Integrand size = 23, antiderivative size = 191 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {11 i b d^3 x}{12 c^2}-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} i b c^2 d^3 x^5-\frac {11 i b d^3 \arctan (c x)}{12 c^3}+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))+\frac {7 b d^3 \log \left (1+c^2 x^2\right )}{15 c^3} \]
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Time = 0.12 (sec) , antiderivative size = 191, 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} \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))-\frac {11 i b d^3 \arctan (c x)}{12 c^3}+\frac {1}{30} i b c^2 d^3 x^5+\frac {11 i b d^3 x}{12 c^2}+\frac {7 b d^3 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac {3}{20} b c d^3 x^4-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3 \]
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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^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-(b c) \int \frac {d^3 x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )}{60 \left (1+c^2 x^2\right )} \, dx \\ & = \frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {1}{60} \left (b c d^3\right ) \int \frac {x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )}{1+c^2 x^2} \, dx \\ & = \frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {1}{60} \left (b c d^3\right ) \int \left (-\frac {55 i}{c^3}+\frac {56 x}{c^2}+\frac {55 i x^2}{c}-36 x^3-10 i c x^4+\frac {55 i-56 c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {11 i b d^3 x}{12 c^2}-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} i b c^2 d^3 x^5+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {\left (b d^3\right ) \int \frac {55 i-56 c x}{1+c^2 x^2} \, dx}{60 c^2} \\ & = \frac {11 i b d^3 x}{12 c^2}-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} i b c^2 d^3 x^5+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))-\frac {\left (11 i b d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{12 c^2}+\frac {\left (14 b d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx}{15 c} \\ & = \frac {11 i b d^3 x}{12 c^2}-\frac {7 b d^3 x^2}{15 c}-\frac {11}{36} i b d^3 x^3+\frac {3}{20} b c d^3 x^4+\frac {1}{30} i b c^2 d^3 x^5-\frac {11 i b d^3 \arctan (c x)}{12 c^3}+\frac {1}{3} d^3 x^3 (a+b \arctan (c x))+\frac {3}{4} i c d^3 x^4 (a+b \arctan (c x))-\frac {3}{5} c^2 d^3 x^5 (a+b \arctan (c x))-\frac {1}{6} i c^3 d^3 x^6 (a+b \arctan (c x))+\frac {7 b d^3 \log \left (1+c^2 x^2\right )}{15 c^3} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.76 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {d^3 \left (3 a c^3 x^3 \left (20+45 i c x-36 c^2 x^2-10 i c^3 x^3\right )+b c x \left (165 i-84 c x-55 i c^2 x^2+27 c^3 x^3+6 i c^4 x^4\right )+3 b \left (-55 i+20 c^3 x^3+45 i c^4 x^4-36 c^5 x^5-10 i c^6 x^6\right ) \arctan (c x)+84 b \log \left (1+c^2 x^2\right )\right )}{180 c^3} \]
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Time = 1.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.81
method | result | size |
parts | \(a \,d^{3} \left (-\frac {1}{6} i c^{3} x^{6}-\frac {3}{5} c^{2} x^{5}+\frac {3}{4} i c \,x^{4}+\frac {1}{3} x^{3}\right )+\frac {b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(154\) |
derivativedivides | \(\frac {a \,d^{3} \left (-\frac {1}{6} i c^{6} x^{6}-\frac {3}{5} c^{5} x^{5}+\frac {3}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(160\) |
default | \(\frac {a \,d^{3} \left (-\frac {1}{6} i c^{6} x^{6}-\frac {3}{5} c^{5} x^{5}+\frac {3}{4} i c^{4} x^{4}+\frac {1}{3} c^{3} x^{3}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{6} x^{6}}{6}-\frac {3 c^{5} x^{5} \arctan \left (c x \right )}{5}+\frac {3 i \arctan \left (c x \right ) c^{4} x^{4}}{4}+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}+\frac {11 i c x}{12}+\frac {i c^{5} x^{5}}{30}+\frac {3 c^{4} x^{4}}{20}-\frac {11 i c^{3} x^{3}}{36}-\frac {7 c^{2} x^{2}}{15}+\frac {7 \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {11 i \arctan \left (c x \right )}{12}\right )}{c^{3}}\) | \(160\) |
parallelrisch | \(-\frac {-165 i b \,d^{3} x c +55 i x^{3} b \,c^{3} d^{3}+165 i b \,d^{3} \arctan \left (c x \right )+108 c^{5} b \,d^{3} \arctan \left (c x \right ) x^{5}-6 i x^{5} b \,c^{5} d^{3}+108 a \,c^{5} d^{3} x^{5}-135 i x^{4} a \,c^{4} d^{3}-27 b \,c^{4} d^{3} x^{4}+30 i x^{6} a \,c^{6} d^{3}-60 x^{3} \arctan \left (c x \right ) b \,d^{3} c^{3}-60 a \,c^{3} d^{3} x^{3}+84 b \,c^{2} d^{3} x^{2}-135 i x^{4} \arctan \left (c x \right ) b \,c^{4} d^{3}+30 i x^{6} \arctan \left (c x \right ) b \,c^{6} d^{3}-84 b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{180 c^{3}}\) | \(209\) |
