Integrand size = 23, antiderivative size = 119 \[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=-\frac {i (a+b \arctan (c+d x))^2}{d e^2}-\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}+\frac {2 b (a+b \arctan (c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2} \]
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Time = 0.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5151, 12, 4946, 5044, 4988, 2497} \[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=-\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}-\frac {i (a+b \arctan (c+d x))^2}{d e^2}+\frac {2 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))}{d e^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right )}{d e^2} \]
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Rule 12
Rule 2497
Rule 4946
Rule 4988
Rule 5044
Rule 5151
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{e^2 x^2} \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}+\frac {(2 b) \text {Subst}\left (\int \frac {a+b \arctan (x)}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^2}{d e^2}-\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}+\frac {(2 i b) \text {Subst}\left (\int \frac {a+b \arctan (x)}{x (i+x)} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^2}{d e^2}-\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}+\frac {2 b (a+b \arctan (c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^2}{d e^2}-\frac {(a+b \arctan (c+d x))^2}{d e^2 (c+d x)}+\frac {2 b (a+b \arctan (c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\frac {-i b^2 (-i+c+d x) \arctan (c+d x)^2+2 b \arctan (c+d x) \left (-a+b (c+d x) \log \left (1-e^{2 i \arctan (c+d x)}\right )\right )+a \left (-a+2 b (c+d x) \log \left (\frac {c+d x}{\sqrt {1+(c+d x)^2}}\right )\right )-i b^2 (c+d x) \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )}{d e^2 (c+d x)} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 324 vs. \(2 (115 ) = 230\).
Time = 1.05 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.73
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}+\frac {b^{2} \left (-\frac {\arctan \left (d x +c \right )^{2}}{d x +c}+2 \ln \left (d x +c \right ) \arctan \left (d x +c \right )-\arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {i \left (\ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (d x +c +i\right )}{2}\right )-\ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )\right )}{2}+\frac {i \left (\ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (d x +c -i\right )}{2}\right )-\ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )\right )}{2}+i \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )-i \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )+i \operatorname {dilog}\left (1+i \left (d x +c \right )\right )-i \operatorname {dilog}\left (1-i \left (d x +c \right )\right )\right )}{e^{2}}+\frac {2 a b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(325\) |
default | \(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}+\frac {b^{2} \left (-\frac {\arctan \left (d x +c \right )^{2}}{d x +c}+2 \ln \left (d x +c \right ) \arctan \left (d x +c \right )-\arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {i \left (\ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (d x +c +i\right )}{2}\right )-\ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )\right )}{2}+\frac {i \left (\ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (d x +c -i\right )}{2}\right )-\ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )\right )}{2}+i \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )-i \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )+i \operatorname {dilog}\left (1+i \left (d x +c \right )\right )-i \operatorname {dilog}\left (1-i \left (d x +c \right )\right )\right )}{e^{2}}+\frac {2 a b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(325\) |
parts | \(-\frac {a^{2}}{e^{2} \left (d x +c \right ) d}+\frac {b^{2} \left (-\frac {\arctan \left (d x +c \right )^{2}}{d x +c}+2 \ln \left (d x +c \right ) \arctan \left (d x +c \right )-\arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {i \left (\ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c -i\right )^{2}}{2}-\operatorname {dilog}\left (-\frac {i \left (d x +c +i\right )}{2}\right )-\ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )\right )}{2}+\frac {i \left (\ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )-\frac {\ln \left (d x +c +i\right )^{2}}{2}-\operatorname {dilog}\left (\frac {i \left (d x +c -i\right )}{2}\right )-\ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )\right )}{2}+i \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )-i \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )+i \operatorname {dilog}\left (1+i \left (d x +c \right )\right )-i \operatorname {dilog}\left (1-i \left (d x +c \right )\right )\right )}{e^{2} d}+\frac {2 a b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2} d}\) | \(330\) |
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\[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\frac {\int \frac {a^{2}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]
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\[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \arctan (c+d x))^2}{(c e+d e x)^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]
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