Integrand size = 21, antiderivative size = 157 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {3 b d^3 x}{5 c}-\frac {3 i b d^3 (i-c x)^2}{20 c^2}-\frac {b d^3 (i-c x)^3}{20 c^2}+\frac {i b d^3 (i-c x)^4}{20 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {6 i b d^3 \log (i+c x)}{5 c^2} \]
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Time = 0.07 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {45, 4992, 12, 78} \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}+\frac {i b d^3 (-c x+i)^4}{20 c^2}-\frac {b d^3 (-c x+i)^3}{20 c^2}-\frac {3 i b d^3 (-c x+i)^2}{20 c^2}+\frac {6 i b d^3 \log (c x+i)}{5 c^2}-\frac {3 b d^3 x}{5 c} \]
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Rule 12
Rule 45
Rule 78
Rule 4992
Rubi steps \begin{align*} \text {integral}& = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-(b c) \int \frac {d^3 (i-c x)^3 (-1+4 i c x)}{20 c^2 (i+c x)} \, dx \\ & = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-\frac {\left (b d^3\right ) \int \frac {(i-c x)^3 (-1+4 i c x)}{i+c x} \, dx}{20 c} \\ & = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-\frac {\left (b d^3\right ) \int \left (12+4 i (i-c x)^3+6 i (-i+c x)-3 (-i+c x)^2-\frac {24 i}{i+c x}\right ) \, dx}{20 c} \\ & = -\frac {3 b d^3 x}{5 c}-\frac {3 i b d^3 (i-c x)^2}{20 c^2}-\frac {b d^3 (i-c x)^3}{20 c^2}+\frac {i b d^3 (i-c x)^4}{20 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {6 i b d^3 \log (i+c x)}{5 c^2} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {d^3 \left (c x \left (b \left (-25-12 i c x+5 c^2 x^2+i c^3 x^3\right )+a c x \left (10+20 i c x-15 c^2 x^2-4 i c^3 x^3\right )\right )+b \left (25+10 c^2 x^2+20 i c^3 x^3-15 c^4 x^4-4 i c^5 x^5\right ) \arctan (c x)+12 i b \log \left (1+c^2 x^2\right )\right )}{20 c^2} \]
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Time = 1.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.92
method | result | size |
parts | \(a \,d^{3} \left (-\frac {1}{5} i c^{3} x^{5}-\frac {3}{4} c^{2} x^{4}+i c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\) | \(145\) |
derivativedivides | \(\frac {a \,d^{3} \left (-\frac {1}{5} i c^{5} x^{5}-\frac {3}{4} c^{4} x^{4}+i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\) | \(151\) |
default | \(\frac {a \,d^{3} \left (-\frac {1}{5} i c^{5} x^{5}-\frac {3}{4} c^{4} x^{4}+i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\) | \(151\) |
parallelrisch | \(\frac {-4 i c^{5} b \,d^{3} \arctan \left (c x \right ) x^{5}-4 i x^{5} a \,c^{5} d^{3}+i x^{4} b \,c^{4} d^{3}-15 x^{4} \arctan \left (c x \right ) b \,c^{4} d^{3}+20 i x^{3} \arctan \left (c x \right ) b \,c^{3} d^{3}-15 a \,c^{4} d^{3} x^{4}+20 i x^{3} a \,c^{3} d^{3}+5 b \,c^{3} d^{3} x^{3}-12 i x^{2} b \,c^{2} d^{3}+10 x^{2} \arctan \left (c x \right ) b \,c^{2} d^{3}+10 x^{2} d^{3} c^{2} a +12 i b \,d^{3} \ln \left (c^{2} x^{2}+1\right )-25 b c \,d^{3} x +25 b \,d^{3} \arctan \left (c x \right )}{20 c^{2}}\) | \(196\) |
risch | \(-\frac {d^{3} b \left (4 c^{3} x^{5}-15 i c^{2} x^{4}-20 c \,x^{3}+10 i x^{2}\right ) \ln \left (i c x +1\right )}{40}+\frac {d^{3} c^{3} b \,x^{5} \ln \left (-i c x +1\right )}{10}-\frac {i a \,c^{3} d^{3} x^{5}}{5}-\frac {3 a \,c^{2} d^{3} x^{4}}{4}-\frac {3 i d^{3} c^{2} x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {d^{3} c b \,x^{3} \ln \left (-i c x +1\right )}{2}+\frac {i b \,c^{2} d^{3} x^{4}}{20}+\frac {b c \,d^{3} x^{3}}{4}+i a c \,d^{3} x^{3}+\frac {a \,d^{3} x^{2}}{2}+\frac {i d^{3} x^{2} b \ln \left (-i c x +1\right )}{4}-\frac {3 i b \,d^{3} x^{2}}{5}-\frac {5 b \,d^{3} x}{4 c}+\frac {5 d^{3} b \arctan \left (c x \right )}{4 c^{2}}+\frac {3 i d^{3} b \ln \left (625 c^{2} x^{2}+625\right )}{5 c^{2}}\) | \(245\) |
