Integrand size = 18, antiderivative size = 124 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=-\frac {b \left (10 c^2 d-3 e\right ) e x^2}{30 c^3}-\frac {b e^2 x^4}{20 c}+d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-\frac {b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (1+c^2 x^2\right )}{30 c^5} \]
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Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {200, 5032, 1608, 1261, 712} \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-\frac {b e x^2 \left (10 c^2 d-3 e\right )}{30 c^3}-\frac {b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (c^2 x^2+1\right )}{30 c^5}-\frac {b e^2 x^4}{20 c} \]
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Rule 200
Rule 712
Rule 1261
Rule 1608
Rule 5032
Rubi steps \begin{align*} \text {integral}& = d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-(b c) \int \frac {d^2 x+\frac {2}{3} d e x^3+\frac {e^2 x^5}{5}}{1+c^2 x^2} \, dx \\ & = d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-(b c) \int \frac {x \left (d^2+\frac {2}{3} d e x^2+\frac {e^2 x^4}{5}\right )}{1+c^2 x^2} \, dx \\ & = d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-\frac {1}{2} (b c) \text {Subst}\left (\int \frac {d^2+\frac {2 d e x}{3}+\frac {e^2 x^2}{5}}{1+c^2 x} \, dx,x,x^2\right ) \\ & = d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-\frac {1}{2} (b c) \text {Subst}\left (\int \left (\frac {\left (10 c^2 d-3 e\right ) e}{15 c^4}+\frac {e^2 x}{5 c^2}+\frac {15 c^4 d^2-10 c^2 d e+3 e^2}{15 c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b \left (10 c^2 d-3 e\right ) e x^2}{30 c^3}-\frac {b e^2 x^4}{20 c}+d^2 x (a+b \arctan (c x))+\frac {2}{3} d e x^3 (a+b \arctan (c x))+\frac {1}{5} e^2 x^5 (a+b \arctan (c x))-\frac {b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (1+c^2 x^2\right )}{30 c^5} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.05 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {c^2 x \left (4 a c^3 \left (15 d^2+10 d e x^2+3 e^2 x^4\right )+b e x \left (6 e-c^2 \left (20 d+3 e x^2\right )\right )\right )+4 b c^5 x \left (15 d^2+10 d e x^2+3 e^2 x^4\right ) \arctan (c x)-2 b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (1+c^2 x^2\right )}{60 c^5} \]
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Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10
method | result | size |
parts | \(a \left (\frac {1}{5} x^{5} e^{2}+\frac {2}{3} x^{3} e d +x \,d^{2}\right )+\frac {b \left (\frac {\arctan \left (c x \right ) c \,e^{2} x^{5}}{5}+\frac {2 \arctan \left (c x \right ) c d e \,x^{3}}{3}+\arctan \left (c x \right ) c x \,d^{2}-\frac {5 d \,c^{4} e \,x^{2}+\frac {3 e^{2} c^{4} x^{4}}{4}-\frac {3 e^{2} c^{2} x^{2}}{2}+\frac {\left (15 c^{4} d^{2}-10 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{15 c^{4}}\right )}{c}\) | \(137\) |
derivativedivides | \(\frac {\frac {a \left (c^{5} x \,d^{2}+\frac {2}{3} d \,c^{5} e \,x^{3}+\frac {1}{5} e^{2} c^{5} x^{5}\right )}{c^{4}}+\frac {b \left (\arctan \left (c x \right ) c^{5} x \,d^{2}+\frac {2 \arctan \left (c x \right ) d \,c^{5} e \,x^{3}}{3}+\frac {\arctan \left (c x \right ) e^{2} c^{5} x^{5}}{5}-\frac {d \,c^{4} e \,x^{2}}{3}-\frac {e^{2} c^{4} x^{4}}{20}+\frac {e^{2} c^{2} x^{2}}{10}-\frac {\left (15 c^{4} d^{2}-10 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}\right )}{c^{4}}}{c}\) | \(153\) |
default | \(\frac {\frac {a \left (c^{5} x \,d^{2}+\frac {2}{3} d \,c^{5} e \,x^{3}+\frac {1}{5} e^{2} c^{5} x^{5}\right )}{c^{4}}+\frac {b \left (\arctan \left (c x \right ) c^{5} x \,d^{2}+\frac {2 \arctan \left (c x \right ) d \,c^{5} e \,x^{3}}{3}+\frac {\arctan \left (c x \right ) e^{2} c^{5} x^{5}}{5}-\frac {d \,c^{4} e \,x^{2}}{3}-\frac {e^{2} c^{4} x^{4}}{20}+\frac {e^{2} c^{2} x^{2}}{10}-\frac {\left (15 c^{4} d^{2}-10 c^{2} d e +3 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}\right )}{c^{4}}}{c}\) | \(153\) |
