\(\int f^{a+b x} \cosh ^2(d+e x+f x^2) \, dx\) [310]

Optimal result

Integrand size = 21, antiderivative size = 161 \[ \int f^{a+b x} \cosh ^2\left (d+e x+f x^2\right ) \, dx=\frac {1}{8} e^{-2 d+\frac {(2 e-b \log (f))^2}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {2 e+4 f x-b \log (f)}{2 \sqrt {2} \sqrt {f}}\right )+\frac {1}{8} e^{2 d-\frac {(2 e+b \log (f))^2}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {2 e+4 f x+b \log (f)}{2 \sqrt {2} \sqrt {f}}\right )+\frac {f^{a+b x}}{2 b \log (f)} \] Output:

1/16*exp(-2*d+1/8*(2*e-b*ln(f))^2/f)*f^(-1/2+a)*2^(1/2)*Pi^(1/2)*erf(1/4*( 
2*e+4*f*x-b*ln(f))*2^(1/2)/f^(1/2))+1/16*exp(2*d-1/8*(2*e+b*ln(f))^2/f)*f^ 
(-1/2+a)*2^(1/2)*Pi^(1/2)*erfi(1/4*(2*e+4*f*x+b*ln(f))*2^(1/2)/f^(1/2))+1/ 
2*f^(b*x+a)/b/ln(f)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.37 \[ \int f^{a+b x} \cosh ^2\left (d+e x+f x^2\right ) \, dx=\frac {e^{-\frac {4 e^2+b^2 \log ^2(f)}{8 f}} f^{a-\frac {b e+f}{2 f}} \left (4 \sqrt {2} e^{\frac {4 e^2+b^2 \log ^2(f)}{8 f}} f^{\frac {1}{2}+b \left (\frac {e}{2 f}+x\right )}+b e^{\frac {4 e^2+b^2 \log ^2(f)}{4 f}} \sqrt {\pi } \text {erf}\left (\frac {2 e+4 f x-b \log (f)}{2 \sqrt {2} \sqrt {f}}\right ) \log (f) (\cosh (2 d)-\sinh (2 d))+b \sqrt {\pi } \text {erfi}\left (\frac {2 e+4 f x+b \log (f)}{2 \sqrt {2} \sqrt {f}}\right ) \log (f) (\cosh (2 d)+\sinh (2 d))\right )}{8 \sqrt {2} b \log (f)} \] Input:

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

Output:

(f^(a - (b*e + f)/(2*f))*(4*Sqrt[2]*E^((4*e^2 + b^2*Log[f]^2)/(8*f))*f^(1/ 
2 + b*(e/(2*f) + x)) + b*E^((4*e^2 + b^2*Log[f]^2)/(4*f))*Sqrt[Pi]*Erf[(2* 
e + 4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*Log[f]*(Cosh[2*d] - Sinh[2*d]) 
+ b*Sqrt[Pi]*Erfi[(2*e + 4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*Log[f]*(Co 
sh[2*d] + Sinh[2*d])))/(8*Sqrt[2]*b*E^((4*e^2 + b^2*Log[f]^2)/(8*f))*Log[f 
])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

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

\(\Big \downarrow \) 6039

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{8} \sqrt {\frac {\pi }{2}} f^{a-\frac {1}{2}} e^{\frac {(2 e-b \log (f))^2}{8 f}-2 d} \text {erf}\left (\frac {-b \log (f)+2 e+4 f x}{2 \sqrt {2} \sqrt {f}}\right )+\frac {1}{8} \sqrt {\frac {\pi }{2}} f^{a-\frac {1}{2}} e^{2 d-\frac {(b \log (f)+2 e)^2}{8 f}} \text {erfi}\left (\frac {b \log (f)+2 e+4 f x}{2 \sqrt {2} \sqrt {f}}\right )+\frac {f^{a+b x}}{2 b \log (f)}\)

Input:

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

Output:

(E^(-2*d + (2*e - b*Log[f])^2/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erf[(2*e + 4* 
f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])])/8 + (E^(2*d - (2*e + b*Log[f])^2/(8* 
f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erfi[(2*e + 4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[ 
f])])/8 + f^(a + b*x)/(2*b*Log[f])
 

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 = 1.68 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\operatorname {erf}\left (-\sqrt {2}\, \sqrt {f}\, x +\frac {\left (b \ln \left (f \right )-2 e \right ) \sqrt {2}}{4 \sqrt {f}}\right ) \sqrt {2}\, \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {b^{2} \ln \left (f \right )^{2}-4 \ln \left (f \right ) b e -16 d f +4 e^{2}}{8 f}}}{16 \sqrt {f}}-\frac {\operatorname {erf}\left (-\sqrt {-2 f}\, x +\frac {2 e +b \ln \left (f \right )}{2 \sqrt {-2 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}+4 \ln \left (f \right ) b e -16 d f +4 e^{2}}{8 f}}}{8 \sqrt {-2 f}}+\frac {f^{a} f^{b x}}{2 b \ln \left (f \right )}\)	\(158\)

Input:

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

Output:

-1/16*erf(-2^(1/2)*f^(1/2)*x+1/4*(b*ln(f)-2*e)*2^(1/2)/f^(1/2))/f^(1/2)*2^ 
(1/2)*Pi^(1/2)*f^a*exp(1/8*(b^2*ln(f)^2-4*ln(f)*b*e-16*d*f+4*e^2)/f)-1/8*e 
rf(-(-2*f)^(1/2)*x+1/2*(2*e+b*ln(f))/(-2*f)^(1/2))/(-2*f)^(1/2)*Pi^(1/2)*f 
^a*exp(-1/8*(b^2*ln(f)^2+4*ln(f)*b*e-16*d*f+4*e^2)/f)+1/2*f^a*f^(b*x)/b/ln 
(f)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (126) = 252\).

