Integrand size = 13, antiderivative size = 74 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {a^2 \left (a+b x^3\right )^{1+p}}{3 b^3 (1+p)}-\frac {2 a \left (a+b x^3\right )^{2+p}}{3 b^3 (2+p)}+\frac {\left (a+b x^3\right )^{3+p}}{3 b^3 (3+p)} \]
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Time = 0.04 (sec) , antiderivative size = 74, 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} \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {a^2 \left (a+b x^3\right )^{p+1}}{3 b^3 (p+1)}-\frac {2 a \left (a+b x^3\right )^{p+2}}{3 b^3 (p+2)}+\frac {\left (a+b x^3\right )^{p+3}}{3 b^3 (p+3)} \]
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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)^p \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{1+p}}{b^2}+\frac {(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^3\right ) \\ & = \frac {a^2 \left (a+b x^3\right )^{1+p}}{3 b^3 (1+p)}-\frac {2 a \left (a+b x^3\right )^{2+p}}{3 b^3 (2+p)}+\frac {\left (a+b x^3\right )^{3+p}}{3 b^3 (3+p)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {\left (a+b x^3\right )^{1+p} \left (2 a^2-2 a b (1+p) x^3+b^2 \left (2+3 p+p^2\right ) x^6\right )}{3 b^3 (1+p) (2+p) (3+p)} \]
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Time = 3.94 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08
method | result | size |
gosper | \(\frac {\left (b \,x^{3}+a \right )^{1+p} \left (b^{2} p^{2} x^{6}+3 b^{2} x^{6} p +2 b^{2} x^{6}-2 a p \,x^{3} b -2 a b \,x^{3}+2 a^{2}\right )}{3 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(80\) |
risch | \(\frac {\left (b^{3} p^{2} x^{9}+3 b^{3} p \,x^{9}+2 b^{3} x^{9}+a \,b^{2} p^{2} x^{6}+a p \,x^{6} b^{2}-2 a^{2} p \,x^{3} b +2 a^{3}\right ) \left (b \,x^{3}+a \right )^{p}}{3 \left (2+p \right ) \left (3+p \right ) \left (1+p \right ) b^{3}}\) | \(93\) |
parallelrisch | \(\frac {x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3} p^{2}+3 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3} p +2 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{3}+x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{2} p^{2}+x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{2} p -2 x^{3} \left (b \,x^{3}+a \right )^{p} a^{3} b p +2 a^{4} \left (b \,x^{3}+a \right )^{p}}{3 \left (3+p \right ) \left (2+p \right ) a \left (1+p \right ) b^{3}}\) | \(157\) |
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Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {{\left ({\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{9} - 2 \, a^{2} b p x^{3} + {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{6} + 2 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1187 vs. \(2 (61) = 122\).
Time = 7.73 (sec) , antiderivative size = 1187, normalized size of antiderivative = 16.04 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\text {Too large to display} \]
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Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{9} + {\left (p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + 2 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78 \[ \int x^8 \left (a+b x^3\right )^p \, dx=\frac {{\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} p - 2 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a p + 2 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} - 6 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a}{3 \, {\left (b^{3} p^{2} + 5 \, b^{3} p + 6 \, b^{3}\right )}} + \frac {{\left (b x^{3} + a\right )}^{p + 1} a^{2}}{3 \, b^{3} {\left (p + 1\right )}} \]
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Time = 5.68 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^8 \left (a+b x^3\right )^p \, dx={\left (b\,x^3+a\right )}^p\,\left (\frac {2\,a^3}{3\,b^3\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {x^9\,\left (p^2+3\,p+2\right )}{3\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {2\,a^2\,p\,x^3}{3\,b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^6\,\left (p+1\right )}{3\,b\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \]
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