\(\int f^{a+b x} \cosh ^3(d+f x^2) \, dx\) [308]

Optimal result

Integrand size = 18, antiderivative size = 239 \[ \int f^{a+b x} \cosh ^3\left (d+f x^2\right ) \, dx=\frac {3}{16} e^{-d+\frac {b^2 \log ^2(f)}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {2 f x-b \log (f)}{2 \sqrt {f}}\right )+\frac {1}{16} e^{-3 d+\frac {b^2 \log ^2(f)}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {6 f x-b \log (f)}{2 \sqrt {3} \sqrt {f}}\right )+\frac {3}{16} e^{d-\frac {b^2 \log ^2(f)}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {2 f x+b \log (f)}{2 \sqrt {f}}\right )+\frac {1}{16} e^{3 d-\frac {b^2 \log ^2(f)}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {6 f x+b \log (f)}{2 \sqrt {3} \sqrt {f}}\right ) \] Output:

3/16*exp(-d+1/4*b^2*ln(f)^2/f)*f^(-1/2+a)*Pi^(1/2)*erf(1/2*(2*f*x-b*ln(f)) 
/f^(1/2))+1/48*exp(-3*d+1/12*b^2*ln(f)^2/f)*f^(-1/2+a)*3^(1/2)*Pi^(1/2)*er 
f(1/6*(6*f*x-b*ln(f))*3^(1/2)/f^(1/2))+3/16*exp(d-1/4*b^2*ln(f)^2/f)*f^(-1 
/2+a)*Pi^(1/2)*erfi(1/2*(2*f*x+b*ln(f))/f^(1/2))+1/48*exp(3*d-1/12*b^2*ln( 
f)^2/f)*f^(-1/2+a)*3^(1/2)*Pi^(1/2)*erfi(1/6*(6*f*x+b*ln(f))*3^(1/2)/f^(1/ 
2))
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.20 \[ \int f^{a+b x} \cosh ^3\left (d+f x^2\right ) \, dx=\frac {1}{16} e^{-\frac {b^2 \log ^2(f)}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{3}} \left (3 \sqrt {3} \cosh (d) \text {erfi}\left (\frac {2 f x+b \log (f)}{2 \sqrt {f}}\right )+e^{\frac {b^2 \log ^2(f)}{6 f}} \cosh (3 d) \text {erfi}\left (\frac {6 f x+b \log (f)}{2 \sqrt {3} \sqrt {f}}\right )+3 \sqrt {3} e^{\frac {b^2 \log ^2(f)}{2 f}} \text {erf}\left (\frac {2 f x-b \log (f)}{2 \sqrt {f}}\right ) (\cosh (d)-\sinh (d))+3 \sqrt {3} \text {erfi}\left (\frac {2 f x+b \log (f)}{2 \sqrt {f}}\right ) \sinh (d)+e^{\frac {b^2 \log ^2(f)}{3 f}} \text {erf}\left (\frac {6 f x-b \log (f)}{2 \sqrt {3} \sqrt {f}}\right ) (\cosh (3 d)-\sinh (3 d))+e^{\frac {b^2 \log ^2(f)}{6 f}} \text {erfi}\left (\frac {6 f x+b \log (f)}{2 \sqrt {3} \sqrt {f}}\right ) \sinh (3 d)\right ) \] Input:

Integrate[f^(a + b*x)*Cosh[d + f*x^2]^3,x]
 

Output:

(f^(-1/2 + a)*Sqrt[Pi/3]*(3*Sqrt[3]*Cosh[d]*Erfi[(2*f*x + b*Log[f])/(2*Sqr 
t[f])] + E^((b^2*Log[f]^2)/(6*f))*Cosh[3*d]*Erfi[(6*f*x + b*Log[f])/(2*Sqr 
t[3]*Sqrt[f])] + 3*Sqrt[3]*E^((b^2*Log[f]^2)/(2*f))*Erf[(2*f*x - b*Log[f]) 
/(2*Sqrt[f])]*(Cosh[d] - Sinh[d]) + 3*Sqrt[3]*Erfi[(2*f*x + b*Log[f])/(2*S 
qrt[f])]*Sinh[d] + E^((b^2*Log[f]^2)/(3*f))*Erf[(6*f*x - b*Log[f])/(2*Sqrt 
[3]*Sqrt[f])]*(Cosh[3*d] - Sinh[3*d]) + E^((b^2*Log[f]^2)/(6*f))*Erfi[(6*f 
*x + b*Log[f])/(2*Sqrt[3]*Sqrt[f])]*Sinh[3*d]))/(16*E^((b^2*Log[f]^2)/(4*f 
)))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6039, 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.

