\(\int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx\) [234]

Optimal result

Integrand size = 16, antiderivative size = 95 \[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\frac {x}{2 b}+\frac {2 x \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}-\frac {\coth (a+b x)}{2 b^2}-\frac {x \coth ^2(a+b x)}{2 b}+\frac {\operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}-\frac {\operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2} \] Output:

1/2*x/b+2*x*arctanh(exp(2*b*x+2*a))/b-1/2*coth(b*x+a)/b^2-1/2*x*coth(b*x+a 
)^2/b+1/2*polylog(2,-exp(2*b*x+2*a))/b^2-1/2*polylog(2,exp(2*b*x+2*a))/b^2
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=-\frac {\coth (a+b x)+b x \text {csch}^2(a+b x)+2 b x \log \left (1-e^{-2 (a+b x)}\right )-2 b x \log \left (1+e^{-2 (a+b x)}\right )+\operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )-\operatorname {PolyLog}\left (2,e^{-2 (a+b x)}\right )}{2 b^2} \] Input:

Integrate[x*Csch[a + b*x]^3*Sech[a + b*x],x]
 

Output:

-1/2*(Coth[a + b*x] + b*x*Csch[a + b*x]^2 + 2*b*x*Log[1 - E^(-2*(a + b*x)) 
] - 2*b*x*Log[1 + E^(-2*(a + b*x))] + PolyLog[2, -E^(-2*(a + b*x))] - Poly 
Log[2, E^(-2*(a + b*x))])/b^2
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5985, 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.

\(\displaystyle \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx\)

\(\Big \downarrow \) 5985

\(\displaystyle -\int \left (-\frac {\coth ^2(a+b x)}{2 b}-\frac {\log (\tanh (a+b x))}{b}\right )dx-\frac {x \coth ^2(a+b x)}{2 b}-\frac {x \log (\tanh (a+b x))}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 x \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}+\frac {\operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}-\frac {\operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {\coth (a+b x)}{2 b^2}-\frac {x \coth ^2(a+b x)}{2 b}+\frac {x}{2 b}\)

Input:

Int[x*Csch[a + b*x]^3*Sech[a + b*x],x]
 

Output:

x/(2*b) + (2*x*ArcTanh[E^(2*a + 2*b*x)])/b - Coth[a + b*x]/(2*b^2) - (x*Co 
th[a + b*x]^2)/(2*b) + PolyLog[2, -E^(2*a + 2*b*x)]/(2*b^2) - PolyLog[2, E 
^(2*a + 2*b*x)]/(2*b^2)
 

Defintions of rubi rules used

rule 2009

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

rule 5985

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + 
(b_.)*(x_)]^(p_.), x_Symbol] :> With[{u = IntHide[Csch[a + b*x]^n*Sech[a + 
b*x]^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, 
 x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n 
, p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(169\) vs. \(2(82)=164\).

Time = 2.26 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.79

method	result	size
risch	\(-\frac {2 b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}-1}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x}{b}-\frac {\operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {x \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b}+\frac {\operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {\operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}}\)	\(170\)

Input:

int(x*csch(b*x+a)^3*sech(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

-(2*b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)-1)/b^2/(exp(2*b*x+2*a)-1)^2-1/b*ln(e 
xp(b*x+a)+1)*x-polylog(2,-exp(b*x+a))/b^2+x*ln(exp(2*b*x+2*a)+1)/b+1/2*pol 
ylog(2,-exp(2*b*x+2*a))/b^2-1/b*ln(1-exp(b*x+a))*x-1/b^2*ln(1-exp(b*x+a))* 
a-polylog(2,exp(b*x+a))/b^2+1/b^2*a*ln(exp(b*x+a)-1)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 1578, normalized size of antiderivative = 16.61 \[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(x*csch(b*x+a)^3*sech(b*x+a),x, algorithm="fricas")
 

Output:

-((2*b*x + 1)*cosh(b*x + a)^2 + 2*(2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a) 
+ (2*b*x + 1)*sinh(b*x + a)^2 + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b* 
x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2 
*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)* 
dilog(cosh(b*x + a) + sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)* 
sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a 
)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a 
) + 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cos 
h(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*s 
inh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*s 
inh(b*x + a) + 1)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (cosh(b*x + 
a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + 
 a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh 
(b*x + a))*sinh(b*x + a) + 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*x 
*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a) 
^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a) 
^2 + b*x + 4*(b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*log( 
cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a 
)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh( 
b*x + a)^2 - a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a...
 