Time = 0.08 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.07 \[ \int f^{a+b x} \cosh ^2\left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } b \sqrt {-f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f + 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \left (f\right ) + 2 \, e\right )} \sqrt {-f}}{4 \, f}\right ) \log \left (f\right ) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt {f}}\right ) \log \left (f\right ) - \sqrt {2} \sqrt {\pi } b \sqrt {-f} \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \left (f\right ) + 2 \, e\right )} \sqrt {-f}}{4 \, f}\right ) \log \left (f\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f + 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt {f}}\right ) \log \left (f\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, e^{2} - 16 \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right ) - 8 \, f \cosh \left ({\left (b x + a\right )} \log \left (f\right )\right ) - 8 \, f \sinh \left ({\left (b x + a\right )} \log \left (f\right )\right )}{16 \, b f \log \left (f\right )} \] Input:

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

Output:

-1/16*(sqrt(2)*sqrt(pi)*b*sqrt(-f)*cosh(1/8*(b^2*log(f)^2 + 4*e^2 - 16*d*f 
 + 4*(b*e - 2*a*f)*log(f))/f)*erf(1/4*sqrt(2)*(4*f*x + b*log(f) + 2*e)*sqr 
t(-f)/f)*log(f) + sqrt(2)*sqrt(pi)*b*sqrt(f)*cosh(1/8*(b^2*log(f)^2 + 4*e^ 
2 - 16*d*f - 4*(b*e - 2*a*f)*log(f))/f)*erf(-1/4*sqrt(2)*(4*f*x - b*log(f) 
 + 2*e)/sqrt(f))*log(f) - sqrt(2)*sqrt(pi)*b*sqrt(-f)*erf(1/4*sqrt(2)*(4*f 
*x + b*log(f) + 2*e)*sqrt(-f)/f)*log(f)*sinh(1/8*(b^2*log(f)^2 + 4*e^2 - 1 
6*d*f + 4*(b*e - 2*a*f)*log(f))/f) + sqrt(2)*sqrt(pi)*b*sqrt(f)*erf(-1/4*s 
qrt(2)*(4*f*x - b*log(f) + 2*e)/sqrt(f))*log(f)*sinh(1/8*(b^2*log(f)^2 + 4 
*e^2 - 16*d*f - 4*(b*e - 2*a*f)*log(f))/f) - 8*f*cosh((b*x + a)*log(f)) - 
8*f*sinh((b*x + a)*log(f)))/(b*f*log(f))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.89 \[ \int f^{a+b x} \cosh ^2\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {-f} x - \frac {\sqrt {2} {\left (b \log \left (f\right ) + 2 \, e\right )}}{4 \, \sqrt {-f}}\right ) e^{\left (2 \, d - \frac {{\left (b \log \left (f\right ) + 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt {-f}} + \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {f} x - \frac {\sqrt {2} {\left (b \log \left (f\right ) - 2 \, e\right )}}{4 \, \sqrt {f}}\right ) e^{\left (-2 \, d + \frac {{\left (b \log \left (f\right ) - 2 \, e\right )}^{2}}{8 \, f}\right )}}{16 \, \sqrt {f}} + \frac {f^{b x + a}}{2 \, b \log \left (f\right )} \] Input:

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

Output:

1/16*sqrt(2)*sqrt(pi)*f^a*erf(sqrt(2)*sqrt(-f)*x - 1/4*sqrt(2)*(b*log(f) + 
 2*e)/sqrt(-f))*e^(2*d - 1/8*(b*log(f) + 2*e)^2/f)/sqrt(-f) + 1/16*sqrt(2) 
*sqrt(pi)*f^a*erf(sqrt(2)*sqrt(f)*x - 1/4*sqrt(2)*(b*log(f) - 2*e)/sqrt(f) 
)*e^(-2*d + 1/8*(b*log(f) - 2*e)^2/f)/sqrt(f) + 1/2*f^(b*x + a)/(b*log(f))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.40 \[ \int f^{a+b x} \cosh ^2\left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*sqrt(-f)*(4*x + (b*log(f) + 2*e)/f 
))*e^(-1/8*(b^2*log(f)^2 + 4*b*e*log(f) - 8*a*f*log(f) + 4*e^2 - 16*d*f)/f 
)/sqrt(-f) - 1/16*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*sqrt(f)*(4*x - (b*log( 
f) - 2*e)/f))*e^(1/8*(b^2*log(f)^2 - 4*b*e*log(f) + 8*a*f*log(f) + 4*e^2 - 
 16*d*f)/f)/sqrt(f) + (2*b*cos(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a* 
sgn(f) + 1/2*pi*a)*log(abs(f))/(4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - pi*b) 
^2) - (pi*b*sgn(f) - pi*b)*sin(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a* 
sgn(f) + 1/2*pi*a)/(4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - pi*b)^2))*e^(b*x* 
log(abs(f)) + a*log(abs(f))) + I*(I*e^(1/2*I*pi*b*x*sgn(f) - 1/2*I*pi*b*x 
+ 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(2*I*pi*b*sgn(f) - 2*I*pi*b + 4*b*log(ab 
s(f))) - I*e^(-1/2*I*pi*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi*a*sgn(f) + 1/ 
2*I*pi*a)/(-2*I*pi*b*sgn(f) + 2*I*pi*b + 4*b*log(abs(f))))*e^(b*x*log(abs( 
f)) + a*log(abs(f)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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