Integrand size = 14, antiderivative size = 155 \[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=-\frac {2 i \sqrt {1+x^2} \cot ^{-1}(x) \arctan \left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}-\frac {i \sqrt {1+x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}+\frac {i \sqrt {1+x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}} \]
[Out]
Time = 0.03 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5011, 5007} \[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=-\frac {2 i \sqrt {x^2+1} \arctan \left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right ) \cot ^{-1}(x)}{\sqrt {a x^2+a}}-\frac {i \sqrt {x^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}}+\frac {i \sqrt {x^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )}{\sqrt {a x^2+a}} \]
[In]
[Out]
Rule 5007
Rule 5011
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+x^2} \int \frac {\cot ^{-1}(x)}{\sqrt {1+x^2}} \, dx}{\sqrt {a+a x^2}} \\ & = -\frac {2 i \sqrt {1+x^2} \cot ^{-1}(x) \arctan \left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}-\frac {i \sqrt {1+x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}+\frac {i \sqrt {1+x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.57 \[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=-\frac {\sqrt {a \left (1+x^2\right )} \left (\cot ^{-1}(x) \left (\log \left (1-e^{i \cot ^{-1}(x)}\right )-\log \left (1+e^{i \cot ^{-1}(x)}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i \cot ^{-1}(x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \cot ^{-1}(x)}\right )\right )}{a \sqrt {1+\frac {1}{x^2}} x} \]
[In]
[Out]
Time = 0.77 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.64
method | result | size |
default | \(-\frac {i \left (i \operatorname {arccot}\left (x \right ) \ln \left (\frac {i+x}{\sqrt {x^{2}+1}}+1\right )-i \operatorname {arccot}\left (x \right ) \ln \left (1-\frac {i+x}{\sqrt {x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i+x}{\sqrt {x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i+x}{\sqrt {x^{2}+1}}\right )\right ) \sqrt {a \left (i+x \right ) \left (x -i\right )}}{\sqrt {x^{2}+1}\, a}\) | \(99\) |
[In]
[Out]
\[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=\int { \frac {\operatorname {arccot}\left (x\right )}{\sqrt {a x^{2} + a}} \,d x } \]
[In]
[Out]
\[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=\int \frac {\operatorname {acot}{\left (x \right )}}{\sqrt {a \left (x^{2} + 1\right )}}\, dx \]
[In]
[Out]
\[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=\int { \frac {\operatorname {arccot}\left (x\right )}{\sqrt {a x^{2} + a}} \,d x } \]
[In]
[Out]
\[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=\int { \frac {\operatorname {arccot}\left (x\right )}{\sqrt {a x^{2} + a}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\cot ^{-1}(x)}{\sqrt {a+a x^2}} \, dx=\int \frac {\mathrm {acot}\left (x\right )}{\sqrt {a\,x^2+a}} \,d x \]
[In]
[Out]