Integrand size = 76, antiderivative size = 96 \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=-\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^{1+p} \left (a+b x^{n/2}\right )^{1+p} \left (-\frac {a^2 d n (1+p)}{b^2 (1+n+n p)}+d x^n\right )^{-\frac {1+n+n p}{n}}}{a^4 d n (1+p)} \]
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Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {533, 389} \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=-\frac {b^2 x (n p+n+1) \left (a^2-b^2 x^n\right ) \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (d x^n-\frac {a^2 d n (p+1)}{b^2 (n p+n+1)}\right )^{-\frac {n p+n+1}{n}}}{a^4 d n (p+1)} \]
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Rule 389
Rule 533
Rubi steps \begin{align*} \text {integral}& = \left (\left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (a^2-b^2 x^n\right )^{-p}\right ) \int \left (a^2-b^2 x^n\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx \\ & = -\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (a^2-b^2 x^n\right ) \left (-\frac {a^2 d n (1+p)}{b^2 (1+n+n p)}+d x^n\right )^{-\frac {1+n+n p}{n}}}{a^4 d n (1+p)} \\ \end{align*}
Time = 1.42 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07 \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=-\frac {b^2 (1+n+n p) x \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (d \left (-\frac {a^2 n (1+p)}{b^2 (1+n+n p)}+x^n\right )\right )^{-\frac {1+n+n p}{n}} \left (a^2-b^2 x^n\right )}{a^4 d n (1+p)} \]
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\[\int \left (a -b \,x^{\frac {n}{2}}\right )^{p} \left (a +b \,x^{\frac {n}{2}}\right )^{p} \left (\frac {a^{2} d \left (1+p \right )}{b^{2} \left (1+\frac {-n p -2 n -1}{n}\right )}+d \,x^{n}\right )^{\frac {-n p -2 n -1}{n}}d x\]
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none
Time = 0.30 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.88 \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=\frac {{\left ({\left (b^{4} n p + b^{4} n + b^{4}\right )} x x^{2 \, n} - {\left (2 \, a^{2} b^{2} n p + 2 \, a^{2} b^{2} n + a^{2} b^{2}\right )} x x^{n} + {\left (a^{4} n p + a^{4} n\right )} x\right )} {\left (b x^{\frac {1}{2} \, n} + a\right )}^{p} {\left (-b x^{\frac {1}{2} \, n} + a\right )}^{p}}{{\left (a^{4} n p + a^{4} n\right )} \left (-\frac {a^{2} d n p + a^{2} d n - {\left (b^{2} d n p + b^{2} d n + b^{2} d\right )} x^{n}}{b^{2} n p + b^{2} n + b^{2}}\right )^{\frac {n p + 2 \, n + 1}{n}}} \]
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Timed out. \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=\text {Timed out} \]
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\[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=\int { \frac {{\left (b x^{\frac {1}{2} \, n} + a\right )}^{p} {\left (-b x^{\frac {1}{2} \, n} + a\right )}^{p}}{{\left (d x^{n} - \frac {a^{2} d {\left (p + 1\right )}}{b^{2} {\left (\frac {n p + 2 \, n + 1}{n} - 1\right )}}\right )}^{\frac {n p + 2 \, n + 1}{n}}} \,d x } \]
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\[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=\int { \frac {{\left (b x^{\frac {1}{2} \, n} + a\right )}^{p} {\left (-b x^{\frac {1}{2} \, n} + a\right )}^{p}}{{\left (d x^{n} - \frac {a^{2} d {\left (p + 1\right )}}{b^{2} {\left (\frac {n p + 2 \, n + 1}{n} - 1\right )}}\right )}^{\frac {n p + 2 \, n + 1}{n}}} \,d x } \]
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Timed out. \[ \int \left (a-b x^{n/2}\right )^p \left (a+b x^{n/2}\right )^p \left (\frac {a^2 d (1+p)}{b^2 \left (1+\frac {-1-2 n-n p}{n}\right )}+d x^n\right )^{\frac {-1-2 n-n p}{n}} \, dx=\int \frac {{\left (a+b\,x^{n/2}\right )}^p\,{\left (a-b\,x^{n/2}\right )}^p}{{\left (d\,x^n-\frac {a^2\,d\,\left (p+1\right )}{b^2\,\left (\frac {2\,n+n\,p+1}{n}-1\right )}\right )}^{\frac {2\,n+n\,p+1}{n}}} \,d x \]
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