Integrand size = 8, antiderivative size = 51 \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=-\frac {i \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b} \]
[Out]
Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2320, 4941, 2438} \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {i \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b}-\frac {i \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b} \]
[In]
[Out]
Rule 2320
Rule 2438
Rule 4941
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{x} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {i \text {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,e^{a+b x}\right )}{2 b}-\frac {i \text {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{x} \, dx,x,e^{a+b x}\right )}{2 b} \\ & = -\frac {i \operatorname {PolyLog}\left (2,-i e^{-a-b x}\right )}{2 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{-a-b x}\right )}{2 b} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=x \cot ^{-1}\left (e^{a+b x}\right )+\frac {i \left (b x \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{a+b x}\right )+\operatorname {PolyLog}\left (2,i e^{a+b x}\right )\right )}{2 b} \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (41 ) = 82\).
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.86
method | result | size |
derivativedivides | \(\frac {\ln \left ({\mathrm e}^{b x +a}\right ) \operatorname {arccot}\left ({\mathrm e}^{b x +a}\right )-\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2}-\frac {i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{2}}{b}\) | \(95\) |
default | \(\frac {\ln \left ({\mathrm e}^{b x +a}\right ) \operatorname {arccot}\left ({\mathrm e}^{b x +a}\right )-\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2}-\frac {i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{2}}{b}\) | \(95\) |
parts | \(x \,\operatorname {arccot}\left ({\mathrm e}^{b x +a}\right )+\frac {-\frac {i \left (b x +a \right ) \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \left (b x +a \right ) \ln \left (1-i {\mathrm e}^{b x +a}\right )}{2}-\frac {i \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{2}-a \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(96\) |
risch | \(\frac {i x \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2}+\frac {\pi x}{2}+\frac {i \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{2 b}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) x}{2}+\frac {i a \ln \left (1+i {\mathrm e}^{b x +a}\right )}{2 b}+\frac {i \ln \left (-i {\mathrm e}^{b x +a}\right ) \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right )}{2 b}-\frac {i \ln \left (-i \left (-{\mathrm e}^{b x +a}+i\right )\right ) a}{2 b}+\frac {i \operatorname {dilog}\left (-i {\mathrm e}^{b x +a}\right )}{2 b}\) | \(147\) |
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (35) = 70\).
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.02 \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {2 \, b x \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) - i \, a \log \left (e^{\left (b x + a\right )} + i\right ) + i \, a \log \left (e^{\left (b x + a\right )} - i\right ) + {\left (-i \, b x - i \, a\right )} \log \left (i \, e^{\left (b x + a\right )} + 1\right ) + {\left (i \, b x + i \, a\right )} \log \left (-i \, e^{\left (b x + a\right )} + 1\right ) + i \, {\rm Li}_2\left (i \, e^{\left (b x + a\right )}\right ) - i \, {\rm Li}_2\left (-i \, e^{\left (b x + a\right )}\right )}{2 \, b} \]
[In]
[Out]
\[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int \operatorname {acot}{\left (e^{a + b x} \right )}\, dx \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24 \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\frac {{\left (b x + a\right )} \operatorname {arccot}\left (e^{\left (b x + a\right )}\right )}{b} + \frac {\pi \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 i \, {\rm Li}_2\left (i \, e^{\left (b x + a\right )} + 1\right ) - 2 i \, {\rm Li}_2\left (-i \, e^{\left (b x + a\right )} + 1\right )}{4 \, b} \]
[In]
[Out]
\[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int { \operatorname {arccot}\left (e^{\left (b x + a\right )}\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int \cot ^{-1}\left (e^{a+b x}\right ) \, dx=\int \mathrm {acot}\left ({\mathrm {e}}^{a+b\,x}\right ) \,d x \]
[In]
[Out]