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

\(\Big \downarrow \) 6039

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3}{16} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {b^2 \log ^2(f)}{4 f}-d} \text {erf}\left (\frac {2 f x-b \log (f)}{2 \sqrt {f}}\right )+\frac {1}{16} \sqrt {\frac {\pi }{3}} f^{a-\frac {1}{2}} e^{\frac {b^2 \log ^2(f)}{12 f}-3 d} \text {erf}\left (\frac {6 f x-b \log (f)}{2 \sqrt {3} \sqrt {f}}\right )+\frac {3}{16} \sqrt {\pi } f^{a-\frac {1}{2}} e^{d-\frac {b^2 \log ^2(f)}{4 f}} \text {erfi}\left (\frac {b \log (f)+2 f x}{2 \sqrt {f}}\right )+\frac {1}{16} \sqrt {\frac {\pi }{3}} f^{a-\frac {1}{2}} e^{3 d-\frac {b^2 \log ^2(f)}{12 f}} \text {erfi}\left (\frac {b \log (f)+6 f x}{2 \sqrt {3} \sqrt {f}}\right )\)

Input:

Int[f^(a + b*x)*Cosh[d + f*x^2]^3,x]
 

Output:

(3*E^(-d + (b^2*Log[f]^2)/(4*f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[(2*f*x - b*Log[ 
f])/(2*Sqrt[f])])/16 + (E^(-3*d + (b^2*Log[f]^2)/(12*f))*f^(-1/2 + a)*Sqrt 
[Pi/3]*Erf[(6*f*x - b*Log[f])/(2*Sqrt[3]*Sqrt[f])])/16 + (3*E^(d - (b^2*Lo 
g[f]^2)/(4*f))*f^(-1/2 + a)*Sqrt[Pi]*Erfi[(2*f*x + b*Log[f])/(2*Sqrt[f])]) 
/16 + (E^(3*d - (b^2*Log[f]^2)/(12*f))*f^(-1/2 + a)*Sqrt[Pi/3]*Erfi[(6*f*x 
 + b*Log[f])/(2*Sqrt[3]*Sqrt[f])])/16
 

Defintions of rubi rules used

rule 2009

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

rule 6039

Int[Cosh[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cosh[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 = 2.76 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-\sqrt {3}\, \sqrt {f}\, x +\frac {\ln \left (f \right ) b \sqrt {3}}{6 \sqrt {f}}\right ) \sqrt {3}\, \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {b^{2} \ln \left (f \right )^{2}-36 d f}{12 f}}}{48 \sqrt {f}}-\frac {\operatorname {erf}\left (-\sqrt {-3 f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-3 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-36 d f}{12 f}}}{16 \sqrt {-3 f}}-\frac {3 \,\operatorname {erf}\left (-\sqrt {f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {b^{2} \ln \left (f \right )^{2}-4 d f}{4 f}}}{16 \sqrt {f}}-\frac {3 \,\operatorname {erf}\left (-\sqrt {-f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-4 d f}{4 f}}}{16 \sqrt {-f}}\)	\(207\)

Input:

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

Output:

-1/48*erf(-3^(1/2)*f^(1/2)*x+1/6*ln(f)*b*3^(1/2)/f^(1/2))/f^(1/2)*3^(1/2)* 
Pi^(1/2)*f^a*exp(1/12*(b^2*ln(f)^2-36*d*f)/f)-1/16*erf(-(-3*f)^(1/2)*x+1/2 
*ln(f)*b/(-3*f)^(1/2))/(-3*f)^(1/2)*Pi^(1/2)*f^a*exp(-1/12*(b^2*ln(f)^2-36 
*d*f)/f)-3/16*erf(-f^(1/2)*x+1/2*ln(f)*b/f^(1/2))/f^(1/2)*Pi^(1/2)*f^a*exp 
(1/4*(b^2*ln(f)^2-4*d*f)/f)-3/16*erf(-(-f)^(1/2)*x+1/2*ln(f)*b/(-f)^(1/2)) 
/(-f)^(1/2)*Pi^(1/2)*f^a*exp(-1/4*(b^2*ln(f)^2-4*d*f)/f)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (181) = 362\).

