Optimal. Leaf size=35 \[ \frac {1}{4} i \text {Li}_2\left (\frac {i}{x+1}\right )-\frac {1}{4} i \text {Li}_2\left (-\frac {i}{x+1}\right ) \]
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Rubi [A] time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5044, 12, 4849, 2391} \[ \frac {1}{4} i \text {PolyLog}\left (2,\frac {i}{x+1}\right )-\frac {1}{4} i \text {PolyLog}\left (2,-\frac {i}{x+1}\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 2391
Rule 4849
Rule 5044
Rubi steps
\begin {align*} \int \frac {\cot ^{-1}(1+x)}{2+2 x} \, dx &=\operatorname {Subst}\left (\int \frac {\cot ^{-1}(x)}{2 x} \, dx,x,1+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\cot ^{-1}(x)}{x} \, dx,x,1+x\right )\\ &=\frac {1}{4} i \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,1+x\right )-\frac {1}{4} i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{x} \, dx,x,1+x\right )\\ &=-\frac {1}{4} i \text {Li}_2\left (-\frac {i}{1+x}\right )+\frac {1}{4} i \text {Li}_2\left (\frac {i}{1+x}\right )\\ \end {align*}
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Mathematica [A] time = 0.01, size = 35, normalized size = 1.00 \[ \frac {1}{4} i \text {Li}_2\left (\frac {i}{x+1}\right )-\frac {1}{4} i \text {Li}_2\left (-\frac {i}{x+1}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 2.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\left (x + 1\right )}{2 \, {\left (x + 1\right )}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 26, normalized size = 0.74 \[ -\frac {1}{4} \, {\left (x + 1\right )}^{2} \arctan \left (\frac {1}{x + 1}\right ) - \frac {1}{4} \, x - \frac {1}{4} \, \arctan \left (\frac {1}{x + 1}\right ) - \frac {1}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 68, normalized size = 1.94 \[ \frac {\ln \left (x +1\right ) \mathrm {arccot}\left (x +1\right )}{2}-\frac {i \ln \left (x +1\right ) \ln \left (1+i \left (x +1\right )\right )}{4}+\frac {i \ln \left (x +1\right ) \ln \left (1-i \left (x +1\right )\right )}{4}-\frac {i \dilog \left (1+i \left (x +1\right )\right )}{4}+\frac {i \dilog \left (1-i \left (x +1\right )\right )}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.46, size = 64, normalized size = 1.83 \[ \frac {1}{4} \, \arctan \left (x + 1, 0\right ) \log \left (x^{2} + 2 \, x + 2\right ) + \frac {1}{2} \, \operatorname {arccot}\left (x + 1\right ) \log \left (x + 1\right ) + \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left (x + 1\right ) - \frac {1}{2} \, \arctan \left (x + 1\right ) \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} i \, {\rm Li}_2\left (i \, x + i + 1\right ) - \frac {1}{4} i \, {\rm Li}_2\left (-i \, x - i + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {acot}\left (x+1\right )}{2\,x+2} \,d x \]
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 \[ \frac {\int \frac {\operatorname {acot}{\left (x + 1 \right )}}{x + 1}\, dx}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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