Integrand size = 12, antiderivative size = 56 \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {b x^2}{10 c^3}-\frac {b x^4}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x))-\frac {b \log \left (1+c^2 x^2\right )}{10 c^5} \]
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Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4946, 272, 45} \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {1}{5} x^5 (a+b \arctan (c x))+\frac {b x^2}{10 c^3}-\frac {b \log \left (c^2 x^2+1\right )}{10 c^5}-\frac {b x^4}{20 c} \]
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Rule 45
Rule 272
Rule 4946
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 (a+b \arctan (c x))-\frac {1}{5} (b c) \int \frac {x^5}{1+c^2 x^2} \, dx \\ & = \frac {1}{5} x^5 (a+b \arctan (c x))-\frac {1}{10} (b c) \text {Subst}\left (\int \frac {x^2}{1+c^2 x} \, dx,x,x^2\right ) \\ & = \frac {1}{5} x^5 (a+b \arctan (c x))-\frac {1}{10} (b c) \text {Subst}\left (\int \left (-\frac {1}{c^4}+\frac {x}{c^2}+\frac {1}{c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {b x^2}{10 c^3}-\frac {b x^4}{20 c}+\frac {1}{5} x^5 (a+b \arctan (c x))-\frac {b \log \left (1+c^2 x^2\right )}{10 c^5} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {b x^2}{10 c^3}-\frac {b x^4}{20 c}+\frac {a x^5}{5}+\frac {1}{5} b x^5 \arctan (c x)-\frac {b \log \left (1+c^2 x^2\right )}{10 c^5} \]
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Time = 0.39 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96
method | result | size |
parts | \(\frac {a \,x^{5}}{5}+\frac {b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}-\frac {\ln \left (c^{2} x^{2}+1\right )}{10}\right )}{c^{5}}\) | \(54\) |
derivativedivides | \(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}-\frac {\ln \left (c^{2} x^{2}+1\right )}{10}\right )}{c^{5}}\) | \(58\) |
default | \(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \arctan \left (c x \right )}{5}-\frac {c^{4} x^{4}}{20}+\frac {c^{2} x^{2}}{10}-\frac {\ln \left (c^{2} x^{2}+1\right )}{10}\right )}{c^{5}}\) | \(58\) |
parallelrisch | \(-\frac {-4 b \arctan \left (c x \right ) x^{5} c^{5}-4 a \,c^{5} x^{5}+b \,c^{4} x^{4}-2 b \,c^{2} x^{2}+2 b \ln \left (c^{2} x^{2}+1\right )+2 b}{20 c^{5}}\) | \(62\) |
risch | \(-\frac {i x^{5} b \ln \left (i c x +1\right )}{10}+\frac {i x^{5} b \ln \left (-i c x +1\right )}{10}+\frac {a \,x^{5}}{5}-\frac {b \,x^{4}}{20 c}+\frac {b \,x^{2}}{10 c^{3}}-\frac {b \ln \left (-c^{2} x^{2}-1\right )}{10 c^{5}}\) | \(73\) |
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Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {4 \, b c^{5} x^{5} \arctan \left (c x\right ) + 4 \, a c^{5} x^{5} - b c^{4} x^{4} + 2 \, b c^{2} x^{2} - 2 \, b \log \left (c^{2} x^{2} + 1\right )}{20 \, c^{5}} \]
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Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.07 \[ \int x^4 (a+b \arctan (c x)) \, dx=\begin {cases} \frac {a x^{5}}{5} + \frac {b x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b x^{4}}{20 c} + \frac {b x^{2}}{10 c^{3}} - \frac {b \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\\frac {a x^{5}}{5} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b \]
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\[ \int x^4 (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} x^{4} \,d x } \]
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Time = 0.47 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int x^4 (a+b \arctan (c x)) \, dx=\frac {a\,x^5}{5}-\frac {\frac {b\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {b\,c^2\,x^2}{10}+\frac {b\,c^4\,x^4}{20}}{c^5}+\frac {b\,x^5\,\mathrm {atan}\left (c\,x\right )}{5} \]
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