\(\int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx\) [108]

Optimal result

Integrand size = 20, antiderivative size = 145 \[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=\frac {2 a^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (\frac {c f}{d}+f x\right )}{d}+\frac {a^2 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}+\frac {3 a^2 \log (c+d x)}{2 d}+\frac {2 a^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d}+\frac {a^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d} \] Output:

2*a^2*cosh(-e+c*f/d)*Chi(c*f/d+f*x)/d+1/2*a^2*cosh(-2*e+2*c*f/d)*Chi(2*c*f 
/d+2*f*x)/d+3/2*a^2*ln(d*x+c)/d-2*a^2*sinh(-e+c*f/d)*Shi(c*f/d+f*x)/d-1/2* 
a^2*sinh(-2*e+2*c*f/d)*Shi(2*c*f/d+2*f*x)/d
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.78 \[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=\frac {a^2 \left (4 \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right )+\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 f (c+d x)}{d}\right )+3 \log (c+d x)+4 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )+\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )\right )}{2 d} \] Input:

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

Output:

(a^2*(4*Cosh[e - (c*f)/d]*CoshIntegral[f*(c/d + x)] + Cosh[2*e - (2*c*f)/d 
]*CoshIntegral[(2*f*(c + d*x))/d] + 3*Log[c + d*x] + 4*Sinh[e - (c*f)/d]*S 
inhIntegral[f*(c/d + x)] + Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*f*(c + d* 
x))/d]))/(2*d)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3799, 3042, 3793, 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.

\(\displaystyle \int \frac {(a \cosh (e+f x)+a)^2}{c+d x} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+a \sin \left (i e+i f x+\frac {\pi }{2}\right )\right )^2}{c+d x}dx\)

\(\Big \downarrow \) 3799

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

\(\Big \downarrow \) 3042

\(\displaystyle 4 a^2 \int \frac {\sin \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{2}\right )^4}{c+d x}dx\)

\(\Big \downarrow \) 3793

\(\displaystyle 4 a^2 \int \left (\frac {\cosh (e+f x)}{2 (c+d x)}+\frac {\cosh (2 e+2 f x)}{8 (c+d x)}+\frac {3}{8 (c+d x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 a^2 \left (\frac {\text {Chi}\left (x f+\frac {c f}{d}\right ) \cosh \left (e-\frac {c f}{d}\right )}{2 d}+\frac {\text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{8 d}+\frac {\sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{2 d}+\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{8 d}+\frac {3 \log (c+d x)}{8 d}\right )\)

Input:

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

Output:

4*a^2*((Cosh[e - (c*f)/d]*CoshIntegral[(c*f)/d + f*x])/(2*d) + (Cosh[2*e - 
 (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(8*d) + (3*Log[c + d*x])/(8*d 
) + (Sinh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])/(2*d) + (Sinh[2*e - (2 
*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(8*d))
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Simp[(2*a)^n   Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) 
+ f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 
2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
 

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.32

method	result	size
risch	\(-\frac {a^{2} {\mathrm e}^{\frac {c f -d e}{d}} \operatorname {expIntegral}_{1}\left (f x +e +\frac {c f -d e}{d}\right )}{d}-\frac {a^{2} {\mathrm e}^{-\frac {c f -d e}{d}} \operatorname {expIntegral}_{1}\left (-f x -e -\frac {c f -d e}{d}\right )}{d}+\frac {3 a^{2} \ln \left (d x +c \right )}{2 d}-\frac {a^{2} {\mathrm e}^{-\frac {2 \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 f x -2 e -\frac {2 \left (c f -d e \right )}{d}\right )}{4 d}-\frac {a^{2} {\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {expIntegral}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{4 d}\)	\(191\)

Input:

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

Output:

