Integrand size = 14, antiderivative size = 112 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\frac {a b x}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 x \arctan (c x)}{2 c^3}-\frac {b x^3 (a+b \arctan (c x))}{6 c}-\frac {(a+b \arctan (c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {b^2 \log \left (1+c^2 x^2\right )}{3 c^4} \]
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Time = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4946, 5036, 272, 45, 4930, 266, 5004} \[ \int x^3 (a+b \arctan (c x))^2 \, dx=-\frac {(a+b \arctan (c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {b x^3 (a+b \arctan (c x))}{6 c}+\frac {a b x}{2 c^3}+\frac {b^2 x \arctan (c x)}{2 c^3}+\frac {b^2 x^2}{12 c^2}-\frac {b^2 \log \left (c^2 x^2+1\right )}{3 c^4} \]
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Rule 45
Rule 266
Rule 272
Rule 4930
Rule 4946
Rule 5004
Rule 5036
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} (b c) \int \frac {x^4 (a+b \arctan (c x))}{1+c^2 x^2} \, dx \\ & = \frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {b \int x^2 (a+b \arctan (c x)) \, dx}{2 c}+\frac {b \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{2 c} \\ & = -\frac {b x^3 (a+b \arctan (c x))}{6 c}+\frac {1}{4} x^4 (a+b \arctan (c x))^2+\frac {1}{6} b^2 \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {b \int (a+b \arctan (c x)) \, dx}{2 c^3}-\frac {b \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{2 c^3} \\ & = \frac {a b x}{2 c^3}-\frac {b x^3 (a+b \arctan (c x))}{6 c}-\frac {(a+b \arctan (c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x))^2+\frac {1}{12} b^2 \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {b^2 \int \arctan (c x) \, dx}{2 c^3} \\ & = \frac {a b x}{2 c^3}+\frac {b^2 x \arctan (c x)}{2 c^3}-\frac {b x^3 (a+b \arctan (c x))}{6 c}-\frac {(a+b \arctan (c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x))^2+\frac {1}{12} b^2 \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac {b^2 \int \frac {x}{1+c^2 x^2} \, dx}{2 c^2} \\ & = \frac {a b x}{2 c^3}+\frac {b^2 x^2}{12 c^2}+\frac {b^2 x \arctan (c x)}{2 c^3}-\frac {b x^3 (a+b \arctan (c x))}{6 c}-\frac {(a+b \arctan (c x))^2}{4 c^4}+\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {b^2 \log \left (1+c^2 x^2\right )}{3 c^4} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\frac {c x \left (6 a b+b^2 c x-2 a b c^2 x^2+3 a^2 c^3 x^3\right )-2 b \left (b c x \left (-3+c^2 x^2\right )+a \left (3-3 c^4 x^4\right )\right ) \arctan (c x)+3 b^2 \left (-1+c^4 x^4\right ) \arctan (c x)^2-4 b^2 \log \left (1+c^2 x^2\right )}{12 c^4} \]
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Time = 1.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05
method | result | size |
parts | \(\frac {a^{2} x^{4}}{4}+\frac {b^{2} \left (\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {c x \arctan \left (c x \right )}{2}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}\right )}{c^{4}}+\frac {2 a b \left (\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {c^{3} x^{3}}{12}+\frac {c x}{4}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(118\) |
derivativedivides | \(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {c x \arctan \left (c x \right )}{2}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}\right )+2 a b \left (\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {c^{3} x^{3}}{12}+\frac {c x}{4}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(119\) |
default | \(\frac {\frac {a^{2} c^{4} x^{4}}{4}+b^{2} \left (\frac {c^{4} x^{4} \arctan \left (c x \right )^{2}}{4}-\frac {c^{3} x^{3} \arctan \left (c x \right )}{6}+\frac {c x \arctan \left (c x \right )}{2}-\frac {\arctan \left (c x \right )^{2}}{4}+\frac {c^{2} x^{2}}{12}-\frac {\ln \left (c^{2} x^{2}+1\right )}{3}\right )+2 a b \left (\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}-\frac {c^{3} x^{3}}{12}+\frac {c x}{4}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\) | \(119\) |
