Integrand size = 12, antiderivative size = 151 \[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{b^2}+\frac {i x \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{b^2}-\frac {i \operatorname {PolyLog}\left (4,-i e^{-a-b x}\right )}{b^3}+\frac {i \operatorname {PolyLog}\left (4,i e^{-a-b x}\right )}{b^3} \]
-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*pol ylog(4,-I*exp(-b*x-a))/b^3+I*polylog(4,I*exp(-b*x-a))/b^3
Time = 0.02 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.88 \[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i \left (b^2 x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )-b^2 x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )+2 \left (b x \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )-b x \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )+\operatorname {PolyLog}\left (4,-i e^{-a-b x}\right )-\operatorname {PolyLog}\left (4,i e^{-a-b x}\right )\right )\right )}{2 b^3} \]
((-1/2*I)*(b^2*x^2*PolyLog[2, (-I)*E^(-a - b*x)] - b^2*x^2*PolyLog[2, I*E^ (-a - b*x)] + 2*(b*x*PolyLog[3, (-I)*E^(-a - b*x)] - b*x*PolyLog[3, I*E^(- a - b*x)] + PolyLog[4, (-I)*E^(-a - b*x)] - PolyLog[4, I*E^(-a - b*x)])))/ b^3
Time = 0.57 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5667, 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.
| \(\displaystyle \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx\) |
| \(\Big \downarrow \) 5667 |
| \(\displaystyle \frac {1}{2} i \int x^2 \log \left (1-i e^{-a-b x}\right )dx-\frac {1}{2} i \int x^2 \log \left (1+i e^{-a-b x}\right )dx\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{b}-\frac {2 \int x \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )dx}{b}\right )-\frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{b}-\frac {2 \int x \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )dx}{b}\right )\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{b}-\frac {2 \left (\frac {\int \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )dx}{b}-\frac {x \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{b}\right )}{b}\right )-\frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{b}-\frac {2 \left (\frac {\int \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )dx}{b}-\frac {x \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{b}\right )}{b}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{b}-\frac {2 \left (-\frac {\int e^{a+b x} \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )de^{-a-b x}}{b^2}-\frac {x \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{b}\right )}{b}\right )-\frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{b}-\frac {2 \left (-\frac {\int e^{a+b x} \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )de^{-a-b x}}{b^2}-\frac {x \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{b}\right )}{b}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{b}-\frac {2 \left (-\frac {\operatorname {PolyLog}\left (4,i e^{-a-b x}\right )}{b^2}-\frac {x \operatorname {PolyLog}\left (3,i e^{-a-b x}\right )}{b}\right )}{b}\right )-\frac {1}{2} i \left (\frac {x^2 \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{b}-\frac {2 \left (-\frac {\operatorname {PolyLog}\left (4,-i e^{-a-b x}\right )}{b^2}-\frac {x \operatorname {PolyLog}\left (3,-i e^{-a-b x}\right )}{b}\right )}{b}\right )\) |
(-1/2*I)*((x^2*PolyLog[2, (-I)*E^(-a - b*x)])/b - (2*(-((x*PolyLog[3, (-I) *E^(-a - b*x)])/b) - PolyLog[4, (-I)*E^(-a - b*x)]/b^2))/b) + (I/2)*((x^2* PolyLog[2, I*E^(-a - b*x)])/b - (2*(-((x*PolyLog[3, I*E^(-a - b*x)])/b) - PolyLog[4, I*E^(-a - b*x)]/b^2))/b)
3.3.23.3.1 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[ArcCot[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] : > Simp[I/2 Int[x^m*Log[1 - I/(a + b*f^(c + d*x))], x], x] - Simp[I/2 In t[x^m*Log[1 + I/(a + 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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (129 ) = 258\).
Time = 0.80 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.74
| method | result | size |
| risch | \(\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{3}}{2 b^{3}}+\frac {\pi \,x^{3}}{6}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a^{2} x}{2 b^{2}}-\frac {i \operatorname {polylog}\left (3, i {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a^{3}}{2 b^{3}}+\frac {i \operatorname {polylog}\left (3, -i {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {i \operatorname {polylog}\left (4, i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {i \ln \left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) x \,a^{2}}{2 b^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x \,a^{2}}{2 b^{2}}-\frac {i \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {i a^{3} \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{3}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2} x}{2 b^{2}}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{3}}{2 b^{3}}+\frac {i \operatorname {polylog}\left (2, -i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {i \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}+\frac {i \operatorname {dilog}\left (-i {\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {i \operatorname {polylog}\left (4, -i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {i \operatorname {dilog}\left (-i \left ({\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{3}}+\frac {i \operatorname {polylog}\left (2, i {\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a^{2}}{2 b^{3}}\) | \(413\) |
1/2*I/b^3*ln(-I*(exp(b*x+a)+I))*a^3+1/6*Pi*x^3+1/2*I/b^2*ln(1+I*exp(b*x+a) )*a^2*x-I/b^2*polylog(3,I*exp(b*x+a))*x-1/2*I/b^3*ln(1-I*exp(b*x+a))*a^3+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^2*ln (-I*(exp(b*x+a)+I))*x*a^2-1/2*I/b^2*ln(1-I*exp(b*x+a))*x*a^2-1/2*I/b*polyl og(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))*a^2*x-1/2*I/b^3*ln(-I*(-exp(b*x+a)+I))*a^3+1/2*I/b^3*polyl og(2,-I*exp(b*x+a))*a^2-1/2*I/b^3*polylog(2,I*exp(b*x+a))*a^2+1/2*I/b^3*di log(-I*exp(b*x+a))*a^2-I*polylog(4,-I*exp(b*x+a))/b^3+1/2*I/b^3*dilog(-I*( exp(b*x+a)+I))*a^2+1/2*I/b*polylog(2,I*exp(b*x+a))*x^2+1/2*I/b^3*ln(-I*exp (b*x+a))*ln(-I*(-exp(b*x+a)+I))*a^2
Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.24 \[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {2 \, b^{3} x^{3} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) + 3 i \, b^{2} x^{2} {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - 3 i \, b^{2} x^{2} {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right ) - i \, a^{3} \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a^{3} \log \left (e^{\left (b x + a\right )} - i\right ) - 6 i \, b x {\rm polylog}\left (3, i \, e^{\left (b x + a\right )}\right ) + 6 i \, b x {\rm polylog}\left (3, -i \, e^{\left (b x + a\right )}\right ) + {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b^{3} x^{3} + i \, a^{3}\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (4, i \, e^{\left (b x + a\right )}\right ) - 6 i \, {\rm polylog}\left (4, -i \, e^{\left (b x + a\right )}\right )}{6 \, b^{3}} \]
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*pol ylog(4, -I*e^(b*x + a)))/b^3
\[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int x^{2} \operatorname {acot}{\left (e^{a} e^{b x} \right )}\, dx \]
\[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int { x^{2} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) \,d x } \]
\[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int { x^{2} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) \,d x } \]
Timed out. \[ \int x^2 \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int x^2\,\mathrm {acot}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]