\(\int f^{a+b x+c x^2} \sinh (d+e x+f x^2) \, dx\) [363]

Optimal result

Integrand size = 24, antiderivative size = 161 \[ \int f^{a+b x+c x^2} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {e^{-d+\frac {(e-b \log (f))^2}{4 (f-c \log (f))}} f^a \sqrt {\pi } \text {erf}\left (\frac {e-b \log (f)+2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{4 \sqrt {f-c \log (f)}}+\frac {e^{d-\frac {(e+b \log (f))^2}{4 (f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+b \log (f)+2 x (f+c \log (f))}{2 \sqrt {f+c \log (f)}}\right )}{4 \sqrt {f+c \log (f)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.98 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.57 \[ \int f^{a+b x+c x^2} \sinh \left (d+e x+f x^2\right ) \, dx=\frac {e^{-\frac {e^2+b^2 \log ^2(f)}{4 (f+c \log (f))}} f^{a+\frac {b e f}{-f^2+c^2 \log ^2(f)}} \sqrt {\pi } \left (-e^{\frac {f \left (e^2+b^2 \log ^2(f)\right )}{2 \left (f^2-c^2 \log ^2(f)\right )}} f^{\frac {b e}{2 (f+c \log (f))}} \text {erf}\left (\frac {e+2 f x-(b+2 c x) \log (f)}{2 \sqrt {f-c \log (f)}}\right ) \sqrt {f-c \log (f)} (f+c \log (f)) (\cosh (d)-\sinh (d))+f^{\frac {b e}{2 f-2 c \log (f)}} \text {erfi}\left (\frac {e+2 f x+(b+2 c x) \log (f)}{2 \sqrt {f+c \log (f)}}\right ) (f-c \log (f)) \sqrt {f+c \log (f)} (\cosh (d)+\sinh (d))\right )}{4 \left (f^2-c^2 \log ^2(f)\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 6038

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/4*(E^(-d + (e - b*Log[f])^2/(4*(f - c*Log[f])))*f^a*Sqrt[Pi]*Erf[(e - b 
*Log[f] + 2*x*(f - c*Log[f]))/(2*Sqrt[f - c*Log[f]])])/Sqrt[f - c*Log[f]] 
+ (E^(d - (e + b*Log[f])^2/(4*(f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[(e + b*Lo 
g[f] + 2*x*(f + c*Log[f]))/(2*Sqrt[f + c*Log[f]])])/(4*Sqrt[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 = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-f}\, x +\frac {e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+2 \ln \left (f \right ) b e -4 d \ln \left (f \right ) c -4 d f +e^{2}}{4 \left (f +c \ln \left (f \right )\right )}}}{4 \sqrt {-c \ln \left (f \right )-f}}+\frac {\operatorname {erf}\left (-x \sqrt {f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-e}{2 \sqrt {f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-2 \ln \left (f \right ) b e +4 d \ln \left (f \right ) c -4 d f +e^{2}}{4 \left (c \ln \left (f \right )-f \right )}}}{4 \sqrt {f -c \ln \left (f \right )}}\)	\(186\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (139) = 278\).

Time = 0.12 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.25 \[ \int f^{a+b x+c x^2} \sinh \left (d+e x+f x^2\right ) \, dx=\frac {{\left (\sqrt {\pi } {\left (c \log \left (f\right ) + f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 4 \, d f + 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right ) + \sqrt {\pi } {\left (c \log \left (f\right ) + f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 4 \, d f + 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )\right )} \sqrt {-c \log \left (f\right ) + f} \operatorname {erf}\left (-\frac {{\left (2 \, f x - {\left (2 \, c x + b\right )} \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right ) + f}}{2 \, {\left (c \log \left (f\right ) - f\right )}}\right ) - {\left (\sqrt {\pi } {\left (c \log \left (f\right ) - f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 4 \, d f - 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right ) + \sqrt {\pi } {\left (c \log \left (f\right ) - f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 4 \, d f - 2 \, {\left (2 \, c d - b e + 2 \, a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )\right )} \sqrt {-c \log \left (f\right ) - f} \operatorname {erf}\left (\frac {{\left (2 \, f x + {\left (2 \, c x + b\right )} \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right ) - f}}{2 \, {\left (c \log \left (f\right ) + f\right )}}\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} - f^{2}\right )}} \] Input:

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

Output:

1/4*((sqrt(pi)*(c*log(f) + f)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 4* 
d*f + 2*(2*c*d - b*e + 2*a*f)*log(f))/(c*log(f) - f)) + sqrt(pi)*(c*log(f) 
 + f)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 4*d*f + 2*(2*c*d - b*e + 2 
*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) + e)*sqrt(-c*log(f) + f)/(c*log(f) - f)) - (sqrt(pi)*(c*log( 
f) - f)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 4*d*f - 2*(2*c*d - b*e + 
 2*a*f)*log(f))/(c*log(f) + f)) + sqrt(pi)*(c*log(f) - f)*sinh(-1/4*((b^2 
- 4*a*c)*log(f)^2 + e^2 - 4*d*f - 2*(2*c*d - b*e + 2*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) + e)*sqrt 
(-c*log(f) - f)/(c*log(f) + f)))/(c^2*log(f)^2 - f^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94 \[ \int f^{a+b x+c x^2} \sinh \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - f} x - \frac {b \log \left (f\right ) + e}{2 \, \sqrt {-c \log \left (f\right ) - f}}\right ) e^{\left (-\frac {{\left (b \log \left (f\right ) + e\right )}^{2}}{4 \, {\left (c \log \left (f\right ) + f\right )}} + d\right )}}{4 \, \sqrt {-c \log \left (f\right ) - f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + f} x - \frac {b \log \left (f\right ) - e}{2 \, \sqrt {-c \log \left (f\right ) + f}}\right ) e^{\left (-\frac {{\left (b \log \left (f\right ) - e\right )}^{2}}{4 \, {\left (c \log \left (f\right ) - f\right )}} - d\right )}}{4 \, \sqrt {-c \log \left (f\right ) + f}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.29 \[ \int f^{a+b x+c x^2} \sinh \left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - f} {\left (2 \, x + \frac {b \log \left (f\right ) + e}{c \log \left (f\right ) + f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) + 2 \, b e \log \left (f\right ) - 4 \, a f \log \left (f\right ) + e^{2} - 4 \, d f}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )}}{4 \, \sqrt {-c \log \left (f\right ) - f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + f} {\left (2 \, x + \frac {b \log \left (f\right ) - e}{c \log \left (f\right ) - f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - 2 \, b e \log \left (f\right ) + 4 \, a f \log \left (f\right ) + e^{2} - 4 \, d f}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )}}{4 \, \sqrt {-c \log \left (f\right ) + f}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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