Time = 0.10 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.85 \[ \int f^{a+b x} \cosh ^3\left (d+f x^2\right ) \, dx=-\frac {\sqrt {3} \sqrt {\pi } \sqrt {-f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} - 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right ) \operatorname {erf}\left (\frac {\sqrt {3} {\left (6 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{6 \, f}\right ) + \sqrt {3} \sqrt {\pi } \sqrt {f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right ) \operatorname {erf}\left (-\frac {\sqrt {3} {\left (6 \, f x - b \log \left (f\right )\right )}}{6 \, \sqrt {f}}\right ) + \sqrt {3} \sqrt {\pi } \sqrt {f} \operatorname {erf}\left (-\frac {\sqrt {3} {\left (6 \, f x - b \log \left (f\right )\right )}}{6 \, \sqrt {f}}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right ) - \sqrt {3} \sqrt {\pi } \sqrt {-f} \operatorname {erf}\left (\frac {\sqrt {3} {\left (6 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{6 \, f}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} - 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right ) + 9 \, \sqrt {\pi } \sqrt {-f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right ) \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{2 \, f}\right ) + 9 \, \sqrt {\pi } \sqrt {f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right ) \operatorname {erf}\left (-\frac {2 \, f x - b \log \left (f\right )}{2 \, \sqrt {f}}\right ) + 9 \, \sqrt {\pi } \sqrt {f} \operatorname {erf}\left (-\frac {2 \, f x - b \log \left (f\right )}{2 \, \sqrt {f}}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right ) - 9 \, \sqrt {\pi } \sqrt {-f} \operatorname {erf}\left (\frac {{\left (2 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{2 \, f}\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right )}{48 \, f} \] Input:

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

Output:

-1/48*(sqrt(3)*sqrt(pi)*sqrt(-f)*cosh(1/12*(b^2*log(f)^2 - 12*a*f*log(f) - 
 36*d*f)/f)*erf(1/6*sqrt(3)*(6*f*x + b*log(f))*sqrt(-f)/f) + sqrt(3)*sqrt( 
pi)*sqrt(f)*cosh(1/12*(b^2*log(f)^2 + 12*a*f*log(f) - 36*d*f)/f)*erf(-1/6* 
sqrt(3)*(6*f*x - b*log(f))/sqrt(f)) + sqrt(3)*sqrt(pi)*sqrt(f)*erf(-1/6*sq 
rt(3)*(6*f*x - b*log(f))/sqrt(f))*sinh(1/12*(b^2*log(f)^2 + 12*a*f*log(f) 
- 36*d*f)/f) - sqrt(3)*sqrt(pi)*sqrt(-f)*erf(1/6*sqrt(3)*(6*f*x + b*log(f) 
)*sqrt(-f)/f)*sinh(1/12*(b^2*log(f)^2 - 12*a*f*log(f) - 36*d*f)/f) + 9*sqr 
t(pi)*sqrt(-f)*cosh(1/4*(b^2*log(f)^2 - 4*a*f*log(f) - 4*d*f)/f)*erf(1/2*( 
2*f*x + b*log(f))*sqrt(-f)/f) + 9*sqrt(pi)*sqrt(f)*cosh(1/4*(b^2*log(f)^2 
+ 4*a*f*log(f) - 4*d*f)/f)*erf(-1/2*(2*f*x - b*log(f))/sqrt(f)) + 9*sqrt(p 
i)*sqrt(f)*erf(-1/2*(2*f*x - b*log(f))/sqrt(f))*sinh(1/4*(b^2*log(f)^2 + 4 
*a*f*log(f) - 4*d*f)/f) - 9*sqrt(pi)*sqrt(-f)*erf(1/2*(2*f*x + b*log(f))*s 
qrt(-f)/f)*sinh(1/4*(b^2*log(f)^2 - 4*a*f*log(f) - 4*d*f)/f))/f
 

Sympy [F]

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

integrate(f**(b*x+a)*cosh(f*x**2+d)**3,x)
 

Output:

