Integrand size = 23, antiderivative size = 152 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4-\frac {i b d^2 \arctan (c x)}{2 c^3}+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))+\frac {4 b d^2 \log \left (1+c^2 x^2\right )}{15 c^3} \]
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Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {45, 4992, 12, 1816, 649, 209, 266} \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))-\frac {i b d^2 \arctan (c x)}{2 c^3}+\frac {i b d^2 x}{2 c^2}+\frac {4 b d^2 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac {1}{20} b c d^2 x^4-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3 \]
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Rule 12
Rule 45
Rule 209
Rule 266
Rule 649
Rule 1816
Rule 4992
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))-(b c) \int \frac {d^2 x^3 \left (10+15 i c x-6 c^2 x^2\right )}{30 \left (1+c^2 x^2\right )} \, dx \\ & = \frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))-\frac {1}{30} \left (b c d^2\right ) \int \frac {x^3 \left (10+15 i c x-6 c^2 x^2\right )}{1+c^2 x^2} \, dx \\ & = \frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))-\frac {1}{30} \left (b c d^2\right ) \int \left (-\frac {15 i}{c^3}+\frac {16 x}{c^2}+\frac {15 i x^2}{c}-6 x^3+\frac {15 i-16 c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))-\frac {\left (b d^2\right ) \int \frac {15 i-16 c x}{1+c^2 x^2} \, dx}{30 c^2} \\ & = \frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))-\frac {\left (i b d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 c^2}+\frac {\left (8 b d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{15 c} \\ & = \frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4-\frac {i b d^2 \arctan (c x)}{2 c^3}+\frac {1}{3} d^2 x^3 (a+b \arctan (c x))+\frac {1}{2} i c d^2 x^4 (a+b \arctan (c x))-\frac {1}{5} c^2 d^2 x^5 (a+b \arctan (c x))+\frac {4 b d^2 \log \left (1+c^2 x^2\right )}{15 c^3} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.76 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\frac {d^2 \left (2 a c^3 x^3 \left (10+15 i c x-6 c^2 x^2\right )+b c x \left (30 i-16 c x-10 i c^2 x^2+3 c^3 x^3\right )+2 b \left (-15 i+10 c^3 x^3+15 i c^4 x^4-6 c^5 x^5\right ) \arctan (c x)+16 b \log \left (1+c^2 x^2\right )\right )}{60 c^3} \]
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Time = 1.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81
method | result | size |
parts | \(a \,d^{2} \left (-\frac {1}{5} c^{2} x^{5}+\frac {1}{2} i c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {b \,d^{2} \left (-\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{2}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{2}+\frac {c^{4} x^{4}}{20}-\frac {i c^{3} x^{3}}{6}-\frac {4 c^{2} x^{2}}{15}+\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {i \arctan \left (c x \right )}{2}\right )}{c^{3}}\) | \(123\) |
derivativedivides | \(\frac {a \,d^{2} \left (-\frac {1}{5} c^{5} x^{5}+\frac {1}{2} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{2} \left (-\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{2}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{2}+\frac {c^{4} x^{4}}{20}-\frac {i c^{3} x^{3}}{6}-\frac {4 c^{2} x^{2}}{15}+\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {i \arctan \left (c x \right )}{2}\right )}{c^{3}}\) | \(129\) |
default | \(\frac {a \,d^{2} \left (-\frac {1}{5} c^{5} x^{5}+\frac {1}{2} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{2} \left (-\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {i \arctan \left (c x \right ) c^{4} x^{4}}{2}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {i c x}{2}+\frac {c^{4} x^{4}}{20}-\frac {i c^{3} x^{3}}{6}-\frac {4 c^{2} x^{2}}{15}+\frac {4 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {i \arctan \left (c x \right )}{2}\right )}{c^{3}}\) | \(129\) |
