\(\int f^{a+b x+c x^2} \sinh ^2(d+f x^2) \, dx\) [361]

Optimal result

Integrand size = 23, antiderivative size = 225 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx=-\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 d+\frac {b^2 \log ^2(f)}{8 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (2 f-c \log (f))}{2 \sqrt {2 f-c \log (f)}}\right )}{8 \sqrt {2 f-c \log (f)}}+\frac {e^{2 d-\frac {b^2 \log ^2(f)}{8 f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (2 f+c \log (f))}{2 \sqrt {2 f+c \log (f)}}\right )}{8 \sqrt {2 f+c \log (f)}} \] 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+b^2*ln(f)^2/(8*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf 
(1/2*(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-b^2*ln(f)^2/(8*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(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] (verified)

Time = 1.46 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.14 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx=\frac {1}{8} f^a \sqrt {\pi } \left (-\frac {2 f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{\sqrt {c} \sqrt {\log (f)}}-\frac {e^{-\frac {b^2 \log ^2(f)}{8 f+4 c \log (f)}} \left (e^{\frac {b^2 f \log ^2(f)}{4 f^2-c^2 \log ^2(f)}} \text {erf}\left (\frac {4 f x-(b+2 c x) \log (f)}{2 \sqrt {2 f-c \log (f)}}\right ) \sqrt {2 f-c \log (f)} (2 f+c \log (f)) (\cosh (2 d)-\sinh (2 d))+\text {erfi}\left (\frac {4 f x+(b+2 c x) \log (f)}{2 \sqrt {2 f+c \log (f)}}\right ) (2 f-c \log (f)) \sqrt {2 f+c \log (f)} (\cosh (2 d)+\sinh (2 d))\right )}{-4 f^2+c^2 \log ^2(f)}\right ) \] Input:

Integrate[f^(a + b*x + c*x^2)*Sinh[d + f*x^2]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 225, 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.

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

\(\Big \downarrow \) 6038

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

\(\Big \downarrow \) 2009

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

Input:

Int[f^(a + b*x + c*x^2)*Sinh[d + 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 + (b^2*Log[f]^2)/(8*f - 4*c*Log[f]) 
)*f^a*Sqrt[Pi]*Erf[(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 - (b^2*Log[f]^2)/(8*f + 4*c*Log[f 
]))*f^a*Sqrt[Pi]*Erfi[(b*Log[f] + 2*x*(2*f + c*Log[f]))/(2*Sqrt[2*f + c*Lo 
g[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 = 0.39 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-x \sqrt {2 f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {2 f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+8 d \ln \left (f \right ) c -16 d f}{4 \left (c \ln \left (f \right )-2 f \right )}}}{8 \sqrt {2 f -c \ln \left (f \right )}}-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-2 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-8 d \ln \left (f \right ) c -16 d f}{4 \left (2 f +c \ln \left (f \right )\right )}}}{8 \sqrt {-c \ln \left (f \right )-2 f}}+\frac {f^{a} \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(217\)

Input:

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

Output:

-1/8*erf(-x*(2*f-c*ln(f))^(1/2)+1/2*ln(f)*b/(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+8*d*ln(f)*c-16*d*f)/(c*ln(f) 
-2*f))-1/8*erf(-(-c*ln(f)-2*f)^(1/2)*x+1/2*ln(f)*b/(-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)^2-8*d*ln(f)*c-16*d*f) 
/(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. 466 vs. \(2 (185) = 370\).

Time = 0.10 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.07 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+d)^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 - 16*d*f + 8*(c*d + a*f)*log(f))/(c*log(f) - 2*f)) + sqrt(pi)*(c^2*l 
og(f)^2 + 2*c*f*log(f))*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 16*d*f + 8*(c* 
d + 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))*sqrt(-c*log(f) + 2*f)/(c*log(f) - 2*f)) + (sqrt(pi)* 
(c^2*log(f)^2 - 2*c*f*log(f))*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 16*d*f - 
 8*(c*d + 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 - 16*d*f - 8*(c*d + 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))*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)*s 
qrt(-c*log(f))/c))/(c^3*log(f)^3 - 4*c*f^2*log(f))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.88 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 2 \, f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) - 2 \, f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) + 2 \, f\right )}} + 2 \, d\right )}}{8 \, \sqrt {-c \log \left (f\right ) - 2 \, f}} + \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 2 \, f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) + 2 \, f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) - 2 \, f\right )}} - 2 \, d\right )}}{8 \, \sqrt {-c \log \left (f\right ) + 2 \, f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{4 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \] Input:

integrate(f^(c*x^2+b*x+a)*sinh(f*x^2+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)/sqrt(-c*log(f) 
 - 2*f))*e^(-1/4*b^2*log(f)^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*log(f)/sqrt(-c*lo 
g(f) + 2*f))*e^(-1/4*b^2*log(f)^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 = 0.15 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.06 \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - 2 \, f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) + 2 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 8 \, c d \log \left (f\right ) - 8 \, a f \log \left (f\right ) - 16 \, d f}{4 \, {\left (c \log \left (f\right ) + 2 \, f\right )}}\right )}}{8 \, \sqrt {-c \log \left (f\right ) - 2 \, f}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + 2 \, f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) - 2 \, f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 8 \, c d \log \left (f\right ) + 8 \, a f \log \left (f\right ) - 16 \, d f}{4 \, {\left (c \log \left (f\right ) - 2 \, f\right )}}\right )}}{8 \, \sqrt {-c \log \left (f\right ) + 2 \, f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \] Input:

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

Output:

-1/8*sqrt(pi)*erf(-1/2*sqrt(-c*log(f) - 2*f)*(2*x + b*log(f)/(c*log(f) + 2 
*f)))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 - 8*c*d*log(f) - 8*a*f*log(f) 
 - 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)/(c*log(f) - 2*f)))*e^(-1/4*(b^2*log 
(f)^2 - 4*a*c*log(f)^2 + 8*c*d*log(f) + 8*a*f*log(f) - 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. \[ \int f^{a+b x+c x^2} \sinh ^2\left (d+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\mathrm {sinh}\left (f\,x^2+d\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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