Integrand size = 8, antiderivative size = 91 \[ \int x^2 \arctan \left (e^x\right ) \, dx=\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,-i e^x\right )-\frac {1}{2} i x^2 \operatorname {PolyLog}\left (2,i e^x\right )-i x \operatorname {PolyLog}\left (3,-i e^x\right )+i x \operatorname {PolyLog}\left (3,i e^x\right )+i \operatorname {PolyLog}\left (4,-i e^x\right )-i \operatorname {PolyLog}\left (4,i e^x\right ) \]
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*e xp(x))
Time = 0.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int x^2 \arctan \left (e^x\right ) \, dx=\frac {1}{2} i \left (x^2 \operatorname {PolyLog}\left (2,-i e^x\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )+2 \left (-x \operatorname {PolyLog}\left (3,-i e^x\right )+x \operatorname {PolyLog}\left (3,i e^x\right )+\operatorname {PolyLog}\left (4,-i e^x\right )-\operatorname {PolyLog}\left (4,i e^x\right )\right )\right ) \]
(I/2)*(x^2*PolyLog[2, (-I)*E^x] - x^2*PolyLog[2, I*E^x] + 2*(-(x*PolyLog[3 , (-I)*E^x]) + x*PolyLog[3, I*E^x] + PolyLog[4, (-I)*E^x] - PolyLog[4, I*E ^x]))
Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5666, 3011, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \arctan \left (e^x\right ) \, dx\) |
\(\Big \downarrow \) 5666 |
\(\displaystyle \frac {1}{2} i \int x^2 \log \left (1-i e^x\right )dx-\frac {1}{2} i \int x^2 \log \left (1+i e^x\right )dx\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} i \left (2 \int x \operatorname {PolyLog}\left (2,i e^x\right )dx-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-\frac {1}{2} i \left (2 \int x \operatorname {PolyLog}\left (2,-i e^x\right )dx-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\int \operatorname {PolyLog}\left (3,i e^x\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-\frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\int \operatorname {PolyLog}\left (3,-i e^x\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,i e^x\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-\frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\int e^{-x} \operatorname {PolyLog}\left (3,-i e^x\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,i e^x\right )-\operatorname {PolyLog}\left (4,i e^x\right )\right )-x^2 \operatorname {PolyLog}\left (2,i e^x\right )\right )-\frac {1}{2} i \left (2 \left (x \operatorname {PolyLog}\left (3,-i e^x\right )-\operatorname {PolyLog}\left (4,-i e^x\right )\right )-x^2 \operatorname {PolyLog}\left (2,-i e^x\right )\right )\) |
(-1/2*I)*(-(x^2*PolyLog[2, (-I)*E^x]) + 2*(x*PolyLog[3, (-I)*E^x] - PolyLo g[4, (-I)*E^x])) + (I/2)*(-(x^2*PolyLog[2, I*E^x]) + 2*(x*PolyLog[3, I*E^x ] - PolyLog[4, I*E^x]))
3.2.12.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] : > Simp[I/2 Int[x^m*Log[1 - I*a - I*b*f^(c + d*x)], x], x] - Simp[I/2 In t[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
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
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] - Simp[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]
Time = 0.64 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {i x^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{x}\right )}{2}-\frac {i x^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{x}\right )}{2}-i x \operatorname {polylog}\left (3, -i {\mathrm e}^{x}\right )+i x \operatorname {polylog}\left (3, i {\mathrm e}^{x}\right )+i \operatorname {polylog}\left (4, -i {\mathrm e}^{x}\right )-i \operatorname {polylog}\left (4, i {\mathrm e}^{x}\right )\) | \(70\) |
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*e xp(x))
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int x^2 \arctan \left (e^x\right ) \, dx=\frac {1}{3} \, x^{3} \arctan \left (e^{x}\right ) + \frac {1}{6} i \, x^{3} \log \left (i \, e^{x} + 1\right ) - \frac {1}{6} i \, x^{3} \log \left (-i \, e^{x} + 1\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (i \, e^{x}\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-i \, e^{x}\right ) + i \, x {\rm polylog}\left (3, i \, e^{x}\right ) - i \, x {\rm polylog}\left (3, -i \, e^{x}\right ) - i \, {\rm polylog}\left (4, i \, e^{x}\right ) + i \, {\rm polylog}\left (4, -i \, e^{x}\right ) \]
1/3*x^3*arctan(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)
\[ \int x^2 \arctan \left (e^x\right ) \, dx=\int x^{2} \operatorname {atan}{\left (e^{x} \right )}\, dx \]
\[ \int x^2 \arctan \left (e^x\right ) \, dx=\int { x^{2} \arctan \left (e^{x}\right ) \,d x } \]
\[ \int x^2 \arctan \left (e^x\right ) \, dx=\int { x^{2} \arctan \left (e^{x}\right ) \,d x } \]
Timed out. \[ \int x^2 \arctan \left (e^x\right ) \, dx=\int x^2\,\mathrm {atan}\left ({\mathrm {e}}^x\right ) \,d x \]