Integrand size = 18, antiderivative size = 160 \[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=-\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \arctan (a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^2} \]
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Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5050, 5000, 4930, 5040, 4964, 2449, 2352, 8} \[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\frac {c \left (a^2 x^2+1\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c x \left (a^2 x^2+1\right ) \arctan (a x)^2}{4 a}+\frac {c \left (a^2 x^2+1\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c \arctan (a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{i a x+1}\right )}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x}{4 a} \]
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Rule 8
Rule 2352
Rule 2449
Rule 4930
Rule 4964
Rule 5000
Rule 5040
Rule 5050
Rubi steps \begin{align*} \text {integral}& = \frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {3 \int \left (c+a^2 c x^2\right ) \arctan (a x)^2 \, dx}{4 a} \\ & = \frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \int 1 \, dx}{4 a}-\frac {c \int \arctan (a x)^2 \, dx}{2 a} \\ & = -\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}+c \int \frac {x \arctan (a x)}{1+a^2 x^2} \, dx \\ & = -\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \int \frac {\arctan (a x)}{i-a x} \, dx}{a} \\ & = -\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \arctan (a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}+\frac {c \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a} \\ & = -\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \arctan (a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {(i c) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{a^2} \\ & = -\frac {c x}{4 a}+\frac {c \left (1+a^2 x^2\right ) \arctan (a x)}{4 a^2}-\frac {i c \arctan (a x)^2}{2 a^2}-\frac {c x \arctan (a x)^2}{2 a}-\frac {c x \left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a}+\frac {c \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{4 a^2}-\frac {c \arctan (a x) \log \left (\frac {2}{1+i a x}\right )}{a^2}-\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{2 a^2} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.63 \[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\frac {c \left (-a x-\left (-2 i+3 a x+a^3 x^3\right ) \arctan (a x)^2+\left (1+a^2 x^2\right )^2 \arctan (a x)^3+\arctan (a x) \left (1+a^2 x^2-4 \log \left (1+e^{2 i \arctan (a x)}\right )\right )+2 i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (a x)}\right )\right )}{4 a^2} \]
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Time = 2.58 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.48
method | result | size |
parts | \(\frac {c \arctan \left (a x \right )^{3} a^{2} x^{4}}{4}+\frac {c \arctan \left (a x \right )^{3} x^{2}}{2}+\frac {c \arctan \left (a x \right )^{3}}{4 a^{2}}-\frac {3 c \left (\frac {a^{3} \arctan \left (a x \right )^{2} x^{3}}{3}+a \arctan \left (a x \right )^{2} x -\frac {a^{2} \arctan \left (a x \right ) x^{2}}{3}-\frac {2 \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3}+\frac {a x}{3}-\frac {\arctan \left (a x \right )}{3}-\frac {i \left (\ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )-\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\frac {\ln \left (a x -i\right )^{2}}{2}\right )}{3}+\frac {i \left (\ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (\frac {i \left (a x -i\right )}{2}\right )-\ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )-\frac {\ln \left (a x +i\right )^{2}}{2}\right )}{3}\right )}{4 a^{2}}\) | \(237\) |
derivativedivides | \(\frac {\frac {c \arctan \left (a x \right )^{3} a^{4} x^{4}}{4}+\frac {a^{2} c \,x^{2} \arctan \left (a x \right )^{3}}{2}+\frac {c \arctan \left (a x \right )^{3}}{4}-\frac {3 c \left (\frac {a^{3} \arctan \left (a x \right )^{2} x^{3}}{3}+a \arctan \left (a x \right )^{2} x -\frac {a^{2} \arctan \left (a x \right ) x^{2}}{3}-\frac {2 \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3}+\frac {a x}{3}-\frac {\arctan \left (a x \right )}{3}-\frac {i \left (\ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )-\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\frac {\ln \left (a x -i\right )^{2}}{2}\right )}{3}+\frac {i \left (\ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (\frac {i \left (a x -i\right )}{2}\right )-\ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )-\frac {\ln \left (a x +i\right )^{2}}{2}\right )}{3}\right )}{4}}{a^{2}}\) | \(238\) |
default | \(\frac {\frac {c \arctan \left (a x \right )^{3} a^{4} x^{4}}{4}+\frac {a^{2} c \,x^{2} \arctan \left (a x \right )^{3}}{2}+\frac {c \arctan \left (a x \right )^{3}}{4}-\frac {3 c \left (\frac {a^{3} \arctan \left (a x \right )^{2} x^{3}}{3}+a \arctan \left (a x \right )^{2} x -\frac {a^{2} \arctan \left (a x \right ) x^{2}}{3}-\frac {2 \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3}+\frac {a x}{3}-\frac {\arctan \left (a x \right )}{3}-\frac {i \left (\ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )-\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\frac {\ln \left (a x -i\right )^{2}}{2}\right )}{3}+\frac {i \left (\ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (\frac {i \left (a x -i\right )}{2}\right )-\ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )-\frac {\ln \left (a x +i\right )^{2}}{2}\right )}{3}\right )}{4}}{a^{2}}\) | \(238\) |
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\[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\int { {\left (a^{2} c x^{2} + c\right )} x \arctan \left (a x\right )^{3} \,d x } \]
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\[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=c \left (\int x \operatorname {atan}^{3}{\left (a x \right )}\, dx + \int a^{2} x^{3} \operatorname {atan}^{3}{\left (a x \right )}\, dx\right ) \]
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\[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\int { {\left (a^{2} c x^{2} + c\right )} x \arctan \left (a x\right )^{3} \,d x } \]
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\[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\int { {\left (a^{2} c x^{2} + c\right )} x \arctan \left (a x\right )^{3} \,d x } \]
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Timed out. \[ \int x \left (c+a^2 c x^2\right ) \arctan (a x)^3 \, dx=\int x\,{\mathrm {atan}\left (a\,x\right )}^3\,\left (c\,a^2\,x^2+c\right ) \,d x \]
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