Integrand size = 24, antiderivative size = 130 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+2 p)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+p)}+\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (3+2 p)} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 130, 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} \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+3)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (p+1)}+\frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (2 p+1)} \]
[In]
[Out]
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^8 \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^2 \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^2 \left (1+\frac {b x}{a}\right )^{2 p}}{b^2}-\frac {2 a^2 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^2}+\frac {a^2 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^2}\right ) \, dx,x,x^3\right ) \\ & = \frac {a^2 \left (a+b x^3\right ) \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+2 p)}-\frac {a \left (a+b x^3\right )^2 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (1+p)}+\frac {\left (a+b x^3\right )^3 \left (a^2+2 a b x^3+b^2 x^6\right )^p}{3 b^3 (3+2 p)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.59 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {\left (a+b x^3\right ) \left (\left (a+b x^3\right )^2\right )^p \left (a^2-a b (1+2 p) x^3+b^2 \left (1+3 p+2 p^2\right ) x^6\right )}{3 b^3 (1+p) (1+2 p) (3+2 p)} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74
method | result | size |
gosper | \(\frac {\left (b \,x^{3}+a \right ) \left (2 b^{2} p^{2} x^{6}+3 b^{2} p \,x^{6}+b^{2} x^{6}-2 a b p \,x^{3}-a b \,x^{3}+a^{2}\right ) \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{3 b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\) | \(96\) |
risch | \(\frac {\left (2 b^{3} p^{2} x^{9}+3 b^{3} p \,x^{9}+b^{3} x^{9}+2 a \,b^{2} p^{2} x^{6}+a \,b^{2} p \,x^{6}-2 a^{2} b p \,x^{3}+a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{p}}{3 \left (1+p \right ) \left (3+2 p \right ) \left (1+2 p \right ) b^{3}}\) | \(98\) |
parallelrisch | \(\frac {2 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p^{2}+3 x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3} p +x^{9} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a \,b^{3}+2 x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p^{2}+x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{2} b^{2} p -2 x^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{3} b p +\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p} a^{4}}{3 \left (3+2 p \right ) \left (1+p \right ) \left (1+2 p \right ) a \,b^{3}}\) | \(238\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.83 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x^{9} - 2 \, a^{2} b p x^{3} + {\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{6} + a^{3}\right )} {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{3 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]
[In]
[Out]
\[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\begin {cases} \frac {x^{9} \left (a^{2}\right )^{p}}{9} & \text {for}\: b = 0 \\\int \frac {x^{8}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {2 a^{2} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a^{2} \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^{3} + 3 b^{4} x^{3}} - \frac {2 a^{2}}{3 a b^{3} + 3 b^{4} x^{3}} + \frac {4 a^{2} \log {\left (2 \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a b x^{3} \log {\left (x - \sqrt [3]{- \frac {a}{b}} \right )}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a 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^{3} + 3 b^{4} x^{3}} + \frac {4 a b x^{3} \log {\left (2 \right )}}{3 a b^{3} + 3 b^{4} x^{3}} + \frac {b^{2} x^{6}}{3 a b^{3} + 3 b^{4} x^{3}} & \text {for}\: p = -1 \\\int \frac {x^{8}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} - \frac {2 a^{2} b p x^{3} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {2 a b^{2} p^{2} x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {a b^{2} p x^{6} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {2 b^{3} p^{2} x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {3 b^{3} p x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} + \frac {b^{3} x^{9} \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{p}}{12 b^{3} p^{3} + 36 b^{3} p^{2} + 33 b^{3} p + 9 b^{3}} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.61 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{9} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + a^{3}\right )} {\left (b x^{3} + a\right )}^{2 \, p}}{3 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.81 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} p^{2} x^{9} + 3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} p x^{9} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} b^{3} x^{9} + 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{2} p^{2} x^{6} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a b^{2} p x^{6} - 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{2} b p x^{3} + {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p} a^{3}}{3 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]
[In]
[Out]
Time = 8.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^p \, dx={\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p\,\left (\frac {x^9\,\left (\frac {2\,p^2}{3}+p+\frac {1}{3}\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {a^3}{3\,b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}-\frac {2\,a^2\,p\,x^3}{3\,b^2\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,p\,x^6\,\left (2\,p+1\right )}{3\,b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \]
[In]
[Out]