parallelrisch | \(\frac {-12 c^{5} b \,d^{2} \arctan \left (c x \right ) x^{5}+30 i x^{4} \arctan \left (c x \right ) b \,c^{4} d^{2}-12 a \,c^{5} d^{2} x^{5}+30 i x^{4} a \,c^{4} d^{2}+3 b \,c^{4} d^{2} x^{4}-10 i x^{3} b \,c^{3} d^{2}+20 x^{3} \arctan \left (c x \right ) b \,d^{2} c^{3}+20 a \,c^{3} d^{2} x^{3}-16 b \,c^{2} d^{2} x^{2}+30 i b \,d^{2} x c -30 i b \,d^{2} \arctan \left (c x \right )+16 b \ln \left (c^{2} x^{2}+1\right ) d^{2}}{60 c^{3}}\) | \(166\) |
risch | \(\frac {i d^{2} b \left (6 c^{2} x^{5}-15 i c \,x^{4}-10 x^{3}\right ) \ln \left (i c x +1\right )}{60}-\frac {i d^{2} c^{2} b \,x^{5} \ln \left (-i c x +1\right )}{10}-\frac {a \,c^{2} d^{2} x^{5}}{5}+\frac {i a c \,d^{2} x^{4}}{2}-\frac {d^{2} c \,x^{4} b \ln \left (-i c x +1\right )}{4}+\frac {i d^{2} b \,x^{3} \ln \left (-i c x +1\right )}{6}+\frac {b c \,d^{2} x^{4}}{20}-\frac {i b \,d^{2} x^{3}}{6}+\frac {a \,d^{2} x^{3}}{3}-\frac {4 b \,d^{2} x^{2}}{15 c}+\frac {i b \,d^{2} x}{2 c^{2}}-\frac {i b \,d^{2} \arctan \left (c x \right )}{2 c^{3}}+\frac {4 b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{15 c^{3}}\) | \(203\) |
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Time = 0.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {12 \, a c^{5} d^{2} x^{5} + 3 \, {\left (-10 i \, a - b\right )} c^{4} d^{2} x^{4} - 10 \, {\left (2 \, a - i \, b\right )} c^{3} d^{2} x^{3} + 16 \, b c^{2} d^{2} x^{2} - 30 i \, b c d^{2} x - 31 \, b d^{2} \log \left (\frac {c x + i}{c}\right ) - b d^{2} \log \left (\frac {c x - i}{c}\right ) - {\left (-6 i \, b c^{5} d^{2} x^{5} - 15 \, b c^{4} d^{2} x^{4} + 10 i \, b c^{3} d^{2} x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{60 \, c^{3}} \]
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Time = 1.92 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.64 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=- \frac {a c^{2} d^{2} x^{5}}{5} - \frac {4 b d^{2} x^{2}}{15 c} + \frac {i b d^{2} x}{2 c^{2}} - \frac {b d^{2} \left (- \frac {\log {\left (47 b c d^{2} x - 47 i b d^{2} \right )}}{60} - \frac {49 \log {\left (47 b c d^{2} x + 47 i b d^{2} \right )}}{120}\right )}{c^{3}} - x^{4} \left (- \frac {i a c d^{2}}{2} - \frac {b c d^{2}}{20}\right ) - x^{3} \left (- \frac {a d^{2}}{3} + \frac {i b d^{2}}{6}\right ) + \left (\frac {i b c^{2} d^{2} x^{5}}{10} + \frac {b c d^{2} x^{4}}{4} - \frac {i b d^{2} x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 12 i b c^{5} d^{2} x^{5} - 30 b c^{4} d^{2} x^{4} + 20 i b c^{3} d^{2} x^{3} + 13 b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{120 c^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.14 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {1}{5} \, a c^{2} d^{2} x^{5} + \frac {1}{2} i \, a c d^{2} x^{4} - \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{6} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} \]
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\[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
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Time = 0.82 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int x^2 (d+i c d x)^2 (a+b \arctan (c x)) \, dx=-\frac {\frac {d^2\,\left (-16\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{60}+\frac {4\,b\,c^2\,d^2\,x^2}{15}-\frac {b\,c\,d^2\,x\,1{}\mathrm {i}}{2}}{c^3}+\frac {d^2\,\left (20\,a\,x^3+20\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,10{}\mathrm {i}\right )}{60}-\frac {c^2\,d^2\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atan}\left (c\,x\right )\right )}{60}+\frac {c\,d^2\,\left (a\,x^4\,30{}\mathrm {i}+3\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{60} \]
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