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

Optimal. Leaf size=59 \[ -\frac {2 \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {2 \text {Li}_2\left (e^{a+b x}\right )}{b^3}-\frac {4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^2 \text {csch}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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]

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*}

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Mathematica [B]  time = 0.79, size = 133, normalized size = 2.25 \[ -\frac {b^2 x^2 \text {csch}(a+b x)-2 \text {Li}_2\left (-e^{-a-b x}\right )+2 \text {Li}_2\left (e^{-a-b x}\right )-2 b x \log \left (1-e^{-a-b x}\right )+2 b x \log \left (e^{-a-b x}+1\right )-2 a \log \left (1-e^{-a-b x}\right )+2 a \log \left (e^{-a-b x}+1\right )+2 a \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{b^3} \]

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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fricas [B]  time = 0.49, size = 367, normalized size = 6.22 \[ -\frac {2 \, {\left (b^{2} x^{2} \cosh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} - a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} - b x - a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )\right )}}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} - b^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh \left (b x + a\right ) \operatorname {csch}\left (b x + a\right )^{2}\,{d x} \]

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)

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maple [B]  time = 0.22, size = 134, normalized size = 2.27 \[ -\frac {2 x^{2} {\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {2 \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 \ln \left (1+{\mathrm e}^{b x +a}\right ) a}{b^{3}}-\frac {2 \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{3}}+\frac {2 \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {4 a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]

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 = 0.90, size = 83, normalized size = 1.41 \[ -\frac {2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac {2 \, {\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac {2 \, {\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh {\left (a + b x \right )} \operatorname {csch}^{2}{\left (a + b x \right )}\, dx \]

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]

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

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