Integrand size = 49, antiderivative size = 23 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx=x^2 \left (a+b x+c x^2+d x^3\right )^{1+p} \]
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Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {1602} \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx=x^2 \left (a+b x+c x^2+d x^3\right )^{p+1} \]
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Rule 1602
Rubi steps \begin{align*} \text {integral}& = x^2 \left (a+b x+c x^2+d x^3\right )^{1+p} \\ \end{align*}
Time = 1.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx=x^2 (a+x (b+x (c+d x)))^{1+p} \]
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{1+p}\) | \(24\) |
risch | \(\left (x^{3} d +c \,x^{2}+b x +a \right )^{p} x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )\) | \(37\) |
norman | \(a \,x^{2} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+b \,x^{3} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+c \,x^{4} {\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}+x^{5} d \,{\mathrm e}^{p \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}\) | \(98\) |
parallelrisch | \(\frac {x^{5} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a d +x^{4} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} a c +a b \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} x^{3}+a^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{p} x^{2}}{a}\) | \(99\) |
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none
Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx={\left (d x^{5} + c x^{4} + b x^{3} + a x^{2}\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \]
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Timed out. \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx=\text {Timed out} \]
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none
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx={\left (d x^{5} + c x^{4} + b x^{3} + a x^{2}\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} \]
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Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (23) = 46\).
Time = 0.39 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.87 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx={\left (d x^{3} + c x^{2} + b x + a\right )}^{p} d x^{5} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} c x^{4} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} b x^{3} + {\left (d x^{3} + c x^{2} + b x + a\right )}^{p} a x^{2} \]
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Time = 9.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int x \left (a+b x+c x^2+d x^3\right )^p \left (2 a+b (3+p) x+c (4+2 p) x^2+d (5+3 p) x^3\right ) \, dx={\left (d\,x^3+c\,x^2+b\,x+a\right )}^p\,\left (d\,x^5+c\,x^4+b\,x^3+a\,x^2\right ) \]
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