\(\int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx\) [166]

Optimal result

Integrand size = 20, antiderivative size = 156 \[ \int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx=\frac {b^2 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}+\frac {a^2 \log (c+d x)}{d}-\frac {b^2 \log (c+d x)}{2 d}+\frac {2 a b \text {Chi}\left (\frac {c f}{d}+f x\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}+\frac {2 a b \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d}+\frac {b^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx=\frac {b^2 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 f (c+d x)}{d}\right )+2 a^2 \log (c+d x)-b^2 \log (c+d x)+4 a b \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right ) \sinh \left (e-\frac {c f}{d}\right )+4 a b \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )+b^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3798, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3798

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b^2*Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(2*d) + (a^2*L 
og[c + d*x])/d - (b^2*Log[c + d*x])/(2*d) + (2*a*b*CoshIntegral[(c*f)/d + 
f*x]*Sinh[e - (c*f)/d])/d + (2*a*b*Cosh[e - (c*f)/d]*SinhIntegral[(c*f)/d 
+ f*x])/d + (b^2*Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2 
*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 3798

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

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.29

method	result	size
risch	\(-\frac {a b \,{\mathrm e}^{-\frac {c f -d e}{d}} \operatorname {expIntegral}_{1}\left (-f x -e -\frac {c f -d e}{d}\right )}{d}+\frac {a^{2} \ln \left (d x +c \right )}{d}-\frac {b^{2} \ln \left (d x +c \right )}{2 d}-\frac {b^{2} {\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {expIntegral}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{4 d}-\frac {b^{2} {\mathrm e}^{-\frac {2 \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 f x -2 e -\frac {2 \left (c f -d e \right )}{d}\right )}{4 d}+\frac {a b \,{\mathrm e}^{\frac {c f -d e}{d}} \operatorname {expIntegral}_{1}\left (f x +e +\frac {c f -d e}{d}\right )}{d}\)	\(201\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx=\frac {4 \, {\left (a b {\rm Ei}\left (\frac {d f x + c f}{d}\right ) - a b {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \cosh \left (-\frac {d e - c f}{d}\right ) + {\left (b^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + b^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \cosh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 2 \, {\left (2 \, a^{2} - b^{2}\right )} \log \left (d x + c\right ) - 4 \, {\left (a b {\rm Ei}\left (\frac {d f x + c f}{d}\right ) + a b {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \sinh \left (-\frac {d e - c f}{d}\right ) - {\left (b^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - b^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right )\right )} \sinh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )}{4 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{1}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {2 \, \log \left (d x + c\right )}{d}\right )} + a b {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{1}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {e^{\left (e - \frac {c f}{d}\right )} E_{1}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac {a^{2} \log \left (d x + c\right )}{d} \] Input:

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

Output:

-1/4*b^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* 
b*(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 = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \sinh (e+f x))^2}{c+d x} \, dx=\frac {b^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} + 4 \, a b {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} - 4 \, a b {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + b^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} + 4 \, a^{2} \log \left (d x + c\right ) - 2 \, b^{2} \log \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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