Integrand size = 14, antiderivative size = 152 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=-\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d} \]
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Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5155, 4966, 2449, 2352, 2497} \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c-a d+i d) (1-i (a+b x))}\right )}{2 d}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d} \]
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Rule 2352
Rule 2449
Rule 2497
Rule 4966
Rule 5155
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\arctan (x)}{\frac {b c-a d}{b}+\frac {d x}{b}} \, dx,x,a+b x\right )}{b} \\ & = -\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a+b x\right )}{d}-\frac {\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (\frac {b c-a d}{b}+\frac {d x}{b}\right )}{\left (\frac {i d}{b}+\frac {b c-a d}{b}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,a+b x\right )}{d} \\ & = -\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i (a+b x)}\right )}{d} \\ & = -\frac {\arctan (a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\arctan (a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.52 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\frac {i \log (1-i (a+b x)) \log \left (-\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}-\frac {i (b c-a d)}{b}}\right )}{2 d}-\frac {i \log (1+i (a+b x)) \log \left (\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}+\frac {i (b c-a d)}{b}}\right )}{2 d}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i d (1-i (a+b x))}{b c-i d-a d}\right )}{2 d}-\frac {i \operatorname {PolyLog}\left (2,\frac {i d (1+i (a+b x))}{b c+i d-a d}\right )}{2 d} \]
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Time = 0.38 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\operatorname {dilog}\left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\) | \(186\) |
default | \(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\operatorname {dilog}\left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\operatorname {dilog}\left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\) | \(186\) |
parts | \(\frac {\ln \left (d x +c \right ) \arctan \left (b x +a \right )}{d}-b \left (-\frac {i \ln \left (d x +c \right ) \left (\ln \left (\frac {i d -a d +b c -b \left (d x +c \right )}{-a d +b c +i d}\right )-\ln \left (\frac {i d +a d -b c +b \left (d x +c \right )}{a d -b c +i d}\right )\right )}{2 d b}-\frac {i \left (\operatorname {dilog}\left (\frac {i d -a d +b c -b \left (d x +c \right )}{-a d +b c +i d}\right )-\operatorname {dilog}\left (\frac {i d +a d -b c +b \left (d x +c \right )}{a d -b c +i d}\right )\right )}{2 d b}\right )\) | \(194\) |
risch | \(\frac {i \operatorname {dilog}\left (\frac {i a d -i b c +\left (-b x i-i a +1\right ) d -d}{i a d -i b c -d}\right )}{2 d}+\frac {i \ln \left (-b x i-i a +1\right ) \ln \left (\frac {i a d -i b c +\left (-b x i-i a +1\right ) d -d}{i a d -i b c -d}\right )}{2 d}-\frac {i \operatorname {dilog}\left (\frac {-i a d +i b c +\left (b x i+i a +1\right ) d -d}{-i a d +i b c -d}\right )}{2 d}-\frac {i \ln \left (b x i+i a +1\right ) \ln \left (\frac {-i a d +i b c +\left (b x i+i a +1\right ) d -d}{-i a d +i b c -d}\right )}{2 d}\) | \(230\) |
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\[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\int { \frac {\arctan \left (b x + a\right )}{d x + c} \,d x } \]
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Timed out. \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\text {Timed out} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (130) = 260\).
Time = 0.36 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.87 \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\frac {\arctan \left (b x + a\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}, \frac {b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) + i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a + 1\right )} d}{-i \, b c + {\left (i \, a + 1\right )} d}\right ) - i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a - 1\right )} d}{-i \, b c + {\left (i \, a - 1\right )} d}\right )}{2 \, d} \]
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\[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\int { \frac {\arctan \left (b x + a\right )}{d x + c} \,d x } \]
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Timed out. \[ \int \frac {\arctan (a+b x)}{c+d x} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{c+d\,x} \,d x \]
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