Sympy [F]

\[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int x \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \] Input:

integrate(x*csch(b*x+a)**3*sech(b*x+a),x)
                                                                                    
                                                                                    
 

Output:

Integral(x*csch(a + b*x)**3*sech(a + b*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.53 \[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=-\frac {{\left (2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 1}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} - \frac {b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} - \frac {b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \] Input:

integrate(x*csch(b*x+a)^3*sech(b*x+a),x, algorithm="maxima")
 

Output:

-((2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x) - 1)/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^ 
(2*b*x + 2*a) + b^2) + 1/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b 
*x + 2*a)))/b^2 - (b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^2 - ( 
b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^2
 

Giac [F]

\[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int { x \operatorname {csch}\left (b x + a\right )^{3} \operatorname {sech}\left (b x + a\right ) \,d x } \] Input:

integrate(x*csch(b*x+a)^3*sech(b*x+a),x, algorithm="giac")
 

Output:

integrate(x*csch(b*x + a)^3*sech(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int \frac {x}{\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \] Input:

int(x/(cosh(a + b*x)*sinh(a + b*x)^3),x)
 

Output:

int(x/(cosh(a + b*x)*sinh(a + b*x)^3), x)
 

Reduce [F]

\[ \int x \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\frac {-16 e^{4 b x +6 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}-2 e^{6 b x +6 a}+2 e^{2 b x +2 a}-1}d x \right ) b^{2}-2 e^{4 b x +4 a} \mathrm {log}\left (e^{b x +a}-1\right )-2 e^{4 b x +4 a} \mathrm {log}\left (e^{b x +a}+1\right )+4 e^{4 b x +4 a} b x -e^{4 b x +4 a}+32 e^{2 b x +4 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}-2 e^{6 b x +6 a}+2 e^{2 b x +2 a}-1}d x \right ) b^{2}+4 e^{2 b x +2 a} \mathrm {log}\left (e^{b x +a}-1\right )+4 e^{2 b x +2 a} \mathrm {log}\left (e^{b x +a}+1\right )-8 e^{2 b x +2 a} b x -16 e^{2 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}-2 e^{6 b x +6 a}+2 e^{2 b x +2 a}-1}d x \right ) b^{2}-2 \,\mathrm {log}\left (e^{b x +a}-1\right )-2 \,\mathrm {log}\left (e^{b x +a}+1\right )+1}{b^{2} \left (e^{4 b x +4 a}-2 e^{2 b x +2 a}+1\right )} \] Input:

int(x*csch(b*x+a)^3*sech(b*x+a),x)
 

Output:

( - 16*e**(6*a + 4*b*x)*int((e**(2*b*x)*x)/(e**(8*a + 8*b*x) - 2*e**(6*a + 
 6*b*x) + 2*e**(2*a + 2*b*x) - 1),x)*b**2 - 2*e**(4*a + 4*b*x)*log(e**(a + 
 b*x) - 1) - 2*e**(4*a + 4*b*x)*log(e**(a + b*x) + 1) + 4*e**(4*a + 4*b*x) 
*b*x - e**(4*a + 4*b*x) + 32*e**(4*a + 2*b*x)*int((e**(2*b*x)*x)/(e**(8*a 
+ 8*b*x) - 2*e**(6*a + 6*b*x) + 2*e**(2*a + 2*b*x) - 1),x)*b**2 + 4*e**(2* 
a + 2*b*x)*log(e**(a + b*x) - 1) + 4*e**(2*a + 2*b*x)*log(e**(a + b*x) + 1 
) - 8*e**(2*a + 2*b*x)*b*x - 16*e**(2*a)*int((e**(2*b*x)*x)/(e**(8*a + 8*b 
*x) - 2*e**(6*a + 6*b*x) + 2*e**(2*a + 2*b*x) - 1),x)*b**2 - 2*log(e**(a + 
 b*x) - 1) - 2*log(e**(a + b*x) + 1) + 1)/(b**2*(e**(4*a + 4*b*x) - 2*e**( 
2*a + 2*b*x) + 1))