Integrand size = 24, antiderivative size = 84 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=-\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (1+2 p)}+\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (1+p)} \]
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Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1370, 272, 45} \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (p+1)}-\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (2 p+1)} \]
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Rule 45
Rule 272
Rule 1370
Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int x^5 \left (1+\frac {b x^3}{a}\right )^{2 p} \, dx \\ & = \frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \text {Subst}\left (\int x \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \text {Subst}\left (\int \left (-\frac {a \left (1+\frac {b x}{a}\right )^{2 p}}{b}+\frac {a \left (1+\frac {b x}{a}\right )^{1+2 p}}{b}\right ) \, dx,x,x^3\right ) \\ & = -\frac {a \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^2 (1+2 p)}+\frac {\left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{6 b^2 (1+p)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (-a+b (1+2 p) x^3\right )}{6 b^2 (1+p) (1+2 p)} \]
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Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {\left (-2 b^{2} p \,x^{6}-b^{2} x^{6}-2 a b p \,x^{3}+a^{2}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{6 b^{2} \left (1+p \right ) \left (1+2 p \right )}\) | \(58\) |
gosper | \(-\frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} \left (-2 x^{3} p b -b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )}{6 b^{2} \left (2 p^{2}+3 p +1\right )}\) | \(60\) |
norman | \(\frac {x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 p +6}-\frac {a^{2} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{6 b^{2} \left (2 p^{2}+3 p +1\right )}+\frac {p a \,x^{3} {\mathrm e}^{p \ln \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}}{3 b \left (2 p^{2}+3 p +1\right )}\) | \(120\) |
parallelrisch | \(\frac {2 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{2} p +x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} b^{2}+2 x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a b p -\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2}}{6 b^{2} \left (2 p^{2}+3 p +1\right )}\) | \(128\) |
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Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, b^{2} p + b^{2}\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
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\[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\begin {cases} \frac {x^{6} \left (a^{2}\right )^{p}}{6} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 a \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {a}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} + \frac {b x^{3} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{3 a b^{2} + 3 b^{3} x^{3}} - \frac {2 b x^{3} \log {\left (2 \right )}}{3 a b^{2} + 3 b^{3} x^{3}} & \text {for}\: p = -1 \\\int \frac {x^{5}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\- \frac {a^{2} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 a b p x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {2 b^{2} p x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} + \frac {b^{2} x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{2} p^{2} + 18 b^{2} p + 6 b^{2}} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{6} + 2 \, a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{6 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]
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Time = 0.32 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.57 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} p x^{6} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{2} x^{6} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b p x^{3} - {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2}}{6 \, {\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]
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Time = 8.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx={\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^6\,\left (2\,p+1\right )}{6\,\left (2\,p^2+3\,p+1\right )}-\frac {a^2}{6\,b^2\,\left (2\,p^2+3\,p+1\right )}+\frac {a\,p\,x^3}{3\,b\,\left (2\,p^2+3\,p+1\right )}\right ) \]
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