3.4.48 \(\int x^2 \tanh ^{-1}(e^x) \, dx\) [348]

Optimal. Leaf size=58 \[ -\frac {1}{2} x^2 \text {PolyLog}\left (2,-e^x\right )+\frac {1}{2} x^2 \text {PolyLog}\left (2,e^x\right )+x \text {PolyLog}\left (3,-e^x\right )-x \text {PolyLog}\left (3,e^x\right )-\text {PolyLog}\left (4,-e^x\right )+\text {PolyLog}\left (4,e^x\right ) \]

[Out]

-1/2*x^2*polylog(2,-exp(x))+1/2*x^2*polylog(2,exp(x))+x*polylog(3,-exp(x))-x*polylog(3,exp(x))-polylog(4,-exp(
x))+polylog(4,exp(x))

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Rubi [A]
time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6348, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {1}{2} x^2 \text {Li}_2\left (-e^x\right )+\frac {1}{2} x^2 \text {Li}_2\left (e^x\right )+x \text {Li}_3\left (-e^x\right )-x \text {Li}_3\left (e^x\right )-\text {Li}_4\left (-e^x\right )+\text {Li}_4\left (e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[E^x],x]

[Out]

-1/2*(x^2*PolyLog[2, -E^x]) + (x^2*PolyLog[2, E^x])/2 + x*PolyLog[3, -E^x] - x*PolyLog[3, E^x] - PolyLog[4, -E
^x] + PolyLog[4, E^x]

Rule 2320

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 2611

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 6348

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

Rule 6724

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 6744

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

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Mathematica [A]
time = 0.02, size = 93, normalized size = 1.60 \begin {gather*} \frac {1}{6} \left (2 x^3 \tanh ^{-1}\left (e^x\right )+x^3 \log \left (1-e^x\right )-x^3 \log \left (1+e^x\right )-3 x^2 \text {PolyLog}\left (2,-e^x\right )+3 x^2 \text {PolyLog}\left (2,e^x\right )+6 x \text {PolyLog}\left (3,-e^x\right )-6 x \text {PolyLog}\left (3,e^x\right )-6 \text {PolyLog}\left (4,-e^x\right )+6 \text {PolyLog}\left (4,e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[E^x],x]

[Out]

(2*x^3*ArcTanh[E^x] + x^3*Log[1 - E^x] - x^3*Log[1 + E^x] - 3*x^2*PolyLog[2, -E^x] + 3*x^2*PolyLog[2, E^x] + 6
*x*PolyLog[3, -E^x] - 6*x*PolyLog[3, E^x] - 6*PolyLog[4, -E^x] + 6*PolyLog[4, E^x])/6

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Maple [A]
time = 0.04, size = 79, normalized size = 1.36

method	result	size
risch	\(-\frac {x^{2} \polylog \left (2, -{\mathrm e}^{x}\right )}{2}+\frac {x^{2} \polylog \left (2, {\mathrm e}^{x}\right )}{2}+x \polylog \left (3, -{\mathrm e}^{x}\right )-x \polylog \left (3, {\mathrm e}^{x}\right )-\polylog \left (4, -{\mathrm e}^{x}\right )+\polylog \left (4, {\mathrm e}^{x}\right )\)	\(49\)
default	\(\frac {x^{3} \arctanh \left ({\mathrm e}^{x}\right )}{3}+\frac {x^{3} \ln \left (1-{\mathrm e}^{x}\right )}{6}+\frac {x^{2} \polylog \left (2, {\mathrm e}^{x}\right )}{2}-x \polylog \left (3, {\mathrm e}^{x}\right )+\polylog \left (4, {\mathrm e}^{x}\right )-\frac {x^{3} \ln \left ({\mathrm e}^{x}+1\right )}{6}-\frac {x^{2} \polylog \left (2, -{\mathrm e}^{x}\right )}{2}+x \polylog \left (3, -{\mathrm e}^{x}\right )-\polylog \left (4, -{\mathrm e}^{x}\right )\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 76, normalized size = 1.31 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {artanh}\left (e^{x}\right ) - \frac {1}{6} \, x^{3} \log \left (e^{x} + 1\right ) + \frac {1}{6} \, x^{3} \log \left (-e^{x} + 1\right ) - \frac {1}{2} \, x^{2} {\rm Li}_2\left (-e^{x}\right ) + \frac {1}{2} \, x^{2} {\rm Li}_2\left (e^{x}\right ) + x {\rm Li}_{3}(-e^{x}) - x {\rm Li}_{3}(e^{x}) - {\rm Li}_{4}(-e^{x}) + {\rm Li}_{4}(e^{x}) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(exp(x)),x, algorithm="maxima")

[Out]

1/3*x^3*arctanh(e^x) - 1/6*x^3*log(e^x + 1) + 1/6*x^3*log(-e^x + 1) - 1/2*x^2*dilog(-e^x) + 1/2*x^2*dilog(e^x)
+ x*polylog(3, -e^x) - x*polylog(3, e^x) - polylog(4, -e^x) + polylog(4, e^x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (46) = 92\).
time = 0.37, size = 120, normalized size = 2.07 \begin {gather*} \frac {1}{6} \, x^{3} \log \left (-\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac {1}{6} \, x^{3} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{6} \, x^{3} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, x^{2} {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac {1}{2} \, x^{2} {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) - x {\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + x {\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) + {\rm polylog}\left (4, \cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\rm polylog}\left (4, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(exp(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {atanh}{\left (e^{x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(exp(x)),x)

[Out]

Integral(x**2*atanh(exp(x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(exp(x)),x, algorithm="giac")

[Out]

integrate(x^2*arctanh(e^x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,\mathrm {atanh}\left ({\mathrm {e}}^x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(exp(x)),x)

[Out]

int(x^2*atanh(exp(x)), x)

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