Integrand size = 20, antiderivative size = 55 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{4} a^2 c x^4+\frac {1}{6} a (2 b c+a d) x^6+\frac {1}{8} b (b c+2 a d) x^8+\frac {1}{10} b^2 d x^{10} \]
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Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 77} \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{4} a^2 c x^4+\frac {1}{8} b x^8 (2 a d+b c)+\frac {1}{6} a x^6 (a d+2 b c)+\frac {1}{10} b^2 d x^{10} \]
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Rule 77
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x (a+b x)^2 (c+d x) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (a^2 c x+a (2 b c+a d) x^2+b (b c+2 a d) x^3+b^2 d x^4\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{4} a^2 c x^4+\frac {1}{6} a (2 b c+a d) x^6+\frac {1}{8} b (b c+2 a d) x^8+\frac {1}{10} b^2 d x^{10} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{4} a^2 c x^4+\frac {1}{6} a (2 b c+a d) x^6+\frac {1}{8} b (b c+2 a d) x^8+\frac {1}{10} b^2 d x^{10} \]
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Time = 2.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {b^{2} d \,x^{10}}{10}+\frac {\left (2 a b d +b^{2} c \right ) x^{8}}{8}+\frac {\left (a^{2} d +2 a b c \right ) x^{6}}{6}+\frac {a^{2} c \,x^{4}}{4}\) | \(52\) |
norman | \(\frac {b^{2} d \,x^{10}}{10}+\left (\frac {1}{4} a b d +\frac {1}{8} b^{2} c \right ) x^{8}+\left (\frac {1}{6} a^{2} d +\frac {1}{3} a b c \right ) x^{6}+\frac {a^{2} c \,x^{4}}{4}\) | \(52\) |
gosper | \(\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{4} x^{8} a b d +\frac {1}{8} x^{8} b^{2} c +\frac {1}{6} x^{6} a^{2} d +\frac {1}{3} x^{6} a b c +\frac {1}{4} a^{2} c \,x^{4}\) | \(54\) |
risch | \(\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{4} x^{8} a b d +\frac {1}{8} x^{8} b^{2} c +\frac {1}{6} x^{6} a^{2} d +\frac {1}{3} x^{6} a b c +\frac {1}{4} a^{2} c \,x^{4}\) | \(54\) |
parallelrisch | \(\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{4} x^{8} a b d +\frac {1}{8} x^{8} b^{2} c +\frac {1}{6} x^{6} a^{2} d +\frac {1}{3} x^{6} a b c +\frac {1}{4} a^{2} c \,x^{4}\) | \(54\) |
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Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{10} \, b^{2} d x^{10} + \frac {1}{8} \, {\left (b^{2} c + 2 \, a b d\right )} x^{8} + \frac {1}{4} \, a^{2} c x^{4} + \frac {1}{6} \, {\left (2 \, a b c + a^{2} d\right )} x^{6} \]
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Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {a^{2} c x^{4}}{4} + \frac {b^{2} d x^{10}}{10} + x^{8} \left (\frac {a b d}{4} + \frac {b^{2} c}{8}\right ) + x^{6} \left (\frac {a^{2} d}{6} + \frac {a b c}{3}\right ) \]
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Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{10} \, b^{2} d x^{10} + \frac {1}{8} \, {\left (b^{2} c + 2 \, a b d\right )} x^{8} + \frac {1}{4} \, a^{2} c x^{4} + \frac {1}{6} \, {\left (2 \, a b c + a^{2} d\right )} x^{6} \]
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Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{10} \, b^{2} d x^{10} + \frac {1}{8} \, b^{2} c x^{8} + \frac {1}{4} \, a b d x^{8} + \frac {1}{3} \, a b c x^{6} + \frac {1}{6} \, a^{2} d x^{6} + \frac {1}{4} \, a^{2} c x^{4} \]
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Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=x^6\,\left (\frac {d\,a^2}{6}+\frac {b\,c\,a}{3}\right )+x^8\,\left (\frac {c\,b^2}{8}+\frac {a\,d\,b}{4}\right )+\frac {a^2\,c\,x^4}{4}+\frac {b^2\,d\,x^{10}}{10} \]
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