3.288 \(\int x^2 \coth ^{-1}(e^{a+b x}) \, dx\)

Optimal. Leaf size=119 \[ \frac {\text {Li}_4\left (-e^{-a-b x}\right )}{b^3}-\frac {\text {Li}_4\left (e^{-a-b x}\right )}{b^3}+\frac {x \text {Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{-a-b x}\right )}{b^2}+\frac {x^2 \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {Li}_2\left (e^{-a-b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6214, 2531, 6609, 2282, 6589} \[ \frac {x \text {PolyLog}\left (3,-e^{-a-b x}\right )}{b^2}-\frac {x \text {PolyLog}\left (3,e^{-a-b x}\right )}{b^2}+\frac {\text {PolyLog}\left (4,-e^{-a-b x}\right )}{b^3}-\frac {\text {PolyLog}\left (4,e^{-a-b x}\right )}{b^3}+\frac {x^2 \text {PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[E^(a + b*x)],x]

[Out]

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6214

Int[ArcCoth[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + 1/(a +
b*f^(c + d*x))], x], x] - Dist[1/2, Int[x^m*Log[1 - 1/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f},
x] && IGtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

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

Rubi steps

\begin {align*} \int x^2 \coth ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac {1}{2} \int x^2 \log \left (1-e^{-a-b x}\right ) \, dx\right )+\frac {1}{2} \int x^2 \log \left (1+e^{-a-b x}\right ) \, dx\\ &=\frac {x^2 \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {Li}_2\left (e^{-a-b x}\right )}{2 b}-\frac {\int x \text {Li}_2\left (-e^{-a-b x}\right ) \, dx}{b}+\frac {\int x \text {Li}_2\left (e^{-a-b x}\right ) \, dx}{b}\\ &=\frac {x^2 \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{-a-b x}\right )}{b^2}-\frac {\int \text {Li}_3\left (-e^{-a-b x}\right ) \, dx}{b^2}+\frac {\int \text {Li}_3\left (e^{-a-b x}\right ) \, dx}{b^2}\\ &=\frac {x^2 \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{-a-b x}\right )}{b^2}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-a-b x}\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{-a-b x}\right )}{b^3}\\ &=\frac {x^2 \text {Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac {x^2 \text {Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac {x \text {Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac {x \text {Li}_3\left (e^{-a-b x}\right )}{b^2}+\frac {\text {Li}_4\left (-e^{-a-b x}\right )}{b^3}-\frac {\text {Li}_4\left (e^{-a-b x}\right )}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 149, normalized size = 1.25 \[ \frac {b^3 x^3 \log \left (1-e^{a+b x}\right )-b^3 x^3 \log \left (e^{a+b x}+1\right )+2 b^3 x^3 \coth ^{-1}\left (e^{a+b x}\right )-3 b^2 x^2 \text {Li}_2\left (-e^{a+b x}\right )+3 b^2 x^2 \text {Li}_2\left (e^{a+b x}\right )+6 b x \text {Li}_3\left (-e^{a+b x}\right )-6 b x \text {Li}_3\left (e^{a+b x}\right )-6 \text {Li}_4\left (-e^{a+b x}\right )+6 \text {Li}_4\left (e^{a+b x}\right )}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[E^(a + b*x)],x]

[Out]

(2*b^3*x^3*ArcCoth[E^(a + b*x)] + b^3*x^3*Log[1 - E^(a + b*x)] - b^3*x^3*Log[1 + E^(a + b*x)] - 3*b^2*x^2*Poly
Log[2, -E^(a + b*x)] + 3*b^2*x^2*PolyLog[2, E^(a + b*x)] + 6*b*x*PolyLog[3, -E^(a + b*x)] - 6*b*x*PolyLog[3, E
^(a + b*x)] - 6*PolyLog[4, -E^(a + b*x)] + 6*PolyLog[4, E^(a + b*x)])/(6*b^3)

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fricas [C]  time = 0.57, size = 247, normalized size = 2.08 \[ \frac {b^{3} x^{3} \log \left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(b*x+a)),x, algorithm="fricas")

[Out]

1/6*(b^3*x^3*log((cosh(b*x + a) + sinh(b*x + a) + 1)/(cosh(b*x + a) + sinh(b*x + a) - 1)) - b^3*x^3*log(cosh(b
*x + a) + sinh(b*x + a) + 1) + 3*b^2*x^2*dilog(cosh(b*x + a) + sinh(b*x + a)) - 3*b^2*x^2*dilog(-cosh(b*x + a)
- sinh(b*x + a)) - a^3*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 6*b*x*polylog(3, cosh(b*x + a) + sinh(b*x + a
)) + 6*b*x*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + (b^3*x^3 + a^3)*log(-cosh(b*x + a) - sinh(b*x + a) + 1
) + 6*polylog(4, cosh(b*x + a) + sinh(b*x + a)) - 6*polylog(4, -cosh(b*x + a) - sinh(b*x + a)))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arccoth(e^(b*x + a)), x)

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maple [A]  time = 0.07, size = 185, normalized size = 1.55 \[ \frac {x^{3} \mathrm {arccoth}\left ({\mathrm e}^{b x +a}\right )}{3}-\frac {x^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{2 b}+\frac {x \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{6}+\frac {x^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b}+\frac {a^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{3 b^{3}}+\frac {\polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {x \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{6}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) a^{3}}{6 b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{6 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(exp(b*x+a)),x)

[Out]

1/3*x^3*arccoth(exp(b*x+a))-1/2*x^2*polylog(2,-exp(b*x+a))/b+x*polylog(3,-exp(b*x+a))/b^2+1/6*ln(1-exp(b*x+a))
*x^3+1/2*x^2*polylog(2,exp(b*x+a))/b+1/3/b^3*a^3*arctanh(exp(b*x+a))+polylog(4,exp(b*x+a))/b^3-polylog(4,-exp(
b*x+a))/b^3-x*polylog(3,exp(b*x+a))/b^2-1/6*ln(exp(b*x+a)+1)*x^3-1/6/b^3*ln(exp(b*x+a)+1)*a^3+1/6/b^3*ln(1-exp
(b*x+a))*a^3

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maxima [A]  time = 0.35, size = 142, normalized size = 1.19 \[ \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (e^{\left (b x + a\right )}\right ) - \frac {1}{6} \, b {\left (\frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} - \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(exp(b*x+a)),x, algorithm="maxima")

[Out]

1/3*x^3*arccoth(e^(b*x + a)) - 1/6*b*((b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*b*x*po
lylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^4 - (b^3*x^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*dilog(e^
(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)))/b^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {acoth}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acoth(exp(a + b*x)),x)

[Out]

int(x^2*acoth(exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(exp(b*x+a)),x)

[Out]

Integral(x**2*acoth(exp(a)*exp(b*x)), x)

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