Integral(f**(a + b*x)*cosh(d + f*x**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.84 \[ \int f^{a+b x} \cosh ^3\left (d+f x^2\right ) \, dx=\frac {3}{16} \, \sqrt {\pi } f^{a - \frac {1}{2}} \operatorname {erf}\left (\sqrt {f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {f}}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2}}{4 \, f} - d\right )} + \frac {\sqrt {3} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {3} \sqrt {f} x - \frac {\sqrt {3} b \log \left (f\right )}{6 \, \sqrt {f}}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2}}{12 \, f} - 3 \, d\right )}}{48 \, \sqrt {f}} + \frac {\sqrt {3} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {3} \sqrt {-f} x - \frac {\sqrt {3} b \log \left (f\right )}{6 \, \sqrt {-f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{12 \, f} + 3 \, d\right )}}{48 \, \sqrt {-f}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-f} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{4 \, f} + d\right )}}{16 \, \sqrt {-f}} \] Input:

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

Output:

3/16*sqrt(pi)*f^(a - 1/2)*erf(sqrt(f)*x - 1/2*b*log(f)/sqrt(f))*e^(1/4*b^2 
*log(f)^2/f - d) + 1/48*sqrt(3)*sqrt(pi)*f^a*erf(sqrt(3)*sqrt(f)*x - 1/6*s 
qrt(3)*b*log(f)/sqrt(f))*e^(1/12*b^2*log(f)^2/f - 3*d)/sqrt(f) + 1/48*sqrt 
(3)*sqrt(pi)*f^a*erf(sqrt(3)*sqrt(-f)*x - 1/6*sqrt(3)*b*log(f)/sqrt(-f))*e 
^(-1/12*b^2*log(f)^2/f + 3*d)/sqrt(-f) + 3/16*sqrt(pi)*f^a*erf(sqrt(-f)*x 
- 1/2*b*log(f)/sqrt(-f))*e^(-1/4*b^2*log(f)^2/f + d)/sqrt(-f)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.93 \[ \int f^{a+b x} \cosh ^3\left (d+f x^2\right ) \, dx=-\frac {\sqrt {3} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{6} \, \sqrt {3} \sqrt {f} {\left (6 \, x - \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2} + 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right )}}{48 \, \sqrt {f}} - \frac {\sqrt {3} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{6} \, \sqrt {3} \sqrt {-f} {\left (6 \, x + \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 12 \, a f \log \left (f\right ) - 36 \, d f}{12 \, f}\right )}}{48 \, \sqrt {-f}} - \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {f} {\left (2 \, x - \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right )}}{16 \, \sqrt {f}} - \frac {3 \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-f} {\left (2 \, x + \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a f \log \left (f\right ) - 4 \, d f}{4 \, f}\right )}}{16 \, \sqrt {-f}} \] Input:

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

Output:

-1/48*sqrt(3)*sqrt(pi)*erf(-1/6*sqrt(3)*sqrt(f)*(6*x - b*log(f)/f))*e^(1/1 
2*(b^2*log(f)^2 + 12*a*f*log(f) - 36*d*f)/f)/sqrt(f) - 1/48*sqrt(3)*sqrt(p 
i)*erf(-1/6*sqrt(3)*sqrt(-f)*(6*x + b*log(f)/f))*e^(-1/12*(b^2*log(f)^2 - 
12*a*f*log(f) - 36*d*f)/f)/sqrt(-f) - 3/16*sqrt(pi)*erf(-1/2*sqrt(f)*(2*x 
- b*log(f)/f))*e^(1/4*(b^2*log(f)^2 + 4*a*f*log(f) - 4*d*f)/f)/sqrt(f) - 3 
/16*sqrt(pi)*erf(-1/2*sqrt(-f)*(2*x + b*log(f)/f))*e^(-1/4*(b^2*log(f)^2 - 
 4*a*f*log(f) - 4*d*f)/f)/sqrt(-f)
 

Mupad [F(-1)]

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

int(f^(a + b*x)*cosh(d + f*x^2)^3,x)
 

Output:

int(f^(a + b*x)*cosh(d + f*x^2)^3, x)
 

Reduce [F]

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

int(f^(b*x+a)*cosh(f*x^2+d)^3,x)
 

Output:

f**a*int(f**(b*x)*cosh(d + f*x**2)**3,x)