parallelrisch | \(-\frac {-12 x^{5} \arctan \left (c x \right ) b \,c^{5} e^{2}-12 a \,c^{5} e^{2} x^{5}-40 x^{3} \arctan \left (c x \right ) b \,c^{5} d e +3 b \,c^{4} e^{2} x^{4}-40 a \,c^{5} d e \,x^{3}-60 x \arctan \left (c x \right ) b \,c^{5} d^{2}+20 b \,c^{4} d e \,x^{2}-60 a \,c^{5} d^{2} x +30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{2}-6 b \,c^{2} e^{2} x^{2}-20 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d e +6 \ln \left (c^{2} x^{2}+1\right ) b \,e^{2}}{60 c^{5}}\) | \(173\) |
risch | \(-\frac {i b \left (3 x^{5} e^{2}+10 x^{3} e d +15 x \,d^{2}\right ) \ln \left (i c x +1\right )}{30}+\frac {i b \,e^{2} x^{5} \ln \left (-i c x +1\right )}{10}+\frac {i b d e \,x^{3} \ln \left (-i c x +1\right )}{3}+\frac {a \,e^{2} x^{5}}{5}+\frac {i b \,d^{2} x \ln \left (-i c x +1\right )}{2}+\frac {2 a d e \,x^{3}}{3}-\frac {b \,e^{2} x^{4}}{20 c}+a \,d^{2} x -\frac {b d e \,x^{2}}{3 c}-\frac {\ln \left (-c^{2} x^{2}-1\right ) b \,d^{2}}{2 c}+\frac {b \,e^{2} x^{2}}{10 c^{3}}+\frac {\ln \left (-c^{2} x^{2}-1\right ) b d e}{3 c^{3}}-\frac {\ln \left (-c^{2} x^{2}-1\right ) b \,e^{2}}{10 c^{5}}\) | \(204\) |
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Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.21 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {12 \, a c^{5} e^{2} x^{5} + 40 \, a c^{5} d e x^{3} - 3 \, b c^{4} e^{2} x^{4} + 60 \, a c^{5} d^{2} x - 2 \, {\left (10 \, b c^{4} d e - 3 \, b c^{2} e^{2}\right )} x^{2} + 4 \, {\left (3 \, b c^{5} e^{2} x^{5} + 10 \, b c^{5} d e x^{3} + 15 \, b c^{5} d^{2} x\right )} \arctan \left (c x\right ) - 2 \, {\left (15 \, b c^{4} d^{2} - 10 \, b c^{2} d e + 3 \, b e^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \]
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Time = 0.39 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.56 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\begin {cases} a d^{2} x + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{5}}{5} + b d^{2} x \operatorname {atan}{\left (c x \right )} + \frac {2 b d e x^{3} \operatorname {atan}{\left (c x \right )}}{3} + \frac {b e^{2} x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {b d e x^{2}}{3 c} - \frac {b e^{2} x^{4}}{20 c} + \frac {b d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3 c^{3}} + \frac {b e^{2} x^{2}}{10 c^{3}} - \frac {b e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{2} x + \frac {2 d e x^{3}}{3} + \frac {e^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {1}{5} \, a e^{2} x^{5} + \frac {2}{3} \, a d e x^{3} + \frac {1}{3} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d e + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{2} + a d^{2} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{2}}{2 \, c} \]
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\[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} \,d x } \]
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Time = 0.75 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.21 \[ \int \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {a\,e^2\,x^5}{5}+a\,d^2\,x-\frac {b\,d^2\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,e^2\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}-\frac {b\,e^2\,x^4}{20\,c}+\frac {b\,e^2\,x^2}{10\,c^3}+\frac {2\,a\,d\,e\,x^3}{3}+b\,d^2\,x\,\mathrm {atan}\left (c\,x\right )+\frac {b\,e^2\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}+\frac {b\,d\,e\,\ln \left (c^2\,x^2+1\right )}{3\,c^3}-\frac {b\,d\,e\,x^2}{3\,c}+\frac {2\,b\,d\,e\,x^3\,\mathrm {atan}\left (c\,x\right )}{3} \]
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