\(\int f^{a+c x^2} \sinh ^3(d+f x^2) \, dx\) [353]

Optimal result

Integrand size = 20, antiderivative size = 171 \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\frac {3 e^{-d} f^a \sqrt {\pi } \text {erf}\left (x \sqrt {f-c \log (f)}\right )}{16 \sqrt {f-c \log (f)}}-\frac {e^{-3 d} f^a \sqrt {\pi } \text {erf}\left (x \sqrt {3 f-c \log (f)}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 e^d f^a \sqrt {\pi } \text {erfi}\left (x \sqrt {f+c \log (f)}\right )}{16 \sqrt {f+c \log (f)}}+\frac {e^{3 d} f^a \sqrt {\pi } \text {erfi}\left (x \sqrt {3 f+c \log (f)}\right )}{16 \sqrt {3 f+c \log (f)}} \] Output:

3/16*f^a*Pi^(1/2)*erf(x*(f-c*ln(f))^(1/2))/exp(d)/(f-c*ln(f))^(1/2)-1/16*f 
^a*Pi^(1/2)*erf(x*(3*f-c*ln(f))^(1/2))/exp(3*d)/(3*f-c*ln(f))^(1/2)-3/16*e 
xp(d)*f^a*Pi^(1/2)*erfi(x*(f+c*ln(f))^(1/2))/(f+c*ln(f))^(1/2)+1/16*exp(3* 
d)*f^a*Pi^(1/2)*erfi(x*(3*f+c*ln(f))^(1/2))/(3*f+c*ln(f))^(1/2)
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.59 \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\frac {f^a \sqrt {\pi } \left (3 \text {erf}\left (x \sqrt {f-c \log (f)}\right ) \sqrt {f-c \log (f)} \left (9 f^3+9 c f^2 \log (f)-c^2 f \log ^2(f)-c^3 \log ^3(f)\right ) (\cosh (d)-\sinh (d))-(f-c \log (f)) \left (\text {erf}\left (x \sqrt {3 f-c \log (f)}\right ) \sqrt {3 f-c \log (f)} \left (3 f^2+4 c f \log (f)+c^2 \log ^2(f)\right ) (\cosh (3 d)-\sinh (3 d))+(3 f-c \log (f)) \left (3 \text {erfi}\left (x \sqrt {f+c \log (f)}\right ) \sqrt {f+c \log (f)} (3 f+c \log (f)) (\cosh (d)+\sinh (d))-\text {erfi}\left (x \sqrt {3 f+c \log (f)}\right ) (f+c \log (f)) \sqrt {3 f+c \log (f)} (\cosh (3 d)+\sinh (3 d))\right )\right )\right )}{16 \left (9 f^4-10 c^2 f^2 \log ^2(f)+c^4 \log ^4(f)\right )} \] Input:

Integrate[f^(a + c*x^2)*Sinh[d + f*x^2]^3,x]
 

Output:

(f^a*Sqrt[Pi]*(3*Erf[x*Sqrt[f - c*Log[f]]]*Sqrt[f - c*Log[f]]*(9*f^3 + 9*c 
*f^2*Log[f] - c^2*f*Log[f]^2 - c^3*Log[f]^3)*(Cosh[d] - Sinh[d]) - (f - c* 
Log[f])*(Erf[x*Sqrt[3*f - c*Log[f]]]*Sqrt[3*f - c*Log[f]]*(3*f^2 + 4*c*f*L 
og[f] + c^2*Log[f]^2)*(Cosh[3*d] - Sinh[3*d]) + (3*f - c*Log[f])*(3*Erfi[x 
*Sqrt[f + c*Log[f]]]*Sqrt[f + c*Log[f]]*(3*f + c*Log[f])*(Cosh[d] + Sinh[d 
]) - Erfi[x*Sqrt[3*f + c*Log[f]]]*(f + c*Log[f])*Sqrt[3*f + c*Log[f]]*(Cos 
h[3*d] + Sinh[3*d])))))/(16*(9*f^4 - 10*c^2*f^2*Log[f]^2 + c^4*Log[f]^4))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.