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Time = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {-8 i \, a c^{5} d^{3} x^{5} - 2 \, {\left (15 \, a - i \, b\right )} c^{4} d^{3} x^{4} - 10 \, {\left (-4 i \, a - b\right )} c^{3} d^{3} x^{3} + 4 \, {\left (5 \, a - 6 i \, b\right )} c^{2} d^{3} x^{2} - 50 \, b c d^{3} x + 49 i \, b d^{3} \log \left (\frac {c x + i}{c}\right ) - i \, b d^{3} \log \left (\frac {c x - i}{c}\right ) + {\left (4 \, b c^{5} d^{3} x^{5} - 15 i \, b c^{4} d^{3} x^{4} - 20 \, b c^{3} d^{3} x^{3} + 10 i \, b c^{2} d^{3} x^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{40 \, c^{2}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (138) = 276\).
Time = 2.16 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.89 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=- \frac {i a c^{3} d^{3} x^{5}}{5} - \frac {5 b d^{3} x}{4 c} - \frac {b d^{3} \left (\frac {i \log {\left (19 b c d^{3} x - 19 i b d^{3} \right )}}{40} - \frac {37 i \log {\left (19 b c d^{3} x + 19 i b d^{3} \right )}}{40}\right )}{c^{2}} - x^{4} \cdot \left (\frac {3 a c^{2} d^{3}}{4} - \frac {i b c^{2} d^{3}}{20}\right ) - x^{3} \left (- i a c d^{3} - \frac {b c d^{3}}{4}\right ) - x^{2} \left (- \frac {a d^{3}}{2} + \frac {3 i b d^{3}}{5}\right ) + \left (- \frac {b c^{3} d^{3} x^{5}}{10} + \frac {3 i b c^{2} d^{3} x^{4}}{8} + \frac {b c d^{3} x^{3}}{2} - \frac {i b d^{3} x^{2}}{4}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (4 b c^{5} d^{3} x^{5} - 15 i b c^{4} d^{3} x^{4} - 20 b c^{3} d^{3} x^{3} + 10 i b c^{2} d^{3} x^{2} + 12 i b d^{3}\right ) \log {\left (- i c x + 1 \right )}}{40 c^{2}} \]
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Time = 0.31 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.41 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {1}{5} i \, a c^{3} d^{3} x^{5} - \frac {3}{4} \, a c^{2} d^{3} x^{4} - \frac {1}{20} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{3} + i \, a c d^{3} x^{3} - \frac {1}{4} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{2} d^{3} + \frac {1}{2} i \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c d^{3} + \frac {1}{2} \, a d^{3} x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{3} \]
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\[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x \,d x } \]
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Time = 0.79 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.02 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {\frac {d^3\,\left (25\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,12{}\mathrm {i}\right )}{20}-\frac {5\,b\,c\,d^3\,x}{4}}{c^2}+\frac {d^3\,\left (10\,a\,x^2+10\,b\,x^2\,\mathrm {atan}\left (c\,x\right )-b\,x^2\,12{}\mathrm {i}\right )}{20}-\frac {c^3\,d^3\,\left (a\,x^5\,4{}\mathrm {i}+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}\right )}{20}+\frac {c\,d^3\,\left (a\,x^3\,20{}\mathrm {i}+5\,b\,x^3+b\,x^3\,\mathrm {atan}\left (c\,x\right )\,20{}\mathrm {i}\right )}{20}-\frac {c^2\,d^3\,\left (15\,a\,x^4+15\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,1{}\mathrm {i}\right )}{20} \]
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