risch | \(-\frac {d^{3} b \left (10 c^{3} x^{6}-36 i c^{2} x^{5}-45 c \,x^{4}+20 i x^{3}\right ) \ln \left (i c x +1\right )}{120}+\frac {11 i b \,d^{3} x}{12 c^{2}}+\frac {d^{3} c^{3} x^{6} b \ln \left (-i c x +1\right )}{12}-\frac {11 i b \,d^{3} \arctan \left (c x \right )}{12 c^{3}}-\frac {11 i b \,d^{3} x^{3}}{36}-\frac {3 a \,c^{2} d^{3} x^{5}}{5}+\frac {i b \,c^{2} d^{3} x^{5}}{30}-\frac {3 d^{3} c \,x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {i a \,c^{3} d^{3} x^{6}}{6}+\frac {3 b c \,d^{3} x^{4}}{20}-\frac {3 i d^{3} c^{2} b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {a \,d^{3} x^{3}}{3}-\frac {7 b \,d^{3} x^{2}}{15 c}+\frac {3 i a c \,d^{3} x^{4}}{4}+\frac {i d^{3} b \,x^{3} \ln \left (-i c x +1\right )}{6}+\frac {7 b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{15 c^{3}}\) | \(257\) |
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Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.99 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {-60 i \, a c^{6} d^{3} x^{6} - 12 \, {\left (18 \, a - i \, b\right )} c^{5} d^{3} x^{5} - 54 \, {\left (-5 i \, a - b\right )} c^{4} d^{3} x^{4} + 10 \, {\left (12 \, a - 11 i \, b\right )} c^{3} d^{3} x^{3} - 168 \, b c^{2} d^{3} x^{2} + 330 i \, b c d^{3} x + 333 \, b d^{3} \log \left (\frac {c x + i}{c}\right ) + 3 \, b d^{3} \log \left (\frac {c x - i}{c}\right ) + 3 \, {\left (10 \, b c^{6} d^{3} x^{6} - 36 i \, b c^{5} d^{3} x^{5} - 45 \, b c^{4} d^{3} x^{4} + 20 i \, b c^{3} d^{3} x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{360 \, c^{3}} \]
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Time = 2.48 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.65 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=- \frac {i a c^{3} d^{3} x^{6}}{6} - \frac {7 b d^{3} x^{2}}{15 c} + \frac {11 i b d^{3} x}{12 c^{2}} - \frac {b d^{3} \left (- \frac {\log {\left (310 b c d^{3} x - 310 i b d^{3} \right )}}{120} - \frac {209 \log {\left (310 b c d^{3} x + 310 i b d^{3} \right )}}{280}\right )}{c^{3}} - x^{5} \cdot \left (\frac {3 a c^{2} d^{3}}{5} - \frac {i b c^{2} d^{3}}{30}\right ) - x^{4} \left (- \frac {3 i a c d^{3}}{4} - \frac {3 b c d^{3}}{20}\right ) - x^{3} \left (- \frac {a d^{3}}{3} + \frac {11 i b d^{3}}{36}\right ) + \left (- \frac {b c^{3} d^{3} x^{6}}{12} + \frac {3 i b c^{2} d^{3} x^{5}}{10} + \frac {3 b c d^{3} x^{4}}{8} - \frac {i b d^{3} x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (70 b c^{6} d^{3} x^{6} - 252 i b c^{5} d^{3} x^{5} - 315 b c^{4} d^{3} x^{4} + 140 i b c^{3} d^{3} x^{3} + 150 b d^{3}\right ) \log {\left (- i c x + 1 \right )}}{840 c^{3}} \]
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Time = 0.31 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.27 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {1}{6} i \, a c^{3} d^{3} x^{6} - \frac {3}{5} \, a c^{2} d^{3} x^{5} + \frac {3}{4} i \, a c d^{3} x^{4} - \frac {1}{90} i \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{3} d^{3} - \frac {3}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{3} + \frac {1}{3} \, a d^{3} x^{3} + \frac {1}{4} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{3} \]
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\[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )} x^{2} \,d x } \]
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Time = 0.92 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91 \[ \int x^2 (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {\frac {d^3\,\left (-84\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,165{}\mathrm {i}\right )}{180}+\frac {7\,b\,c^2\,d^3\,x^2}{15}-\frac {b\,c\,d^3\,x\,11{}\mathrm {i}}{12}}{c^3}+\frac {d^3\,\left (60\,a\,x^3+60\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,55{}\mathrm {i}\right )}{180}-\frac {c^3\,d^3\,\left (a\,x^6\,30{}\mathrm {i}+b\,x^6\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{180}+\frac {c\,d^3\,\left (a\,x^4\,135{}\mathrm {i}+27\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,135{}\mathrm {i}\right )}{180}-\frac {c^2\,d^3\,\left (108\,a\,x^5+108\,b\,x^5\,\mathrm {atan}\left (c\,x\right )-b\,x^5\,6{}\mathrm {i}\right )}{180} \]
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