Integrand size = 24, antiderivative size = 174 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=-\frac {a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (1+2 p)}+\frac {3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (1+p)}-\frac {3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (3+2 p)}+\frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (2+p)} \]
[Out]
Time = 0.08 (sec) , antiderivative size = 174, 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 = {1127, 272, 45} \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+2)}-\frac {3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+3)}+\frac {3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+1)}-\frac {a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+1)} \]
[In]
[Out]
Rule 45
Rule 272
Rule 1127
Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int x^7 \left (1+\frac {b x^2}{a}\right )^{2 p} \, dx \\ & = \frac {1}{2} \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \text {Subst}\left (\int x^3 \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \text {Subst}\left (\int \left (-\frac {a^3 \left (1+\frac {b x}{a}\right )^{2 p}}{b^3}+\frac {3 a^3 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^3}-\frac {3 a^3 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^3}+\frac {a^3 \left (1+\frac {b x}{a}\right )^{3+2 p}}{b^3}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (1+2 p)}+\frac {3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (1+p)}-\frac {3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (3+2 p)}+\frac {\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (2+p)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.63 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (-3 a^3+3 a^2 b (1+2 p) x^2-3 a b^2 \left (1+3 p+2 p^2\right ) x^4+b^3 \left (3+11 p+12 p^2+4 p^3\right ) x^6\right )}{4 b^4 (1+p) (2+p) (1+2 p) (3+2 p)} \]
[In]
[Out]
Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.86
method | result | size |
gosper | \(-\frac {\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} \left (-4 b^{3} p^{3} x^{6}-12 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+6 a \,b^{2} p^{2} x^{4}-3 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+3 b^{2} x^{4} a -6 a^{2} b p \,x^{2}-3 a^{2} b \,x^{2}+3 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\) | \(150\) |
risch | \(-\frac {\left (-4 b^{4} p^{3} x^{8}-12 b^{4} p^{2} x^{8}-4 a \,b^{3} p^{3} x^{6}-11 b^{4} p \,x^{8}-6 a \,b^{3} p^{2} x^{6}-3 b^{4} x^{8}-2 a p \,x^{6} b^{3}+6 a^{2} b^{2} p^{2} x^{4}+3 a^{2} p \,x^{4} b^{2}-6 a^{3} p \,x^{2} b +3 a^{4}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{p}}{4 \left (3+2 p \right ) \left (2+p \right ) \left (1+p \right ) \left (1+2 p \right ) b^{4}}\) | \(156\) |
norman | \(\frac {x^{8} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 p +8}-\frac {3 a^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{4} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}+\frac {a p \,x^{6} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b \left (2 p^{2}+7 p +6\right )}-\frac {3 a^{2} p \,x^{4} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{4 b^{2} \left (2 p^{3}+9 p^{2}+13 p +6\right )}+\frac {3 p \,a^{3} x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}}{2 b^{3} \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right )}\) | \(237\) |
parallelrisch | \(\frac {4 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p^{3}+12 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p^{2}+11 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4} p +4 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p^{3}+3 x^{8} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a \,b^{4}+6 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p^{2}+2 x^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{2} b^{3} p -6 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3} b^{2} p^{2}-3 x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{3} b^{2} p +6 x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p} a^{4} b p -3 a^{5} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (2+p \right ) \left (2 p^{2}+5 p +3\right ) \left (1+2 p \right ) b^{4} a}\) | \(378\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{8} + 6 \, a^{3} b p x^{2} + 2 \, {\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{6} - 3 \, {\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} - 3 \, a^{4}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
[In]
[Out]
\[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\text {Too large to display} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.66 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{8} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{6} - 3 \, {\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 3 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{4 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (166) = 332\).
Time = 0.31 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.16 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{3} x^{8} + 12 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{2} x^{8} + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{3} x^{6} + 11 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p x^{8} + 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{2} x^{6} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} x^{8} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p x^{6} - 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{4} - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p x^{4} + 6 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3} b p x^{2} - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{4}}{4 \, {\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \]
[In]
[Out]
Time = 13.08 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx={\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^8\,\left (p^3+3\,p^2+\frac {11\,p}{4}+\frac {3}{4}\right )}{4\,p^4+20\,p^3+35\,p^2+25\,p+6}-\frac {3\,a^4}{4\,b^4\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {3\,a^3\,p\,x^2}{2\,b^3\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a\,p\,x^6\,\left (2\,p^2+3\,p+1\right )}{2\,b\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {3\,a^2\,p\,x^4\,\left (2\,p+1\right )}{4\,b^2\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}\right ) \]
[In]
[Out]