Integrand size = 13, antiderivative size = 43 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^9}{9}+\frac {1}{4} a^2 b x^{12}+\frac {1}{5} a b^2 x^{15}+\frac {b^3 x^{18}}{18} \]
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Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^9}{9}+\frac {1}{4} a^2 b x^{12}+\frac {1}{5} a b^2 x^{15}+\frac {b^3 x^{18}}{18} \]
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Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x^2 (a+b x)^3 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (a^3 x^2+3 a^2 b x^3+3 a b^2 x^4+b^3 x^5\right ) \, dx,x,x^3\right ) \\ & = \frac {a^3 x^9}{9}+\frac {1}{4} a^2 b x^{12}+\frac {1}{5} a b^2 x^{15}+\frac {b^3 x^{18}}{18} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^9}{9}+\frac {1}{4} a^2 b x^{12}+\frac {1}{5} a b^2 x^{15}+\frac {b^3 x^{18}}{18} \]
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Time = 3.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84
method | result | size |
gosper | \(\frac {1}{9} a^{3} x^{9}+\frac {1}{4} a^{2} b \,x^{12}+\frac {1}{5} a \,b^{2} x^{15}+\frac {1}{18} b^{3} x^{18}\) | \(36\) |
default | \(\frac {1}{9} a^{3} x^{9}+\frac {1}{4} a^{2} b \,x^{12}+\frac {1}{5} a \,b^{2} x^{15}+\frac {1}{18} b^{3} x^{18}\) | \(36\) |
norman | \(\frac {1}{9} a^{3} x^{9}+\frac {1}{4} a^{2} b \,x^{12}+\frac {1}{5} a \,b^{2} x^{15}+\frac {1}{18} b^{3} x^{18}\) | \(36\) |
risch | \(\frac {1}{9} a^{3} x^{9}+\frac {1}{4} a^{2} b \,x^{12}+\frac {1}{5} a \,b^{2} x^{15}+\frac {1}{18} b^{3} x^{18}\) | \(36\) |
parallelrisch | \(\frac {1}{9} a^{3} x^{9}+\frac {1}{4} a^{2} b \,x^{12}+\frac {1}{5} a \,b^{2} x^{15}+\frac {1}{18} b^{3} x^{18}\) | \(36\) |
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Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {1}{18} \, b^{3} x^{18} + \frac {1}{5} \, a b^{2} x^{15} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{9} \, a^{3} x^{9} \]
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Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {a^{3} x^{9}}{9} + \frac {a^{2} b x^{12}}{4} + \frac {a b^{2} x^{15}}{5} + \frac {b^{3} x^{18}}{18} \]
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Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {1}{18} \, b^{3} x^{18} + \frac {1}{5} \, a b^{2} x^{15} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{9} \, a^{3} x^{9} \]
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Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {1}{18} \, b^{3} x^{18} + \frac {1}{5} \, a b^{2} x^{15} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{9} \, a^{3} x^{9} \]
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Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^8 \left (a+b x^3\right )^3 \, dx=\frac {a^3\,x^9}{9}+\frac {a^2\,b\,x^{12}}{4}+\frac {a\,b^2\,x^{15}}{5}+\frac {b^3\,x^{18}}{18} \]
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