3.3.20 \(\int x^2 \cot ^{-1}(e^x) \, dx\) [220]

Optimal. Leaf size=103 \[ -\frac {1}{2} i x^2 \text {PolyLog}\left (2,-i e^{-x}\right )+\frac {1}{2} i x^2 \text {PolyLog}\left (2,i e^{-x}\right )-i x \text {PolyLog}\left (3,-i e^{-x}\right )+i x \text {PolyLog}\left (3,i e^{-x}\right )-i \text {PolyLog}\left (4,-i e^{-x}\right )+i \text {PolyLog}\left (4,i e^{-x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 103, 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 = {5252, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {1}{2} i x^2 \text {Li}_2\left (-i e^{-x}\right )+\frac {1}{2} i x^2 \text {Li}_2\left (i e^{-x}\right )-i x \text {Li}_3\left (-i e^{-x}\right )+i x \text {Li}_3\left (i e^{-x}\right )-i \text {Li}_4\left (-i e^{-x}\right )+i \text {Li}_4\left (i e^{-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/2*I)*x^2*PolyLog[2, (-I)/E^x] + (I/2)*x^2*PolyLog[2, I/E^x] - I*x*PolyLog[3, (-I)/E^x] + I*x*PolyLog[3, I/
E^x] - I*PolyLog[4, (-I)/E^x] + I*PolyLog[4, I/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 5252

Int[ArcCot[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I/(a +
b*f^(c + d*x))], x], x] - Dist[I/2, Int[x^m*Log[1 + I/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f}, x
] && IntegerQ[m] && 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 \cot ^{-1}\left (e^x\right ) \, dx &=\frac {1}{2} i \int x^2 \log \left (1-i e^{-x}\right ) \, dx-\frac {1}{2} i \int x^2 \log \left (1+i e^{-x}\right ) \, dx\\ &=-\frac {1}{2} i x^2 \text {Li}_2\left (-i e^{-x}\right )+\frac {1}{2} i x^2 \text {Li}_2\left (i e^{-x}\right )+i \int x \text {Li}_2\left (-i e^{-x}\right ) \, dx-i \int x \text {Li}_2\left (i e^{-x}\right ) \, dx\\ &=-\frac {1}{2} i x^2 \text {Li}_2\left (-i e^{-x}\right )+\frac {1}{2} i x^2 \text {Li}_2\left (i e^{-x}\right )-i x \text {Li}_3\left (-i e^{-x}\right )+i x \text {Li}_3\left (i e^{-x}\right )+i \int \text {Li}_3\left (-i e^{-x}\right ) \, dx-i \int \text {Li}_3\left (i e^{-x}\right ) \, dx\\ &=-\frac {1}{2} i x^2 \text {Li}_2\left (-i e^{-x}\right )+\frac {1}{2} i x^2 \text {Li}_2\left (i e^{-x}\right )-i x \text {Li}_3\left (-i e^{-x}\right )+i x \text {Li}_3\left (i e^{-x}\right )-i \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{-x}\right )+i \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{-x}\right )\\ &=-\frac {1}{2} i x^2 \text {Li}_2\left (-i e^{-x}\right )+\frac {1}{2} i x^2 \text {Li}_2\left (i e^{-x}\right )-i x \text {Li}_3\left (-i e^{-x}\right )+i x \text {Li}_3\left (i e^{-x}\right )-i \text {Li}_4\left (-i e^{-x}\right )+i \text {Li}_4\left (i e^{-x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 103, normalized size = 1.00 \begin {gather*} -\frac {1}{2} i x^2 \text {PolyLog}\left (2,-i e^{-x}\right )+\frac {1}{2} i x^2 \text {PolyLog}\left (2,i e^{-x}\right )-i x \text {PolyLog}\left (3,-i e^{-x}\right )+i x \text {PolyLog}\left (3,i e^{-x}\right )-i \text {PolyLog}\left (4,-i e^{-x}\right )+i \text {PolyLog}\left (4,i e^{-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 76, normalized size = 0.74

method	result	size
risch	\(\frac {\pi \,x^{3}}{6}+\frac {i \polylog \left (2, i {\mathrm e}^{x}\right ) x^{2}}{2}-i x \polylog \left (3, i {\mathrm e}^{x}\right )+i \polylog \left (4, i {\mathrm e}^{x}\right )-\frac {i \polylog \left (2, -i {\mathrm e}^{x}\right ) x^{2}}{2}+i \polylog \left (3, -i {\mathrm e}^{x}\right ) x -i \polylog \left (4, -i {\mathrm e}^{x}\right )\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(e^(-x)) + integrate(1/3*x^3*e^x/(e^(2*x) + 1), x)

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Fricas [A]
time = 7.06, size = 87, normalized size = 0.84 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {arccot}\left (e^{x}\right ) - \frac {1}{6} i \, x^{3} \log \left (i \, e^{x} + 1\right ) + \frac {1}{6} i \, x^{3} \log \left (-i \, e^{x} + 1\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (i \, e^{x}\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-i \, e^{x}\right ) - i \, x {\rm polylog}\left (3, i \, e^{x}\right ) + i \, x {\rm polylog}\left (3, -i \, e^{x}\right ) + i \, {\rm polylog}\left (4, i \, e^{x}\right ) - i \, {\rm polylog}\left (4, -i \, e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*acot(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*arccot(exp(x)),x, algorithm="giac")

[Out]

integrate(x^2*arccot(e^x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {acot}\left ({\mathrm {e}}^x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*acot(exp(x)), x)

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