Integrand size = 30, antiderivative size = 103 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=-\frac {a^2 \left (1+\frac {b x^n}{a}\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{b^2 n (1+2 p)}+\frac {a^2 \left (1+\frac {b x^n}{a}\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 b^2 n (1+p)} \]
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Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1370, 272, 45} \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\frac {a^2 \left (\frac {b x^n}{a}+1\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 b^2 n (p+1)}-\frac {a^2 \left (\frac {b x^n}{a}+1\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{b^2 n (2 p+1)} \]
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Rule 45
Rule 272
Rule 1370
Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^n}{a}\right )^{-2 p} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p\right ) \int x^{-1+2 n} \left (1+\frac {b x^n}{a}\right )^{2 p} \, dx \\ & = \frac {\left (\left (1+\frac {b x^n}{a}\right )^{-2 p} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p\right ) \text {Subst}\left (\int x \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^n\right )}{n} \\ & = \frac {\left (\left (1+\frac {b x^n}{a}\right )^{-2 p} \left (a^2+2 a b x^n+b^2 x^{2 n}\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^n\right )}{n} \\ & = -\frac {a^2 \left (1+\frac {b x^n}{a}\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{b^2 n (1+2 p)}+\frac {a^2 \left (1+\frac {b x^n}{a}\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 b^2 n (1+p)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.52 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\frac {\left (a+b x^n\right ) \left (\left (a+b x^n\right )^2\right )^p \left (-a+b (1+2 p) x^n\right )}{2 b^2 n (1+p) (1+2 p)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 14.74 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.10
method | result | size |
risch | \(-\frac {\left (-2 b^{2} p \,x^{2 n}-2 a p \,x^{n} b -b^{2} x^{2 n}+a^{2}\right ) \left (a +b \,x^{n}\right )^{2 p} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i \left (a +b \,x^{n}\right )^{2}\right ) \pi p {\left (-\operatorname {csgn}\left (i \left (a +b \,x^{n}\right )^{2}\right )+\operatorname {csgn}\left (i \left (a +b \,x^{n}\right )\right )\right )}^{2}}{2}}}{2 \left (1+2 p \right ) \left (1+p \right ) n \,b^{2}}\) | \(113\) |
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Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\frac {{\left (2 \, a b p x^{n} - a^{2} + {\left (2 \, b^{2} p + b^{2}\right )} x^{2 \, n}\right )} {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{p}}{2 \, {\left (2 \, b^{2} n p^{2} + 3 \, b^{2} n p + b^{2} n\right )}} \]
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Timed out. \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\text {Timed out} \]
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Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2 \, n} + 2 \, a b p x^{n} - a^{2}\right )} {\left (b x^{n} + a\right )}^{2 \, p}}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2} n} \]
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\[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\int { {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{p} x^{2 \, n - 1} \,d x } \]
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Timed out. \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\int x^{2\,n-1}\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^p \,d x \]
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