\(\displaystyle \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx\)

\(\Big \downarrow \) 6038

\(\displaystyle \int \left (-\frac {1}{8} e^{-3 d-3 f x^2} f^{a+c x^2}+\frac {3}{8} e^{-d-f x^2} f^{a+c x^2}-\frac {3}{8} e^{d+f x^2} f^{a+c x^2}+\frac {1}{8} e^{3 d+3 f x^2} f^{a+c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \sqrt {\pi } e^{-d} f^a \text {erf}\left (x \sqrt {f-c \log (f)}\right )}{16 \sqrt {f-c \log (f)}}-\frac {\sqrt {\pi } e^{-3 d} f^a \text {erf}\left (x \sqrt {3 f-c \log (f)}\right )}{16 \sqrt {3 f-c \log (f)}}-\frac {3 \sqrt {\pi } e^d f^a \text {erfi}\left (x \sqrt {c \log (f)+f}\right )}{16 \sqrt {c \log (f)+f}}+\frac {\sqrt {\pi } e^{3 d} f^a \text {erfi}\left (x \sqrt {c \log (f)+3 f}\right )}{16 \sqrt {c \log (f)+3 f}}\)

Input:

Int[f^(a + c*x^2)*Sinh[d + f*x^2]^3,x]
 

Output:

(3*f^a*Sqrt[Pi]*Erf[x*Sqrt[f - c*Log[f]]])/(16*E^d*Sqrt[f - c*Log[f]]) - ( 
f^a*Sqrt[Pi]*Erf[x*Sqrt[3*f - c*Log[f]]])/(16*E^(3*d)*Sqrt[3*f - c*Log[f]] 
) - (3*E^d*f^a*Sqrt[Pi]*Erfi[x*Sqrt[f + c*Log[f]]])/(16*Sqrt[f + c*Log[f]] 
) + (E^(3*d)*f^a*Sqrt[Pi]*Erfi[x*Sqrt[3*f + c*Log[f]]])/(16*Sqrt[3*f + c*L 
og[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 = 0.80 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {f^{a} {\mathrm e}^{3 d} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )-3 f}\, x \right )}{16 \sqrt {-c \ln \left (f \right )-3 f}}-\frac {f^{a} {\mathrm e}^{-3 d} \sqrt {\pi }\, \operatorname {erf}\left (x \sqrt {3 f -c \ln \left (f \right )}\right )}{16 \sqrt {3 f -c \ln \left (f \right )}}+\frac {3 f^{a} {\mathrm e}^{-d} \sqrt {\pi }\, \operatorname {erf}\left (x \sqrt {f -c \ln \left (f \right )}\right )}{16 \sqrt {f -c \ln \left (f \right )}}-\frac {3 f^{a} {\mathrm e}^{d} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )-f}\, x \right )}{16 \sqrt {-c \ln \left (f \right )-f}}\)	\(144\)

Input:

int(f^(c*x^2+a)*sinh(f*x^2+d)^3,x,method=_RETURNVERBOSE)
 

Output:

1/16*f^a*exp(3*d)*Pi^(1/2)/(-c*ln(f)-3*f)^(1/2)*erf((-c*ln(f)-3*f)^(1/2)*x 
)-1/16*f^a*exp(-3*d)*Pi^(1/2)/(3*f-c*ln(f))^(1/2)*erf(x*(3*f-c*ln(f))^(1/2 
))+3/16*f^a*exp(-d)*Pi^(1/2)/(f-c*ln(f))^(1/2)*erf(x*(f-c*ln(f))^(1/2))-3/ 
16*f^a*exp(d)*Pi^(1/2)/(-c*ln(f)-f)^(1/2)*erf((-c*ln(f)-f)^(1/2)*x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 492 vs. \(2 (135) = 270\).

Time = 0.12 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.88 \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate(f^(c*x^2+a)*sinh(f*x^2+d)^3,x, algorithm="fricas")
 

Output:

1/16*((sqrt(pi)*(c^3*log(f)^3 + 3*c^2*f*log(f)^2 - c*f^2*log(f) - 3*f^3)*c 
osh(a*log(f) - 3*d) + sqrt(pi)*(c^3*log(f)^3 + 3*c^2*f*log(f)^2 - c*f^2*lo 
g(f) - 3*f^3)*sinh(a*log(f) - 3*d))*sqrt(-c*log(f) + 3*f)*erf(sqrt(-c*log( 
f) + 3*f)*x) - 3*(sqrt(pi)*(c^3*log(f)^3 + c^2*f*log(f)^2 - 9*c*f^2*log(f) 
 - 9*f^3)*cosh(a*log(f) - d) + sqrt(pi)*(c^3*log(f)^3 + c^2*f*log(f)^2 - 9 
*c*f^2*log(f) - 9*f^3)*sinh(a*log(f) - d))*sqrt(-c*log(f) + f)*erf(sqrt(-c 
*log(f) + f)*x) + 3*(sqrt(pi)*(c^3*log(f)^3 - c^2*f*log(f)^2 - 9*c*f^2*log 
(f) + 9*f^3)*cosh(a*log(f) + d) + sqrt(pi)*(c^3*log(f)^3 - c^2*f*log(f)^2 
- 9*c*f^2*log(f) + 9*f^3)*sinh(a*log(f) + d))*sqrt(-c*log(f) - f)*erf(sqrt 
(-c*log(f) - f)*x) - (sqrt(pi)*(c^3*log(f)^3 - 3*c^2*f*log(f)^2 - c*f^2*lo 
g(f) + 3*f^3)*cosh(a*log(f) + 3*d) + sqrt(pi)*(c^3*log(f)^3 - 3*c^2*f*log( 
f)^2 - c*f^2*log(f) + 3*f^3)*sinh(a*log(f) + 3*d))*sqrt(-c*log(f) - 3*f)*e 
rf(sqrt(-c*log(f) - 3*f)*x))/(c^4*log(f)^4 - 10*c^2*f^2*log(f)^2 + 9*f^4)
 

Sympy [F]

\[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\int f^{a + c x^{2}} \sinh ^{3}{\left (d + f x^{2} \right )}\, dx \] Input:

integrate(f**(c*x**2+a)*sinh(f*x**2+d)**3,x)
 

Output:

Integral(f**(a + c*x**2)*sinh(d + f*x**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84 \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 3 \, f} x\right ) e^{\left (3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + f} x\right ) e^{\left (-d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 3 \, f} x\right ) e^{\left (-3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} - \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - f} x\right ) e^{d}}{16 \, \sqrt {-c \log \left (f\right ) - f}} \] Input:

integrate(f^(c*x^2+a)*sinh(f*x^2+d)^3,x, algorithm="maxima")
 

Output:

1/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - 3*f)*x)*e^(3*d)/sqrt(-c*log(f) - 3* 
f) + 3/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + f)*x)*e^(-d)/sqrt(-c*log(f) + 
f) - 1/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) + 3*f)*x)*e^(-3*d)/sqrt(-c*log(f 
) + 3*f) - 3/16*sqrt(pi)*f^a*erf(sqrt(-c*log(f) - f)*x)*e^d/sqrt(-c*log(f) 
 - f)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.91 \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right ) - 3 \, f} x\right ) e^{\left (a \log \left (f\right ) + 3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) - 3 \, f}} + \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right ) - f} x\right ) e^{\left (a \log \left (f\right ) + d\right )}}{16 \, \sqrt {-c \log \left (f\right ) - f}} - \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right ) + f} x\right ) e^{\left (a \log \left (f\right ) - d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + f}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {-c \log \left (f\right ) + 3 \, f} x\right ) e^{\left (a \log \left (f\right ) - 3 \, d\right )}}{16 \, \sqrt {-c \log \left (f\right ) + 3 \, f}} \] Input:

integrate(f^(c*x^2+a)*sinh(f*x^2+d)^3,x, algorithm="giac")
 

Output:

-1/16*sqrt(pi)*erf(-sqrt(-c*log(f) - 3*f)*x)*e^(a*log(f) + 3*d)/sqrt(-c*lo 
g(f) - 3*f) + 3/16*sqrt(pi)*erf(-sqrt(-c*log(f) - f)*x)*e^(a*log(f) + d)/s 
qrt(-c*log(f) - f) - 3/16*sqrt(pi)*erf(-sqrt(-c*log(f) + f)*x)*e^(a*log(f) 
 - d)/sqrt(-c*log(f) + f) + 1/16*sqrt(pi)*erf(-sqrt(-c*log(f) + 3*f)*x)*e^ 
(a*log(f) - 3*d)/sqrt(-c*log(f) + 3*f)
 

Mupad [F(-1)]

Timed out. \[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=\int f^{c\,x^2+a}\,{\mathrm {sinh}\left (f\,x^2+d\right )}^3 \,d x \] Input:

int(f^(a + c*x^2)*sinh(d + f*x^2)^3,x)
 

Output:

int(f^(a + c*x^2)*sinh(d + f*x^2)^3, x)
 

Reduce [F]

\[ \int f^{a+c x^2} \sinh ^3\left (d+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}} \sinh \left (f \,x^{2}+d \right )^{3}d x \right ) \] Input:

int(f^(c*x^2+a)*sinh(f*x^2+d)^3,x)
 

Output:

f**a*int(f**(c*x**2)*sinh(d + f*x**2)**3,x)