\(\int f^{a+b x+c x^2} \sinh (d+f x^2) \, dx\) [360]

Optimal result

Integrand size = 21, antiderivative size = 154 \[ \int f^{a+b x+c x^2} \sinh \left (d+f x^2\right ) \, dx=\frac {e^{-d+\frac {b^2 \log ^2(f)}{4 f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (f-c \log (f))}{2 \sqrt {f-c \log (f)}}\right )}{4 \sqrt {f-c \log (f)}}+\frac {e^{d-\frac {b^2 \log ^2(f)}{4 (f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (f+c \log (f))}{2 \sqrt {f+c \log (f)}}\right )}{4 \sqrt {f+c \log (f)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.16 \[ \int f^{a+b x+c x^2} \sinh \left (d+f x^2\right ) \, dx=\frac {e^{-\frac {b^2 \log ^2(f)}{4 (f+c \log (f))}} f^a \sqrt {\pi } \left (-e^{\frac {b^2 f \log ^2(f)}{2 f^2-2 c^2 \log ^2(f)}} \text {erf}\left (\frac {2 f x-(b+2 c x) \log (f)}{2 \sqrt {f-c \log (f)}}\right ) \sqrt {f+c \log (f)} (\cosh (d)-\sinh (d))+\text {erfi}\left (\frac {2 f x+(b+2 c x) \log (f)}{2 \sqrt {f+c \log (f)}}\right ) \sqrt {f-c \log (f)} (\cosh (d)+\sinh (d))\right )}{4 \sqrt {f-c \log (f)} \sqrt {f+c \log (f)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 154, 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.

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

\(\Big \downarrow \) 6038

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(E^(-d + (b^2*Log[f]^2)/(4*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(b*Log[f] - 2 
*x*(f - c*Log[f]))/(2*Sqrt[f - c*Log[f]])])/(4*Sqrt[f - c*Log[f]]) + (E^(d 
 - (b^2*Log[f]^2)/(4*(f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[(b*Log[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 = 0.15 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-4 d \ln \left (f \right ) c -4 d f}{4 \left (f +c \ln \left (f \right )\right )}}}{4 \sqrt {-c \ln \left (f \right )-f}}+\frac {\operatorname {erf}\left (-x \sqrt {f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {f -c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+4 d \ln \left (f \right ) c -4 d f}{4 \left (c \ln \left (f \right )-f \right )}}}{4 \sqrt {f -c \ln \left (f \right )}}\)	\(160\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (131) = 262\).

Time = 0.12 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.11 \[ \int f^{a+b x+c x^2} \sinh \left (d+f x^2\right ) \, dx=\frac {{\left (\sqrt {\pi } {\left (c \log \left (f\right ) + f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f + 4 \, {\left (c d + a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right ) + \sqrt {\pi } {\left (c \log \left (f\right ) + f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f + 4 \, {\left (c d + a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )\right )} \sqrt {-c \log \left (f\right ) + f} \operatorname {erf}\left (-\frac {{\left (2 \, f x - {\left (2 \, c x + b\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + f}}{2 \, {\left (c \log \left (f\right ) - f\right )}}\right ) - {\left (\sqrt {\pi } {\left (c \log \left (f\right ) - f\right )} \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f - 4 \, {\left (c d + a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right ) + \sqrt {\pi } {\left (c \log \left (f\right ) - f\right )} \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - 4 \, d f - 4 \, {\left (c d + a f\right )} \log \left (f\right )}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )\right )} \sqrt {-c \log \left (f\right ) - f} \operatorname {erf}\left (\frac {{\left (2 \, f x + {\left (2 \, c x + b\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - f}}{2 \, {\left (c \log \left (f\right ) + f\right )}}\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} - f^{2}\right )}} \] Input:

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

Output:

1/4*((sqrt(pi)*(c*log(f) + f)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 - 4*d*f + 
4*(c*d + 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 - 4*d*f + 4*(c*d + a*f)*log(f))/(c*log(f) - f)))*s 
qrt(-c*log(f) + f)*erf(-1/2*(2*f*x - (2*c*x + b)*log(f))*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 - 4*d*f - 4*(c*d + 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 - 4*d*f - 4*(c*d + 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))*sq 
rt(-c*log(f) - f)/(c*log(f) + f)))/(c^2*log(f)^2 - f^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90 \[ \int f^{a+b x+c x^2} \sinh \left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) - f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) + f\right )}} + d\right )}}{4 \, \sqrt {-c \log \left (f\right ) - f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right ) + f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, {\left (c \log \left (f\right ) - f\right )}} - d\right )}}{4 \, \sqrt {-c \log \left (f\right ) + f}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18 \[ \int f^{a+b x+c x^2} \sinh \left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) - f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) + f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, {\left (c \log \left (f\right ) + f\right )}}\right )}}{4 \, \sqrt {-c \log \left (f\right ) - f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right ) + f} {\left (2 \, x + \frac {b \log \left (f\right )}{c \log \left (f\right ) - f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, {\left (c \log \left (f\right ) - f\right )}}\right )}}{4 \, \sqrt {-c \log \left (f\right ) + f}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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