Integrand size = 12, antiderivative size = 196 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)} \]
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Time = 0.12 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2320, 5156, 4967, 2449, 2352, 2497} \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{2 d \log (f)}-\frac {\log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}+\frac {\log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \cot ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)} \]
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Rule 2320
Rule 2352
Rule 2449
Rule 2497
Rule 4967
Rule 5156
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\cot ^{-1}(a+b x)}{x} \, dx,x,f^{c+d x}\right )}{d \log (f)} \\ & = \frac {\text {Subst}\left (\int \frac {\cot ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b f^{c+d x}\right )}{b d \log (f)} \\ & = -\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}-\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a+b f^{c+d x}\right )}{d \log (f)}+\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (-\frac {a}{b}+\frac {x}{b}\right )}{\left (\frac {i}{b}-\frac {a}{b}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,a+b f^{c+d x}\right )}{d \log (f)} \\ & = -\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)}-\frac {i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)} \\ & = -\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\cot ^{-1}\left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.85 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=x \cot ^{-1}\left (a+b f^{c+d x}\right )+\frac {b \left (d x \log (f) \left (\log \left (1+\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\log \left (1+\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )+\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )}{2 \sqrt {-b^2} d \log (f)} \]
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Time = 1.25 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\ln \left (-b \,f^{d x +c}\right ) \operatorname {arccot}\left (a +b \,f^{d x +c}\right )+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\) | \(162\) |
default | \(\frac {\ln \left (-b \,f^{d x +c}\right ) \operatorname {arccot}\left (a +b \,f^{d x +c}\right )+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\) | \(162\) |
risch | \(\frac {i x \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{2}+\frac {\pi x}{2}-\frac {i \ln \left (-i b \,f^{d x +c}-i a +1\right ) \ln \left (-\frac {i f^{d x +c} b}{i a -1}\right )}{2 d \ln \left (f \right )}-\frac {i \operatorname {dilog}\left (-\frac {i f^{d x +c} b}{i a -1}\right )}{2 d \ln \left (f \right )}-\frac {i \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 \ln \left (f \right ) d}-\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2}-\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) c}{2 d}+\frac {i c \ln \left (i f^{d x} f^{c} b +i a +1\right )}{2 d}\) | \(218\) |
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Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d x \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \left (f\right ) - i \, c \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right ) + i \, c \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right ) + {\left (-i \, d x - i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d x + i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) + i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right )}{2 \, d \log \left (f\right )} \]
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\[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int \operatorname {acot}{\left (a + b f^{c + d x} \right )}\, dx \]
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Time = 0.34 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96 \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {{\left (d x + c\right )} \operatorname {arccot}\left (b f^{d x + c} + a\right )}{d} + \frac {2 \, {\left (d x + c\right )} \arctan \left (\frac {b^{2} f^{d x + c} + a b}{b}\right ) \log \left (f\right ) + {\left (\pi - \arctan \left (\frac {1}{a}\right )\right )} \log \left (b^{2} f^{2 \, d x + 2 \, c} + 2 \, a b f^{d x + c} + a^{2} + 1\right ) - \arctan \left (b f^{d x + c} + a\right ) \log \left (\frac {b^{2} f^{2 \, d x + 2 \, c}}{a^{2} + 1}\right ) + i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a + 1}{i \, a + 1}\right ) - i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a - 1}{i \, a - 1}\right )}{2 \, d \log \left (f\right )} \]
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\[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \]
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Timed out. \[ \int \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int \mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \]
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