### 3.426 $$\int x^2 \coth (a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=59 $-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}$

[Out]

(-4*x*ArcTanh[E^(a + b*x)])/b^2 - (x^2*Csch[a + b*x])/b - (2*PolyLog[2, -E^(a + b*x)])/b^3 + (2*PolyLog[2, E^(
a + b*x)])/b^3

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Rubi [A]  time = 0.0605293, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {5419, 4182, 2279, 2391} $-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

(-4*x*ArcTanh[E^(a + b*x)])/b^2 - (x^2*Csch[a + b*x])/b - (2*PolyLog[2, -E^(a + b*x)])/b^3 + (2*PolyLog[2, E^(
a + b*x)])/b^3

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

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^2 \coth (a+b x) \text{csch}(a+b x) \, dx &=-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{2 \int x \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \int \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{2 \int \log \left (1+e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}\\ \end{align*}

Mathematica [B]  time = 0.830012, size = 133, normalized size = 2.25 $-\frac{-2 \text{PolyLog}\left (2,-e^{-a-b x}\right )+2 \text{PolyLog}\left (2,e^{-a-b x}\right )+b^2 x^2 \text{csch}(a+b x)-2 b x \log \left (1-e^{-a-b x}\right )+2 b x \log \left (e^{-a-b x}+1\right )-2 a \log \left (1-e^{-a-b x}\right )+2 a \log \left (e^{-a-b x}+1\right )+2 a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

-((b^2*x^2*Csch[a + b*x] - 2*a*Log[1 - E^(-a - b*x)] - 2*b*x*Log[1 - E^(-a - b*x)] + 2*a*Log[1 + E^(-a - b*x)]
+ 2*b*x*Log[1 + E^(-a - b*x)] + 2*a*Log[Tanh[(a + b*x)/2]] - 2*PolyLog[2, -E^(-a - b*x)] + 2*PolyLog[2, E^(-a
- b*x)])/b^3)

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Maple [B]  time = 0.027, size = 134, normalized size = 2.3 \begin{align*} -2\,{\frac{{x}^{2}{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-2\,{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{a{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(b*x+a)*csch(b*x+a)^2,x)

[Out]

-2*x^2*exp(b*x+a)/b/(exp(2*b*x+2*a)-1)-2/b^2*ln(1+exp(b*x+a))*x-2/b^3*ln(1+exp(b*x+a))*a-2*polylog(2,-exp(b*x+
a))/b^3+2/b^2*ln(1-exp(b*x+a))*x+2/b^3*ln(1-exp(b*x+a))*a+2*polylog(2,exp(b*x+a))/b^3+4/b^3*a*arctanh(exp(b*x+
a))

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Maxima [A]  time = 1.36937, size = 112, normalized size = 1.9 \begin{align*} -\frac{2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{2 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac{2 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

-2*x^2*e^(b*x + a)/(b*e^(2*b*x + 2*a) - b) - 2*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^3 + 2*(b*x*l
og(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^3

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Fricas [B]  time = 2.13463, size = 1022, normalized size = 17.32 \begin{align*} -\frac{2 \,{\left (b^{2} x^{2} \cosh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right ) -{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) +{\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) +{\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} - a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) -{\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x + a\right )} \sinh \left (b x + a\right )^{2} - b x - a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )\right )}}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} - b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

-2*(b^2*x^2*cosh(b*x + a) + b^2*x^2*sinh(b*x + a) - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*
x + a)^2 - 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) + (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b
*x + a)^2 - 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a
) + b*x*sinh(b*x + a)^2 - b*x)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^2 + 2*a*cosh(b*x + a)
*sinh(b*x + a) + a*sinh(b*x + a)^2 - a)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - ((b*x + a)*cosh(b*x + a)^2 +
2*(b*x + a)*cosh(b*x + a)*sinh(b*x + a) + (b*x + a)*sinh(b*x + a)^2 - b*x - a)*log(-cosh(b*x + a) - sinh(b*x +
a) + 1))/(b^3*cosh(b*x + a)^2 + 2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*x + a)^2 - b^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cosh(b*x+a)*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^2*cosh(b*x + a)*csch(b*x + a)^2, x)