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

Optimal. Leaf size=133 \[ -\frac{i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^2}+\frac{i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^2}+\frac{i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^3}-\frac{i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^3}+\frac{i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac{i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b} \]

[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*PolyLog[4, I
*E^(a + b*x)])/b^3

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Rubi [A]  time = 0.0899561, antiderivative size = 133, normalized size of antiderivative = 1., 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 = {5143, 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*ArcTan[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*PolyLog[4, I
*E^(a + b*x)])/b^3

Rule 5143

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

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 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]

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 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]

Rubi steps

\begin{align*} \int x^2 \tan ^{-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*}

Mathematica [A]  time = 0.0103102, size = 133, normalized size = 1. \[ -\frac{i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^2}+\frac{i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^2}+\frac{i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^3}-\frac{i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^3}+\frac{i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac{i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.182, size = 407, normalized size = 3.1 \begin{align*}{\frac{-{\frac{i}{2}}{x}^{2}{\it polylog} \left ( 2,i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{i{\it polylog} \left ( 3,-i{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ){a}^{3}}{{b}^{3}}}-{\frac{{\frac{i}{2}}{\it polylog} \left ( 2,-i{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}+{\frac{i{\it polylog} \left ( 4,-i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{ix{\it polylog} \left ( 3,i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{{\frac{i}{2}}{a}^{3}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{{\frac{i}{2}}{x}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{bx+a}} \right ) }{b}}-{\frac{i{\it polylog} \left ( 4,i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{{\frac{i}{2}}{\it polylog} \left ( 2,i{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}+{\frac{{\frac{i}{2}}\ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ){a}^{3}}{{b}^{3}}}+{\frac{{\frac{i}{2}}{a}^{3}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( -i{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}-{\frac{{\frac{i}{2}}\ln \left ( -i{{\rm e}^{bx+a}} \right ) \ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ){a}^{2}}{{b}^{3}}}-{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x{a}^{2}}{{b}^{2}}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ){a}^{2}}{{b}^{3}}}+{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x{a}^{2}}{{b}^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( -i \left ({{\rm e}^{bx+a}}+i \right ) \right ) x{a}^{2}}{{b}^{2}}}+{\frac{{\frac{i}{2}}\ln \left ( -i \left ( -{{\rm e}^{bx+a}}+i \right ) \right ) x{a}^{2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(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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Fricas [C]  time = 2.35137, size = 540, normalized size = 4.06 \begin{align*} \frac{2 \, b^{3} x^{3} \arctan \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*x^3*arctan(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{atan}{\left (e^{a} e^{b x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \arctan \left (e^{\left (b x + a\right )}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arctan(e^(b*x + a)), x)