parallelrisch | \(-\frac {-3 x^{4} \arctan \left (c x \right )^{2} b^{2} c^{4}-6 x^{4} \arctan \left (c x \right ) a b \,c^{4}-3 a^{2} c^{4} x^{4}+2 b^{2} \arctan \left (c x \right ) x^{3} c^{3}+2 a b \,c^{3} x^{3}-b^{2} c^{2} x^{2}-6 b^{2} \arctan \left (c x \right ) x c -6 a b c x +3 b^{2} \arctan \left (c x \right )^{2}+4 b^{2} \ln \left (c^{2} x^{2}+1\right )+6 a b \arctan \left (c x \right )+b^{2}}{12 c^{4}}\) | \(139\) |
risch | \(-\frac {b^{2} \left (c^{4} x^{4}-1\right ) \ln \left (i c x +1\right )^{2}}{16 c^{4}}-\frac {i b \left (6 a \,c^{4} x^{4}+3 i b \,c^{4} x^{4} \ln \left (-i c x +1\right )-2 b \,c^{3} x^{3}+6 x b c -3 i b \ln \left (-i c x +1\right )\right ) \ln \left (i c x +1\right )}{24 c^{4}}-\frac {b^{2} x^{4} \ln \left (-i c x +1\right )^{2}}{16}+\frac {i a b \,x^{4} \ln \left (-i c x +1\right )}{4}-\frac {i b^{2} x^{3} \ln \left (-i c x +1\right )}{12 c}+\frac {a^{2} x^{4}}{4}-\frac {a b \,x^{3}}{6 c}+\frac {i b^{2} x \ln \left (-i c x +1\right )}{4 c^{3}}+\frac {b^{2} x^{2}}{12 c^{2}}+\frac {b^{2} \ln \left (-i c x +1\right )^{2}}{16 c^{4}}+\frac {a b x}{2 c^{3}}-\frac {a b \arctan \left (c x \right )}{2 c^{4}}-\frac {b^{2} \ln \left (c^{2} x^{2}+1\right )}{3 c^{4}}\) | \(254\) |
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Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\frac {3 \, a^{2} c^{4} x^{4} - 2 \, a b c^{3} x^{3} + b^{2} c^{2} x^{2} + 6 \, a b c x + 3 \, {\left (b^{2} c^{4} x^{4} - b^{2}\right )} \arctan \left (c x\right )^{2} - 4 \, b^{2} \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (3 \, a b c^{4} x^{4} - b^{2} c^{3} x^{3} + 3 \, b^{2} c x - 3 \, a b\right )} \arctan \left (c x\right )}{12 \, c^{4}} \]
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Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.38 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atan}{\left (c x \right )}}{2} - \frac {a b x^{3}}{6 c} + \frac {a b x}{2 c^{3}} - \frac {a b \operatorname {atan}{\left (c x \right )}}{2 c^{4}} + \frac {b^{2} x^{4} \operatorname {atan}^{2}{\left (c x \right )}}{4} - \frac {b^{2} x^{3} \operatorname {atan}{\left (c x \right )}}{6 c} + \frac {b^{2} x^{2}}{12 c^{2}} + \frac {b^{2} x \operatorname {atan}{\left (c x \right )}}{2 c^{3}} - \frac {b^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3 c^{4}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{4 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]
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Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \arctan \left (c x\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{6} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a b - \frac {1}{12} \, {\left (2 \, c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )} \arctan \left (c x\right ) - \frac {c^{2} x^{2} + 3 \, \arctan \left (c x\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )} b^{2} \]
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\[ \int x^3 (a+b \arctan (c x))^2 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3} \,d x } \]
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Time = 0.34 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.20 \[ \int x^3 (a+b \arctan (c x))^2 \, dx=\frac {3\,a^2\,c^4\,x^4-4\,b^2\,\ln \left (c^2\,x^2+1\right )-3\,b^2\,{\mathrm {atan}\left (c\,x\right )}^2+b^2\,c^2\,x^2-6\,a\,b\,\mathrm {atan}\left (c\,x\right )-2\,b^2\,c^3\,x^3\,\mathrm {atan}\left (c\,x\right )+6\,b^2\,c\,x\,\mathrm {atan}\left (c\,x\right )+3\,b^2\,c^4\,x^4\,{\mathrm {atan}\left (c\,x\right )}^2-2\,a\,b\,c^3\,x^3+6\,a\,b\,c\,x+6\,a\,b\,c^4\,x^4\,\mathrm {atan}\left (c\,x\right )}{12\,c^4} \]
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