\(\int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx\) [25]

Optimal result

Integrand size = 20, antiderivative size = 297 \[ \int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx=-\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\cosh \left (4 e-\frac {4 c f}{d}\right ) \text {Chi}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}-\frac {\text {Chi}\left (\frac {4 c f}{d}+4 f x\right ) \sinh \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\text {Chi}\left (\frac {2 c f}{d}+2 f x\right ) \sinh \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\cosh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+\frac {\sinh \left (4 e-\frac {4 c f}{d}\right ) \text {Shi}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d} \] Output:

-1/2*cosh(-2*e+2*c*f/d)*Chi(2*c*f/d+2*f*x)/a^2/d+1/4*cosh(-4*e+4*c*f/d)*Ch 
i(4*c*f/d+4*f*x)/a^2/d+1/4*ln(d*x+c)/a^2/d+1/4*Chi(4*c*f/d+4*f*x)*sinh(-4* 
e+4*c*f/d)/a^2/d-1/2*Chi(2*c*f/d+2*f*x)*sinh(-2*e+2*c*f/d)/a^2/d+1/2*cosh( 
-2*e+2*c*f/d)*Shi(2*c*f/d+2*f*x)/a^2/d+1/2*sinh(-2*e+2*c*f/d)*Shi(2*c*f/d+ 
2*f*x)/a^2/d-1/4*cosh(-4*e+4*c*f/d)*Shi(4*c*f/d+4*f*x)/a^2/d-1/4*sinh(-4*e 
+4*c*f/d)*Shi(4*c*f/d+4*f*x)/a^2/d
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx=\frac {\left (\cosh \left (2 e-\frac {2 c f}{d}\right )-\sinh \left (2 e-\frac {2 c f}{d}\right )\right ) \left (-2 \text {Chi}\left (\frac {2 f (c+d x)}{d}\right )+\cosh \left (2 e-\frac {2 c f}{d}\right ) \log (f (c+d x))+\text {Chi}\left (\frac {4 f (c+d x)}{d}\right ) \left (\cosh \left (2 e-\frac {2 c f}{d}\right )-\sinh \left (2 e-\frac {2 c f}{d}\right )\right )+\log (f (c+d x)) \sinh \left (2 e-\frac {2 c f}{d}\right )+2 \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )-\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {4 f (c+d x)}{d}\right )+\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {4 f (c+d x)}{d}\right )\right )}{4 a^2 d} \] Input:

Integrate[1/((c + d*x)*(a + a*Coth[e + f*x])^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 297, 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, 4211, 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.

\(\displaystyle \int \frac {1}{(c+d x) (a \coth (e+f x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4211

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

\(\Big \downarrow \) 2009

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

Input:

Int[1/((c + d*x)*(a + a*Coth[e + f*x])^2),x]
 

Output:

-1/2*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/(a^2*d) + (Co 
sh[4*e - (4*c*f)/d]*CoshIntegral[(4*c*f)/d + 4*f*x])/(4*a^2*d) + Log[c + d 
*x]/(4*a^2*d) - (CoshIntegral[(4*c*f)/d + 4*f*x]*Sinh[4*e - (4*c*f)/d])/(4 
*a^2*d) + (CoshIntegral[(2*c*f)/d + 2*f*x]*Sinh[2*e - (2*c*f)/d])/(2*a^2*d 
) + (Cosh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) - (S 
inh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) - (Cosh[4* 
e - (4*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(4*a^2*d) + (Sinh[4*e - (4 
*c*f)/d]*SinhIntegral[(4*c*f)/d + 4*f*x])/(4*a^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 4211

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.36

method	result	size
risch	\(\frac {\ln \left (d x +c \right )}{4 a^{2} d}-\frac {{\mathrm e}^{\frac {4 c f -4 d e}{d}} \operatorname {expIntegral}_{1}\left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right )}{4 a^{2} d}+\frac {{\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {expIntegral}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{2 a^{2} d}\)	\(106\)

Input:

int(1/(d*x+c)/(a+a*coth(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*ln(d*x+c)/a^2/d-1/4/a^2/d*exp(4*(c*f-d*e)/d)*Ei(1,4*f*x+4*e+4*(c*f-d*e 
)/d)+1/2/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 = 137, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx=-\frac {2 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \cosh \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + 2 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) - \log \left (d x + c\right )}{4 \, a^{2} d} \] Input:

integrate(1/(d*x+c)/(a+a*coth(f*x+e))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(1/(d*x+c)/(a+a*coth(f*x+e))**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.27 \[ \int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx=-\frac {e^{\left (-4 \, e + \frac {4 \, c f}{d}\right )} E_{1}\left (\frac {4 \, {\left (d x + c\right )} f}{d}\right )}{4 \, a^{2} d} + \frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{2 \, a^{2} d} + \frac {\log \left (d x + c\right )}{4 \, a^{2} d} \] Input:

integrate(1/(d*x+c)/(a+a*coth(f*x+e))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.26 \[ \int \frac {1}{(c+d x) (a+a \coth (e+f x))^2} \, dx=-\frac {{\left (2 \, {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (2 \, e + \frac {2 \, c f}{d}\right )} - {\rm Ei}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {4 \, c f}{d}\right )} - e^{\left (4 \, e\right )} \log \left (d x + c\right )\right )} e^{\left (-4 \, e\right )}}{4 \, a^{2} d} \] Input:

integrate(1/(d*x+c)/(a+a*coth(f*x+e))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(1/((a + a*coth(e + f*x))^2*(c + d*x)),x)
 

Output:

int(1/((a + a*coth(e + f*x))^2*(c + d*x)), x)
 

Reduce [F]

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

int(1/(d*x+c)/(a+a*coth(f*x+e))^2,x)
 

Output:

int(1/(coth(e + f*x)**2*c + coth(e + f*x)**2*d*x + 2*coth(e + f*x)*c + 2*c 
oth(e + f*x)*d*x + c + d*x),x)/a**2