Integrand size = 10, antiderivative size = 91 \[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\frac {i x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b}-\frac {i x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{2 b^2}+\frac {i \operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{2 b^2} \]
1/2*I*x*polylog(2,-I*exp(b*x+a))/b-1/2*I*x*polylog(2,I*exp(b*x+a))/b-1/2*I *polylog(3,-I*exp(b*x+a))/b^2+1/2*I*polylog(3,I*exp(b*x+a))/b^2
Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\frac {i \left (b x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )-b x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )-\operatorname {PolyLog}\left (3,-i e^{a+b x}\right )+\operatorname {PolyLog}\left (3,i e^{a+b x}\right )\right )}{2 b^2} \]
((I/2)*(b*x*PolyLog[2, (-I)*E^(a + b*x)] - b*x*PolyLog[2, I*E^(a + b*x)] - PolyLog[3, (-I)*E^(a + b*x)] + PolyLog[3, I*E^(a + b*x)]))/b^2
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5666, 3011, 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.
\(\displaystyle \int x \arctan \left (e^{a+b x}\right ) \, dx\) |
\(\Big \downarrow \) 5666 |
\(\displaystyle \frac {1}{2} i \int x \log \left (1-i e^{a+b x}\right )dx-\frac {1}{2} i \int x \log \left (1+i e^{a+b x}\right )dx\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} i \left (\frac {\int \operatorname {PolyLog}\left (2,i e^{a+b x}\right )dx}{b}-\frac {x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )-\frac {1}{2} i \left (\frac {\int \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )dx}{b}-\frac {x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} i \left (\frac {\int e^{-a-b x} \operatorname {PolyLog}\left (2,i e^{a+b x}\right )de^{a+b x}}{b^2}-\frac {x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )-\frac {1}{2} i \left (\frac {\int e^{-a-b x} \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )de^{a+b x}}{b^2}-\frac {x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} i \left (\frac {\operatorname {PolyLog}\left (3,i e^{a+b x}\right )}{b^2}-\frac {x \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b}\right )-\frac {1}{2} i \left (\frac {\operatorname {PolyLog}\left (3,-i e^{a+b x}\right )}{b^2}-\frac {x \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b}\right )\) |
(-1/2*I)*(-((x*PolyLog[2, (-I)*E^(a + b*x)])/b) + PolyLog[3, (-I)*E^(a + b *x)]/b^2) + (I/2)*(-((x*PolyLog[2, I*E^(a + b*x)])/b) + PolyLog[3, I*E^(a + b*x)]/b^2)
3.2.14.3.1 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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (71 ) = 142\).
Time = 0.38 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.84
method | result | size |
risch | \(-\frac {i a^{2} \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i x \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \operatorname {dilog}\left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}}+\frac {i \operatorname {dilog}\left (-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i x \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right )}{2 b}-\frac {i \operatorname {polylog}\left (3, -i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a x}{2 b}+\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a x}{2 b}+\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a}{2 b^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a x}{2 b}+\frac {i \operatorname {polylog}\left (3, i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a x}{2 b}-\frac {i \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}+\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{2}}+\frac {i a^{2} \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}\) | \(349\) |
-1/2*I/b^2*a^2*ln(1-I*exp(b*x+a))+1/2*I*x*polylog(2,-I*exp(b*x+a))/b+1/2*I /b^2*dilog(-I*(exp(b*x+a)+I))*a+1/2*I/b^2*dilog(-I*exp(b*x+a))*a-1/2*I*x*p olylog(2,I*exp(b*x+a))/b-1/2*I*polylog(3,-I*exp(b*x+a))/b^2-1/2*I/b*ln(-I* (-exp(b*x+a)+I))*a*x+1/2*I/b*ln(-I*(exp(b*x+a)+I))*a*x+1/2*I/b^2*ln(-I*exp (b*x+a))*ln(-I*(-exp(b*x+a)+I))*a-1/2*I/b*ln(1-I*exp(b*x+a))*a*x+1/2*I*pol ylog(3,I*exp(b*x+a))/b^2+1/2*I/b^2*polylog(2,-I*exp(b*x+a))*a+1/2*I/b*ln(1 +I*exp(b*x+a))*a*x-1/2*I/b^2*polylog(2,I*exp(b*x+a))*a-1/2*I/b^2*ln(-I*(-e xp(b*x+a)+I))*a^2+1/2*I/b^2*ln(-I*(exp(b*x+a)+I))*a^2+1/2*I/b^2*a^2*ln(1+I *exp(b*x+a))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (61) = 122\).
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.66 \[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\frac {2 \, b^{2} x^{2} \arctan \left (e^{\left (b x + a\right )}\right ) - 2 i \, b x {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) + 2 i \, b x {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) - i \, a^{2} \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a^{2} \log \left (e^{\left (b x + a\right )} - i\right ) + {\left (i \, b^{2} x^{2} - i \, a^{2}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + 2 i \, {\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) - 2 i \, {\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right )}{4 \, b^{2}} \]
1/4*(2*b^2*x^2*arctan(e^(b*x + a)) - 2*I*b*x*dilog(I*e^(b*x + a)) + 2*I*b* x*dilog(-I*e^(b*x + a)) - I*a^2*log(e^(b*x + a) + I) + I*a^2*log(e^(b*x + a) - I) + (I*b^2*x^2 - I*a^2)*log(I*e^(b*x + a) + 1) + (-I*b^2*x^2 + I*a^2 )*log(-I*e^(b*x + a) + 1) + 2*I*polylog(3, I*e^(b*x + a)) - 2*I*polylog(3, -I*e^(b*x + a)))/b^2
\[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\int x \operatorname {atan}{\left (e^{a} e^{b x} \right )}\, dx \]
\[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\int { x \arctan \left (e^{\left (b x + a\right )}\right ) \,d x } \]
\[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\int { x \arctan \left (e^{\left (b x + a\right )}\right ) \,d x } \]
Timed out. \[ \int x \arctan \left (e^{a+b x}\right ) \, dx=\int x\,\mathrm {atan}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]