Optimal. Leaf size=188 \[ -\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )+\frac{1}{4} i \text{PolyLog}\left (2,1-\frac{2 i c (-x+i)}{(1-c) (1-i c x)}\right )+\frac{1}{4} i \text{PolyLog}\left (2,1+\frac{2 i c (x+i)}{(c+1) (1-i c x)}\right )+\log \left (\frac{2}{1-i c x}\right ) \left (-\cot ^{-1}(c x)\right )+\frac{1}{2} \log \left (\frac{2 i c (-x+i)}{(1-c) (1-i c x)}\right ) \cot ^{-1}(c x)+\frac{1}{2} \log \left (-\frac{2 i c (x+i)}{(c+1) (1-i c x)}\right ) \cot ^{-1}(c x) \]
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Rubi [A] time = 0.18417, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {4929, 4857, 2402, 2315, 2447} \[ -\frac{1}{2} i \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )+\frac{1}{4} i \text{PolyLog}\left (2,1-\frac{2 i c (-x+i)}{(1-c) (1-i c x)}\right )+\frac{1}{4} i \text{PolyLog}\left (2,1+\frac{2 i c (x+i)}{(c+1) (1-i c x)}\right )+\log \left (\frac{2}{1-i c x}\right ) \left (-\cot ^{-1}(c x)\right )+\frac{1}{2} \log \left (\frac{2 i c (-x+i)}{(1-c) (1-i c x)}\right ) \cot ^{-1}(c x)+\frac{1}{2} \log \left (-\frac{2 i c (x+i)}{(c+1) (1-i c x)}\right ) \cot ^{-1}(c x) \]
Antiderivative was successfully verified.
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Rule 4929
Rule 4857
Rule 2402
Rule 2315
Rule 2447
Rubi steps
\begin{align*} \int \frac{x \cot ^{-1}(c x)}{1+x^2} \, dx &=\int \left (-\frac{\cot ^{-1}(c x)}{2 (i-x)}+\frac{\cot ^{-1}(c x)}{2 (i+x)}\right ) \, dx\\ &=-\left (\frac{1}{2} \int \frac{\cot ^{-1}(c x)}{i-x} \, dx\right )+\frac{1}{2} \int \frac{\cot ^{-1}(c x)}{i+x} \, dx\\ &=-\cot ^{-1}(c x) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (\frac{2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (-\frac{2 i c (i+x)}{(1+c) (1-i c x)}\right )-2 \left (\frac{1}{2} c \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx\right )+\frac{1}{2} c \int \frac{\log \left (\frac{2 c (i-x)}{(-i+i c) (1-i c x)}\right )}{1+c^2 x^2} \, dx+\frac{1}{2} c \int \frac{\log \left (\frac{2 c (i+x)}{(i+i c) (1-i c x)}\right )}{1+c^2 x^2} \, dx\\ &=-\cot ^{-1}(c x) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (\frac{2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (-\frac{2 i c (i+x)}{(1+c) (1-i c x)}\right )+\frac{1}{4} i \text{Li}_2\left (1-\frac{2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac{1}{4} i \text{Li}_2\left (1+\frac{2 i c (i+x)}{(1+c) (1-i c x)}\right )-2 \left (\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )\right )\\ &=-\cot ^{-1}(c x) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (\frac{2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac{1}{2} \cot ^{-1}(c x) \log \left (-\frac{2 i c (i+x)}{(1+c) (1-i c x)}\right )-\frac{1}{2} i \text{Li}_2\left (1-\frac{2}{1-i c x}\right )+\frac{1}{4} i \text{Li}_2\left (1-\frac{2 i c (i-x)}{(1-c) (1-i c x)}\right )+\frac{1}{4} i \text{Li}_2\left (1+\frac{2 i c (i+x)}{(1+c) (1-i c x)}\right )\\ \end{align*}
Mathematica [A] time = 0.0831109, size = 343, normalized size = 1.82 \[ -\frac{1}{4} i \text{PolyLog}\left (2,\frac{i c (-x+i)}{1-c}\right )+\frac{1}{4} i \text{PolyLog}\left (2,-\frac{i c (-x+i)}{c+1}\right )+\frac{1}{4} i \text{PolyLog}\left (2,\frac{i c (x+i)}{1-c}\right )-\frac{1}{4} i \text{PolyLog}\left (2,-\frac{i c (x+i)}{c+1}\right )-\frac{1}{4} i \log (-x+i) \log \left (-\frac{i (-c x+i)}{1-c}\right )-\frac{1}{4} i \log (x+i) \log \left (-\frac{i (-c x+i)}{c+1}\right )+\frac{1}{4} i \log (-x+i) \log \left (-\frac{-c x+i}{c x}\right )+\frac{1}{4} i \log (x+i) \log \left (-\frac{-c x+i}{c x}\right )+\frac{1}{4} i \log (x+i) \log \left (-\frac{i (c x+i)}{1-c}\right )+\frac{1}{4} i \log (-x+i) \log \left (-\frac{i (c x+i)}{c+1}\right )-\frac{1}{4} i \log (-x+i) \log \left (\frac{c x+i}{c x}\right )-\frac{1}{4} i \log (x+i) \log \left (\frac{c x+i}{c x}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.174, size = 284, normalized size = 1.5 \begin{align*}{\frac{\ln \left ({c}^{2}{x}^{2}+{c}^{2} \right ){\rm arccot} \left (cx\right )}{2}}+{\frac{i}{4}}\ln \left ( cx-i \right ) \ln \left ({\frac{i \left ( cx-i \right ) -c-1}{-c-1}} \right ) +{\frac{i}{4}}\ln \left ( cx-i \right ) \ln \left ({\frac{i \left ( cx-i \right ) +c-1}{-1+c}} \right ) -{\frac{i}{4}}\ln \left ( cx-i \right ) \ln \left ({c}^{2}{x}^{2}+{c}^{2} \right ) +{\frac{i}{4}}{\it dilog} \left ({\frac{i \left ( cx-i \right ) -c-1}{-c-1}} \right ) +{\frac{i}{4}}{\it dilog} \left ({\frac{i \left ( cx-i \right ) +c-1}{-1+c}} \right ) -{\frac{i}{4}}\ln \left ( cx+i \right ) \ln \left ({\frac{-i \left ( cx+i \right ) -c-1}{-c-1}} \right ) -{\frac{i}{4}}\ln \left ( cx+i \right ) \ln \left ({\frac{-i \left ( cx+i \right ) +c-1}{-1+c}} \right ) +{\frac{i}{4}}\ln \left ( cx+i \right ) \ln \left ({c}^{2}{x}^{2}+{c}^{2} \right ) -{\frac{i}{4}}{\it dilog} \left ({\frac{-i \left ( cx+i \right ) -c-1}{-c-1}} \right ) -{\frac{i}{4}}{\it dilog} \left ({\frac{-i \left ( cx+i \right ) +c-1}{-1+c}} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \operatorname{arccot}\left (c x\right )}{x^{2} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \operatorname{arccot}\left (c x\right )}{x^{2} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \operatorname{acot}{\left (c x \right )}}{x^{2} + 1}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \operatorname{arccot}\left (c x\right )}{x^{2} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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