### 3.469 $$\int x \text{csch}(a+b x) \text{sech}(a+b x) \, dx$$

Optimal. Leaf size=58 $-\frac{\text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac{\text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0557309, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {5461, 4182, 2279, 2391} $-\frac{\text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac{\text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Csch[a + b*x]*Sech[a + b*x],x]

[Out]

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x \text{csch}(a+b x) \text{sech}(a+b x) \, dx &=2 \int x \text{csch}(2 a+2 b x) \, dx\\ &=-\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\int \log \left (1-e^{2 a+2 b x}\right ) \, dx}{b}+\frac{\int \log \left (1+e^{2 a+2 b x}\right ) \, dx}{b}\\ &=-\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{2 b^2}+\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{2 b^2}\\ &=-\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac{\text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.079429, size = 110, normalized size = 1.9 $\frac{\text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-\text{PolyLog}\left (2,e^{-2 (a+b x)}\right )+2 a \log \left (1-e^{-2 (a+b x)}\right )+2 b x \log \left (1-e^{-2 (a+b x)}\right )-2 a \log \left (e^{-2 (a+b x)}+1\right )-2 b x \log \left (e^{-2 (a+b x)}+1\right )-2 a \log (\tanh (a+b x))}{2 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.039, size = 125, normalized size = 2.2 \begin{align*}{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.1433, size = 117, normalized size = 2.02 \begin{align*} -\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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

________________________________________________________________________________________

Fricas [C]  time = 2.4579, size = 686, normalized size = 11.83 \begin{align*} \frac{b x \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) -{\left (b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) -{\left (b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) +{\left (b x + a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) +{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) -{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) -{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*log(cosh(b*x + a) + sinh(b*x + a) + 1) + a*log(cosh(b*x + a) + sinh(b*x + a) + I) + a*log(cosh(b*x + a) +
sinh(b*x + a) - I) - a*log(cosh(b*x + a) + sinh(b*x + a) - 1) - (b*x + a)*log(I*cosh(b*x + a) + I*sinh(b*x +
a) + 1) - (b*x + a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (b*x + a)*log(-cosh(b*x + a) - sinh(b*x + a)
+ 1) + dilog(cosh(b*x + a) + sinh(b*x + a)) - dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - dilog(-I*cosh(b*x +
a) - I*sinh(b*x + a)) + dilog(-cosh(b*x + a) - sinh(b*x + a)))/b^2

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}\left (b x + a\right ) \operatorname{sech}\left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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