Integrand size = 27, antiderivative size = 29 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{2 (1+p)} \left (b x+c x^4\right )^{1+p}}{3 (1+p)} \]
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Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1604} \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{2 (p+1)} \left (b x+c x^4\right )^{p+1}}{3 (p+1)} \]
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Rule 1604
Rubi steps \begin{align*} \text {integral}& = \frac {x^{2 (1+p)} \left (b x+c x^4\right )^{1+p}}{3 (1+p)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.41 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {x^{3+2 p} \left (x \left (b+c x^3\right )\right )^p \left (1+\frac {c x^3}{b}\right )^{-p} \left (b (2+p) \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,-\frac {c x^3}{b}\right )+2 c (1+p) x^3 \operatorname {Hypergeometric2F1}\left (-p,2+p,3+p,-\frac {c x^3}{b}\right )\right )}{3 (1+p) (2+p)} \]
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Time = 2.51 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
gosper | \(\frac {x^{3+2 p} \left (c \,x^{3}+b \right ) \left (c \,x^{4}+b x \right )^{p}}{3+3 p}\) | \(33\) |
parallelrisch | \(\frac {x^{4} x^{2+2 p} {\left (x \left (c \,x^{3}+b \right )\right )}^{p} c^{2}+x \,x^{2+2 p} {\left (x \left (c \,x^{3}+b \right )\right )}^{p} b c}{3 c \left (1+p \right )}\) | \(59\) |
risch | \(\frac {\left (c \,x^{3}+b \right ) x \,x^{2+2 p} \left (c \,x^{3}+b \right )^{p} x^{p} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right ) p \left (-\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right )+\operatorname {csgn}\left (i \left (c \,x^{3}+b \right )\right )\right ) \left (-\operatorname {csgn}\left (i x \left (c \,x^{3}+b \right )\right )+\operatorname {csgn}\left (i x \right )\right )}{2}}}{3+3 p}\) | \(99\) |
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none
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {{\left (c x^{4} + b x\right )} {\left (c x^{4} + b x\right )}^{p} x^{2 \, p + 2}}{3 \, {\left (p + 1\right )}} \]
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Timed out. \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\text {Timed out} \]
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none
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx=\frac {c x^{4} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )} + b x e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \]
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Time = 9.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int x^{2 (1+p)} \left (b+2 c x^3\right ) \left (b x+c x^4\right )^p \, dx={\left (c\,x^4+b\,x\right )}^p\,\left (\frac {c\,x^{2\,p+2}\,x^4}{3\,p+3}+\frac {b\,x\,x^{2\,p+2}}{3\,p+3}\right ) \]
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