Integrand size = 10, antiderivative size = 106 \[ \int x^3 \arctan (a+b x) \, dx=\frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}-\frac {\left (1-6 a^2+a^4\right ) \arctan (a+b x)}{4 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {a \left (1-a^2\right ) \log \left (1+(a+b x)^2\right )}{2 b^4} \]
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Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5155, 4972, 716, 649, 209, 266} \[ \int x^3 \arctan (a+b x) \, dx=-\frac {a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}+\frac {\left (1-6 a^2\right ) x}{4 b^3}-\frac {\left (a^4-6 a^2+1\right ) \arctan (a+b x)}{4 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {(a+b x)^3}{12 b^4}+\frac {a (a+b x)^2}{2 b^4} \]
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Rule 209
Rule 266
Rule 649
Rule 716
Rule 4972
Rule 5155
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \arctan (x) \, dx,x,a+b x\right )}{b} \\ & = \frac {1}{4} x^4 \arctan (a+b x)-\frac {1}{4} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{1+x^2} \, dx,x,a+b x\right ) \\ & = \frac {1}{4} x^4 \arctan (a+b x)-\frac {1}{4} \text {Subst}\left (\int \left (-\frac {1-6 a^2}{b^4}-\frac {4 a x}{b^4}+\frac {x^2}{b^4}+\frac {1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{b^4 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right ) \\ & = \frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {\text {Subst}\left (\int \frac {1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{4 b^4} \\ & = \frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {\left (a \left (1-a^2\right )\right ) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b^4}-\frac {\left (1-6 a^2+a^4\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{4 b^4} \\ & = \frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}-\frac {\left (1-6 a^2+a^4\right ) \arctan (a+b x)}{4 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {a \left (1-a^2\right ) \log \left (1+(a+b x)^2\right )}{2 b^4} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int x^3 \arctan (a+b x) \, dx=\frac {6 \left (1-6 a^2\right ) b x+12 a (a+b x)^2-2 (a+b x)^3+6 b^4 x^4 \arctan (a+b x)+3 i (-i+a)^4 \log (i-a-b x)-3 i (i+a)^4 \log (i+a+b x)}{24 b^4} \]
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Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {3 \arctan \left (b x +a \right ) x^{4} b^{4}-b^{3} x^{3}+3 a \,b^{2} x^{2}-3 \arctan \left (b x +a \right ) a^{4}+6 a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-9 a^{2} b x +18 \arctan \left (b x +a \right ) a^{2}+15 a^{3}-6 a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )+3 b x -3 \arctan \left (b x +a \right )-9 a}{12 b^{4}}\) | \(131\) |
parts | \(\frac {x^{4} \arctan \left (b x +a \right )}{4}-\frac {b \left (\frac {\frac {1}{3} b^{2} x^{3}-a b \,x^{2}+3 a^{2} x -x}{b^{4}}+\frac {\frac {\left (-4 a^{3} b +4 a b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (-3 a^{4}-2 a^{2}+1-\frac {\left (-4 a^{3} b +4 a b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{4}}\right )}{4}\) | \(135\) |
derivativedivides | \(\frac {\frac {\arctan \left (b x +a \right ) a^{4}}{4}-\arctan \left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \arctan \left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\arctan \left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\frac {3 a^{2} \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2} a}{2}-\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}-\frac {\left (-4 a^{3}+4 a \right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{8}-\frac {\left (a^{4}-6 a^{2}+1\right ) \arctan \left (b x +a \right )}{4}}{b^{4}}\) | \(157\) |
default | \(\frac {\frac {\arctan \left (b x +a \right ) a^{4}}{4}-\arctan \left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \arctan \left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\arctan \left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\frac {3 a^{2} \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2} a}{2}-\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}-\frac {\left (-4 a^{3}+4 a \right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{8}-\frac {\left (a^{4}-6 a^{2}+1\right ) \arctan \left (b x +a \right )}{4}}{b^{4}}\) | \(157\) |
risch | \(-\frac {i x^{4} \ln \left (1+i \left (b x +a \right )\right )}{8}+\frac {i x^{4} \ln \left (1-i \left (b x +a \right )\right )}{8}-\frac {x^{3}}{12 b}-\frac {a^{4} \arctan \left (b x +a \right )}{4 b^{4}}+\frac {a \,x^{2}}{4 b^{2}}+\frac {a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}-\frac {3 a^{2} x}{4 b^{3}}+\frac {3 a^{2} \arctan \left (b x +a \right )}{2 b^{4}}-\frac {a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}+\frac {x}{4 b^{3}}-\frac {\arctan \left (b x +a \right )}{4 b^{4}}\) | \(157\) |
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Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int x^3 \arctan (a+b x) \, dx=-\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} b x - 3 \, {\left (b^{4} x^{4} - a^{4} + 6 \, a^{2} - 1\right )} \arctan \left (b x + a\right ) - 6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{12 \, b^{4}} \]
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Time = 0.69 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46 \[ \int x^3 \arctan (a+b x) \, dx=\begin {cases} - \frac {a^{4} \operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} + \frac {a^{3} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} - \frac {3 a^{2} x}{4 b^{3}} + \frac {3 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 b^{4}} + \frac {a x^{2}}{4 b^{2}} - \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} + \frac {x^{4} \operatorname {atan}{\left (a + b x \right )}}{4} - \frac {x^{3}}{12 b} + \frac {x}{4 b^{3}} - \frac {\operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atan}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]
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Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98 \[ \int x^3 \arctan (a+b x) \, dx=\frac {1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac {1}{12} \, b {\left (\frac {b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} x}{b^{4}} + \frac {3 \, {\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{5}} - \frac {6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}}\right )} \]
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\[ \int x^3 \arctan (a+b x) \, dx=\int { x^{3} \arctan \left (b x + a\right ) \,d x } \]
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Time = 0.69 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.25 \[ \int x^3 \arctan (a+b x) \, dx=\frac {x^4\,\mathrm {atan}\left (a+b\,x\right )}{4}-\frac {\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {x}{4\,b^3}-\frac {x^3}{12\,b}+\frac {a^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4}+\frac {3\,a^2\,\mathrm {atan}\left (a+b\,x\right )}{2\,b^4}-\frac {a^4\,\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {a\,x^2}{4\,b^2}-\frac {3\,a^2\,x}{4\,b^3}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4} \]
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