c*x + b) 
*log(f) + 2*e)*sqrt(-c*log(f) - 2*f)/(c*log(f) + 2*f)) - 2*(sqrt(pi)*(c^2* 
log(f)^2 - 4*f^2)*cosh(-1/4*(b^2 - 4*a*c)*log(f)/c) + sqrt(pi)*(c^2*log(f) 
^2 - 4*f^2)*sinh(-1/4*(b^2 - 4*a*c)*log(f)/c))*sqrt(-c*log(f))*erf(1/2*(2* 
c*x + b)*sqrt(-c*log(f))/c))/(c^3*log(f)^3 - 4*c*f^2*log(f))
 

Sympy [F]

\[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sinh ^{2}{\left (d + e x + f x^{2} \right )}\, dx \] Input:

integrate(f**(c*x**2+b*x+a)*sinh(f*x**2+e*x+d)**2,x)
 

Output:

Integral(f**(a + b*x + c*x**2)*sinh(d + e*x + f*x**2)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.90 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 2 \, f} x - \frac {b \log \left (f\right ) + 2 \, e}{2 \, \sqrt {-c \log \left (f\right ) - 2 \, f}}\right ) e^{\left (-\frac {{\left (b \log \left (f\right ) + 2 \, e\right )}^{2}}{4 \, {\left (c \log \left (f\right ) + 2 \, f\right )}} + 2 \, d\right )}}{8 \, \sqrt {-c \log \left (f\right ) - 2 \, f}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 2 \, f} x - \frac {b \log \left (f\right ) - 2 \, e}{2 \, \sqrt {-c \log \left (f\right ) + 2 \, f}}\right ) e^{\left (-\frac {{\left (b \log \left (f\right ) - 2 \, e\right )}^{2}}{4 \, {\left (c \log \left (f\right ) - 2 \, f\right )}} - 2 \, d\right )}}{8 \, \sqrt {-c \log \left (f\right ) + 2 \, f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{4 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \] Input:

integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+e*x+d)^2,x, algorithm="maxima")
 

Output:

1/8*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - 2*f)*x - 1/2*(b*log(f) + 2*e)/sqrt(- 
c*log(f) - 2*f))*e^(-1/4*(b*log(f) + 2*e)^2/(c*log(f) + 2*f) + 2*d)/sqrt(- 
c*log(f) - 2*f) + 1/8*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + 2*f)*x - 1/2*(b*lo 
g(f) - 2*e)/sqrt(-c*log(f) + 2*f))*e^(-1/4*(b*log(f) - 2*e)^2/(c*log(f) - 
2*f) - 2*d)/sqrt(-c*log(f) + 2*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))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.13 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - 2 \, f} {\left (2 \, x + \frac {b \log \left (f\right ) + 2 \, e}{c \log \left (f\right ) + 2 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 8 \, c d \log \left (f\right ) + 4 \, b e \log \left (f\right ) - 8 \, a f \log \left (f\right ) + 4 \, e^{2} - 16 \, d f}{4 \, {\left (c \log \left (f\right ) + 2 \, f\right )}}\right )}}{8 \, \sqrt {-c \log \left (f\right ) - 2 \, f}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + 2 \, f} {\left (2 \, x + \frac {b \log \left (f\right ) - 2 \, e}{c \log \left (f\right ) - 2 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 8 \, c d \log \left (f\right ) - 4 \, b e \log \left (f\right ) + 8 \, a f \log \left (f\right ) + 4 \, e^{2} - 16 \, d f}{4 \, {\left (c \log \left (f\right ) - 2 \, f\right )}}\right )}}{8 \, \sqrt {-c \log \left (f\right ) + 2 \, f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \] Input:

integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+e*x+d)^2,x, algorithm="giac")
 

Output:

-1/8*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) - 2*f)*(2*x + (b*log(f) + 2*e)/(c*lo 
g(f) + 2*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) - 8*a*f*log(f) + 4*e^2 - 16*d*f)/(c*log(f) + 2*f))/sqrt(-c*log(f) 
 - 2*f) - 1/8*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) + 2*f)*(2*x + (b*log(f) - 2 
*e)/(c*log(f) - 2*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) + 8*a*f*log(f) + 4*e^2 - 16*d*f)/(c*log(f) - 2*f))/sqrt( 
-c*log(f) + 2*f) + 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))
 

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\mathrm {sinh}\left (f\,x^2+e\,x+d\right )}^2 \,d x \] Input:

int(f^(a + b*x + c*x^2)*sinh(d + e*x + f*x^2)^2,x)
 

Output:

int(f^(a + b*x + c*x^2)*sinh(d + e*x + f*x^2)^2, x)
 

Reduce [F]

\[ \int f^{a+b x+c x^2} \sinh ^2\left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \sinh \left (f \,x^{2}+e x +d \right )^{2}d x \right ) \] Input:

int(f^(c*x^2+b*x+a)*sinh(f*x^2+e*x+d)^2,x)
 

Output:

f**a*int(f**(b*x + c*x**2)*sinh(d + e*x + f*x**2)**2,x)