-a^2/d*exp((c*f-d*e)/d)*Ei(1,f*x+e+(c*f-d*e)/d)-a^2/d*exp(-(c*f-d*e)/d)*Ei 
(1,-f*x-e-(c*f-d*e)/d)+3/2*a^2*ln(d*x+c)/d-1/4*a^2/d*exp(-2*(c*f-d*e)/d)*E 
i(1,-2*f*x-2*e-2*(c*f-d*e)/d)-1/4*a^2/d*exp(2*(c*f-d*e)/d)*Ei(1,2*f*x+2*e+ 
2*(c*f-d*e)/d)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.57 \[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=\frac {6 \, a^{2} \log \left (d x + c\right ) + 4 \, {\left (a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) + a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \cosh \left (-\frac {d e - c f}{d}\right ) + {\left (a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cosh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - 4 \, {\left (a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) - a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \sinh \left (-\frac {d e - c f}{d}\right ) - {\left (a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \sinh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )}{4 \, d} \] Input:

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

Output:

1/4*(6*a^2*log(d*x + c) + 4*(a^2*Ei((d*f*x + c*f)/d) + a^2*Ei(-(d*f*x + c* 
f)/d))*cosh(-(d*e - c*f)/d) + (a^2*Ei(2*(d*f*x + c*f)/d) + a^2*Ei(-2*(d*f* 
x + c*f)/d))*cosh(-2*(d*e - c*f)/d) - 4*(a^2*Ei((d*f*x + c*f)/d) - a^2*Ei( 
-(d*f*x + c*f)/d))*sinh(-(d*e - c*f)/d) - (a^2*Ei(2*(d*f*x + c*f)/d) - a^2 
*Ei(-2*(d*f*x + c*f)/d))*sinh(-2*(d*e - c*f)/d))/d
 

Sympy [F]

\[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=a^{2} \left (\int \frac {2 \cosh {\left (e + f x \right )}}{c + d x}\, dx + \int \frac {\cosh ^{2}{\left (e + f x \right )}}{c + d x}\, dx + \int \frac {1}{c + d x}\, dx\right ) \] Input:

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

Output:

a**2*(Integral(2*cosh(e + f*x)/(c + d*x), x) + Integral(cosh(e + f*x)**2/( 
c + d*x), x) + Integral(1/(c + d*x), x))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=-\frac {1}{4} \, a^{2} {\left (\frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{1}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} - \frac {2 \, \log \left (d x + c\right )}{d}\right )} - a^{2} {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{1}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} + \frac {e^{\left (e - \frac {c f}{d}\right )} E_{1}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac {a^{2} \log \left (d x + c\right )}{d} \] Input:

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

Output:

-1/4*a^2*(e^(-2*e + 2*c*f/d)*exp_integral_e(1, 2*(d*x + c)*f/d)/d + e^(2*e 
 - 2*c*f/d)*exp_integral_e(1, -2*(d*x + c)*f/d)/d - 2*log(d*x + c)/d) - a^ 
2*(e^(-e + c*f/d)*exp_integral_e(1, (d*x + c)*f/d)/d + e^(e - c*f/d)*exp_i 
ntegral_e(1, -(d*x + c)*f/d)/d) + a^2*log(d*x + c)/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.93 \[ \int \frac {(a+a \cosh (e+f x))^2}{c+d x} \, dx=\frac {a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} + 4 \, a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + 4 \, a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} + 6 \, a^{2} \log \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

1/4*(a^2*Ei(2*(d*f*x + c*f)/d)*e^(2*e - 2*c*f/d) + 4*a^2*Ei((d*f*x + c*f)/ 
d)*e^(e - c*f/d) + 4*a^2*Ei(-(d*f*x + c*f)/d)*e^(-e + c*f/d) + a^2*Ei(-2*( 
d*f*x + c*f)/d)*e^(-2*e + 2*c*f/d) + 6*a^2*log(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(a**2*(2*int(cosh(e + f*x)/(c + d*x),x)*d + int(cosh(e + f*x)**2/(c + d*x) 
,x)*d + log(c + d*x)))/d