Optimal. Leaf size=183 \[ -\frac {1}{4} \text {Li}_2\left (\frac {2 i (i-c x)}{(1-c) (1-i x)}+1\right )+\frac {1}{4} \text {Li}_2\left (\frac {2 i (c x+i)}{(c+1) (1-i x)}+1\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (c x+i)}{(c+1) (1-i x)}\right ) \]
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Rubi [A] time = 0.46, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 13, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {4909, 203, 2470, 260, 6688, 12, 4876, 4848, 2391, 4856, 2402, 2315, 2447} \[ -\frac {1}{4} \text {PolyLog}\left (2,1+\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac {1}{4} \text {PolyLog}\left (2,1+\frac {2 i (c x+i)}{(c+1) (1-i x)}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (-c x+i)}{(1-c) (1-i x)}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (c x+i)}{(c+1) (1-i x)}\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 203
Rule 260
Rule 2315
Rule 2391
Rule 2402
Rule 2447
Rule 2470
Rule 4848
Rule 4856
Rule 4876
Rule 4909
Rule 6688
Rubi steps
\begin {align*} \int \frac {\cot ^{-1}(c x)}{1+x^2} \, dx &=\frac {1}{2} i \int \frac {\log \left (1-\frac {i}{c x}\right )}{1+x^2} \, dx-\frac {1}{2} i \int \frac {\log \left (1+\frac {i}{c x}\right )}{1+x^2} \, dx\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {\int \frac {\tan ^{-1}(x)}{\left (1-\frac {i}{c x}\right ) x^2} \, dx}{2 c}+\frac {\int \frac {\tan ^{-1}(x)}{\left (1+\frac {i}{c x}\right ) x^2} \, dx}{2 c}\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {\int \frac {c \tan ^{-1}(x)}{x (-i+c x)} \, dx}{2 c}+\frac {\int \frac {c \tan ^{-1}(x)}{x (i+c x)} \, dx}{2 c}\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {1}{2} \int \frac {\tan ^{-1}(x)}{x (-i+c x)} \, dx+\frac {1}{2} \int \frac {\tan ^{-1}(x)}{x (i+c x)} \, dx\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )+\frac {1}{2} \int \left (\frac {i \tan ^{-1}(x)}{x}-\frac {i c \tan ^{-1}(x)}{-i+c x}\right ) \, dx+\frac {1}{2} \int \left (-\frac {i \tan ^{-1}(x)}{x}+\frac {i c \tan ^{-1}(x)}{i+c x}\right ) \, dx\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )-\frac {1}{2} (i c) \int \frac {\tan ^{-1}(x)}{-i+c x} \, dx+\frac {1}{2} (i c) \int \frac {\tan ^{-1}(x)}{i+c x} \, dx\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (i+c x)}{(1+c) (1-i x)}\right )+\frac {1}{2} i \int \frac {\log \left (\frac {2 (-i+c x)}{(-i+i c) (1-i x)}\right )}{1+x^2} \, dx-\frac {1}{2} i \int \frac {\log \left (\frac {2 (i+c x)}{(i+i c) (1-i x)}\right )}{1+x^2} \, dx\\ &=\frac {1}{2} i \tan ^{-1}(x) \log \left (1-\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (1+\frac {i}{c x}\right )-\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac {1}{2} i \tan ^{-1}(x) \log \left (-\frac {2 i (i+c x)}{(1+c) (1-i x)}\right )-\frac {1}{4} \text {Li}_2\left (1+\frac {2 i (i-c x)}{(1-c) (1-i x)}\right )+\frac {1}{4} \text {Li}_2\left (1+\frac {2 i (i+c x)}{(1+c) (1-i x)}\right )\\ \end {align*}
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Mathematica [A] time = 0.09, size = 319, normalized size = 1.74 \[ -\frac {1}{4} \text {Li}_2\left (\frac {i c (i-x)}{1-c}\right )+\frac {1}{4} \text {Li}_2\left (-\frac {i c (i-x)}{c+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {i c (x+i)}{1-c}\right )+\frac {1}{4} \text {Li}_2\left (-\frac {i c (x+i)}{c+1}\right )-\frac {1}{4} \log (-x+i) \log \left (-\frac {i (-c x+i)}{1-c}\right )+\frac {1}{4} \log (x+i) \log \left (-\frac {i (-c x+i)}{c+1}\right )+\frac {1}{4} \log (-x+i) \log \left (-\frac {-c x+i}{c x}\right )-\frac {1}{4} \log (x+i) \log \left (-\frac {-c x+i}{c x}\right )-\frac {1}{4} \log (x+i) \log \left (-\frac {i (c x+i)}{1-c}\right )+\frac {1}{4} \log (-x+i) \log \left (-\frac {i (c x+i)}{c+1}\right )-\frac {1}{4} \log (-x+i) \log \left (\frac {c x+i}{c x}\right )+\frac {1}{4} \log (x+i) \log \left (\frac {c x+i}{c x}\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\left (c x\right )}{x^{2} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\left (c x\right )}{x^{2} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.68, size = 304, normalized size = 1.66 \[ \mathrm {arccot}\left (c x \right ) \arctan \relax (x )+\arctan \left (c x \right ) \arctan \relax (x )+\frac {i \arctan \left (c x \right ) \ln \left (1-\frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{2}+\frac {\arctan \left (c x \right )^{2}}{2}+\frac {\polylog \left (2, \frac {\left (1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (1-c \right )}\right )}{4}-\frac {i c \arctan \left (c x \right ) \ln \left (1-\frac {\left (-1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{2 \left (1+c \right )}-\frac {i \ln \left (1-\frac {\left (-1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right ) \arctan \left (c x \right )}{2 \left (1+c \right )}-\frac {c \arctan \left (c x \right )^{2}}{2 \left (1+c \right )}-\frac {c \polylog \left (2, \frac {\left (-1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4 \left (1+c \right )}-\frac {\arctan \left (c x \right )^{2}}{2 \left (1+c \right )}-\frac {\polylog \left (2, \frac {\left (-1+c \right ) \left (i c x +1\right )^{2}}{\left (c^{2} x^{2}+1\right ) \left (-c -1\right )}\right )}{4 \left (1+c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 187, normalized size = 1.02 \[ -\frac {1}{8} \, c {\left (\frac {8 \, \arctan \left (c x\right ) \arctan \relax (x)}{c} - \frac {4 \, \arctan \left (c x\right ) \arctan \relax (x) - 4 \, \arctan \relax (x) \arctan \left (\frac {c x}{c - 1}, -\frac {1}{c - 1}\right ) + \log \left (x^{2} + 1\right ) \log \left (\frac {c^{2} x^{2} + 1}{c^{2} + 2 \, c + 1}\right ) - \log \left (x^{2} + 1\right ) \log \left (\frac {c^{2} x^{2} + 1}{c^{2} - 2 \, c + 1}\right ) + 2 \, {\rm Li}_2\left (\frac {i \, c x + c}{c + 1}\right ) + 2 \, {\rm Li}_2\left (-\frac {i \, c x - c}{c + 1}\right ) - 2 \, {\rm Li}_2\left (\frac {i \, c x + c}{c - 1}\right ) - 2 \, {\rm Li}_2\left (-\frac {i \, c x - c}{c - 1}\right )}{c}\right )} + \operatorname {arccot}\left (c x\right ) \arctan \relax (x) + \arctan \left (c x\right ) \arctan \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acot}\left (c\,x\right )}{x^2+1} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acot}{\left (c x \right )}}{x^{2} + 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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