Integrand size = 21, antiderivative size = 105 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {i b d x}{4 c^2}-\frac {b d x^2}{6 c}-\frac {1}{12} i b d x^3-\frac {i b d \arctan (c x)}{4 c^3}+\frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))+\frac {b d \log \left (1+c^2 x^2\right )}{6 c^3} \]
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Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {45, 4992, 12, 815, 649, 209, 266} \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{4} i c d x^4 (a+b \arctan (c x))+\frac {1}{3} d x^3 (a+b \arctan (c x))-\frac {i b d \arctan (c x)}{4 c^3}+\frac {i b d x}{4 c^2}+\frac {b d \log \left (c^2 x^2+1\right )}{6 c^3}-\frac {b d x^2}{6 c}-\frac {1}{12} i b d x^3 \]
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Rule 12
Rule 45
Rule 209
Rule 266
Rule 649
Rule 815
Rule 4992
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))-(b c) \int \frac {d x^3 (4+3 i c x)}{12 \left (1+c^2 x^2\right )} \, dx \\ & = \frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))-\frac {1}{12} (b c d) \int \frac {x^3 (4+3 i c x)}{1+c^2 x^2} \, dx \\ & = \frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))-\frac {1}{12} (b c d) \int \left (-\frac {3 i}{c^3}+\frac {4 x}{c^2}+\frac {3 i x^2}{c}+\frac {3 i-4 c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {i b d x}{4 c^2}-\frac {b d x^2}{6 c}-\frac {1}{12} i b d x^3+\frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))-\frac {(b d) \int \frac {3 i-4 c x}{1+c^2 x^2} \, dx}{12 c^2} \\ & = \frac {i b d x}{4 c^2}-\frac {b d x^2}{6 c}-\frac {1}{12} i b d x^3+\frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))-\frac {(i b d) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^2}+\frac {(b d) \int \frac {x}{1+c^2 x^2} \, dx}{3 c} \\ & = \frac {i b d x}{4 c^2}-\frac {b d x^2}{6 c}-\frac {1}{12} i b d x^3-\frac {i b d \arctan (c x)}{4 c^3}+\frac {1}{3} d x^3 (a+b \arctan (c x))+\frac {1}{4} i c d x^4 (a+b \arctan (c x))+\frac {b d \log \left (1+c^2 x^2\right )}{6 c^3} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.84 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {d \left (a c^3 x^3 (4+3 i c x)+b c x \left (3 i-2 c x-i c^2 x^2\right )+b \left (-3 i+4 c^3 x^3+3 i c^4 x^4\right ) \arctan (c x)+2 b \log \left (1+c^2 x^2\right )\right )}{12 c^3} \]
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Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87
method | result | size |
parts | \(a d \left (\frac {1}{4} i c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {b d \left (\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{4}-\frac {i c^{3} x^{3}}{12}-\frac {c^{2} x^{2}}{6}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{4}\right )}{c^{3}}\) | \(91\) |
derivativedivides | \(\frac {a d \left (\frac {1}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{4}-\frac {i c^{3} x^{3}}{12}-\frac {c^{2} x^{2}}{6}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{4}\right )}{c^{3}}\) | \(97\) |
default | \(\frac {a d \left (\frac {1}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{4}-\frac {i c^{3} x^{3}}{12}-\frac {c^{2} x^{2}}{6}+\frac {\ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i \arctan \left (c x \right )}{4}\right )}{c^{3}}\) | \(97\) |
parallelrisch | \(\frac {3 i x^{4} \arctan \left (c x \right ) b \,c^{4} d +3 i x^{4} a \,c^{4} d -i x^{3} b \,c^{3} d +4 x^{3} \arctan \left (c x \right ) b d \,c^{3}+4 a \,c^{3} d \,x^{3}-2 b \,c^{2} d \,x^{2}+3 i b d x c -3 i b d \arctan \left (c x \right )+2 b d \ln \left (c^{2} x^{2}+1\right )}{12 c^{3}}\) | \(108\) |
risch | \(\frac {d b \left (3 c \,x^{4}-4 i x^{3}\right ) \ln \left (i c x +1\right )}{24}+\frac {i a c d \,x^{4}}{4}-\frac {d c \,x^{4} b \ln \left (-i c x +1\right )}{8}+\frac {i d b \,x^{3} \ln \left (-i c x +1\right )}{6}-\frac {i b d \,x^{3}}{12}+\frac {x^{3} d a}{3}-\frac {b d \,x^{2}}{6 c}+\frac {i b d x}{4 c^{2}}-\frac {i b d \arctan \left (c x \right )}{4 c^{3}}+\frac {b d \ln \left (c^{2} x^{2}+1\right )}{6 c^{3}}\) | \(131\) |
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Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {6 i \, a c^{4} d x^{4} + 2 \, {\left (4 \, a - i \, b\right )} c^{3} d x^{3} - 4 \, b c^{2} d x^{2} + 6 i \, b c d x + 7 \, b d \log \left (\frac {c x + i}{c}\right ) + b d \log \left (\frac {c x - i}{c}\right ) - {\left (3 \, b c^{4} d x^{4} - 4 i \, b c^{3} d x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{24 \, c^{3}} \]
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Time = 1.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.59 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {i a c d x^{4}}{4} - \frac {b d x^{2}}{6 c} + \frac {i b d x}{4 c^{2}} + \frac {b d \left (\frac {\log {\left (11 b c d x - 11 i b d \right )}}{24} + \frac {9 \log {\left (11 b c d x + 11 i b d \right )}}{40}\right )}{c^{3}} + x^{3} \left (\frac {a d}{3} - \frac {i b d}{12}\right ) + \left (\frac {b c d x^{4}}{8} - \frac {i b d x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 15 b c^{4} d x^{4} + 20 i b c^{3} d x^{3} + 8 b d\right ) \log {\left (- i c x + 1 \right )}}{120 c^{3}} \]
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Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{4} i \, a c d x^{4} + \frac {1}{3} \, a d x^{3} + \frac {1}{12} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d \]
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\[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
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Time = 0.77 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94 \[ \int x^2 (d+i c d x) (a+b \arctan (c x)) \, dx=-\frac {\frac {d\,\left (-2\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i}\right )}{12}+\frac {b\,c^2\,d\,x^2}{6}-\frac {b\,c\,d\,x\,1{}\mathrm {i}}{4}}{c^3}+\frac {d\,\left (4\,a\,x^3+4\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,1{}\mathrm {i}\right )}{12}+\frac {c\,d\,\left (a\,x^4\,3{}\mathrm {i}+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i}\right )}{12} \]
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