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

Optimal. Leaf size=151 \[ -\frac {i \text {Li}_4\left (-i e^{-a-b x}\right )}{b^3}+\frac {i \text {Li}_4\left (i e^{-a-b x}\right )}{b^3}-\frac {i x \text {Li}_3\left (-i e^{-a-b x}\right )}{b^2}+\frac {i x \text {Li}_3\left (i e^{-a-b x}\right )}{b^2}-\frac {i x^2 \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x^2 \text {Li}_2\left (i e^{-a-b x}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 151, 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 = {5144, 2531, 6609, 2282, 6589} \[ -\frac {i x \text {PolyLog}\left (3,-i e^{-a-b x}\right )}{b^2}+\frac {i x \text {PolyLog}\left (3,i e^{-a-b x}\right )}{b^2}-\frac {i \text {PolyLog}\left (4,-i e^{-a-b x}\right )}{b^3}+\frac {i \text {PolyLog}\left (4,i e^{-a-b x}\right )}{b^3}-\frac {i x^2 \text {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x^2 \text {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCot[E^(a + b*x)],x]

[Out]

((-I/2)*x^2*PolyLog[2, (-I)*E^(-a - b*x)])/b + ((I/2)*x^2*PolyLog[2, I*E^(-a - b*x)])/b - (I*x*PolyLog[3, (-I)
*E^(-a - b*x)])/b^2 + (I*x*PolyLog[3, I*E^(-a - b*x)])/b^2 - (I*PolyLog[4, (-I)*E^(-a - b*x)])/b^3 + (I*PolyLo
g[4, I*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 5144

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

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Mathematica [A]  time = 0.01, size = 151, normalized size = 1.00 \[ -\frac {i \text {Li}_4\left (-i e^{-a-b x}\right )}{b^3}+\frac {i \text {Li}_4\left (i e^{-a-b x}\right )}{b^3}-\frac {i x \text {Li}_3\left (-i e^{-a-b x}\right )}{b^2}+\frac {i x \text {Li}_3\left (i e^{-a-b x}\right )}{b^2}-\frac {i x^2 \text {Li}_2\left (-i e^{-a-b x}\right )}{2 b}+\frac {i x^2 \text {Li}_2\left (i e^{-a-b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCot[E^(a + b*x)],x]

[Out]

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

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fricas [C]  time = 0.82, size = 187, normalized size = 1.24 \[ \frac {2 \, b^{3} x^{3} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) + 3 i \, b^{2} x^{2} {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - 3 i \, b^{2} x^{2} {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) - i \, a^{3} \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a^{3} \log \left (e^{\left (b x + a\right )} - i\right ) - 6 i \, b x {\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) + 6 i \, b x {\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right ) + {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b^{3} x^{3} + i \, a^{3}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (4, i \, e^{\left (b x + a\right )}\right ) - 6 i \, {\rm polylog}\left (4, -i \, e^{\left (b x + a\right )}\right )}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccot(e^(b*x + a)), x)

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maple [B]  time = 0.34, size = 413, normalized size = 2.74 \[ \frac {\pi \,x^{3}}{6}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x \,a^{2}}{2 b^{2}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{3}}{2 b^{3}}-\frac {i \polylog \left (4, -i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {i \dilog \left (-i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {i a^{3} \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{3}}+\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{3}}{2 b^{3}}-\frac {i \polylog \left (2, i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}+\frac {i \polylog \left (2, -i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {i \polylog \left (3, i {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {i \dilog \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{3}}+\frac {i \polylog \left (2, i {\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) x \,a^{2}}{2 b^{2}}-\frac {i \polylog \left (2, -i {\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{3}}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x \,a^{2}}{2 b^{2}}+\frac {i \polylog \left (4, i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {i \polylog \left (3, -i {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {i a^{3} \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{3}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) x \,a^{2}}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*Pi*x^3-1/2*I/b^2*ln(1-I*exp(b*x+a))*x*a^2-1/2*I/b^3*ln(-I*(-exp(b*x+a)+I))*a^3+1/2*I/b*polylog(2,I*exp(b*x
+a))*x^2+1/2*I/b^3*dilog(-I*(exp(b*x+a)+I))*a^2-1/2*I/b^3*a^3*ln(1-I*exp(b*x+a))+1/2*I/b^3*ln(-I*(exp(b*x+a)+I
))*a^3+I/b^3*polylog(4,I*exp(b*x+a))+I/b^2*polylog(3,-I*exp(b*x+a))*x-I/b^3*polylog(4,-I*exp(b*x+a))+1/2*I/b^3
*dilog(-I*exp(b*x+a))*a^2-I/b^2*polylog(3,I*exp(b*x+a))*x+1/2*I/b^2*ln(-I*(exp(b*x+a)+I))*x*a^2+1/2*I/b^3*poly
log(2,-I*exp(b*x+a))*a^2+1/2*I/b^3*ln(-I*exp(b*x+a))*ln(-I*(-exp(b*x+a)+I))*a^2+1/2*I/b^2*ln(1+I*exp(b*x+a))*x
*a^2-1/2*I/b^3*polylog(2,I*exp(b*x+a))*a^2-1/2*I/b*polylog(2,-I*exp(b*x+a))*x^2+1/2*I/b^3*a^3*ln(1+I*exp(b*x+a
))-1/2*I/b^2*ln(-I*(-exp(b*x+a)+I))*x*a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, x^{3} \arctan \left (e^{\left (-b x - a\right )}\right ) + b \int \frac {x^{3} e^{\left (b x + a\right )}}{3 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acot(exp(a + b*x)),x)

[Out]

int(x^2*acot(exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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