Integrand size = 19, antiderivative size = 160 \[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=-\frac {(b c-a d) x \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2}+\frac {(b c-a d) (a d (1-2 n)-b c (1+n)) x}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac {\left (2 a b c d (1-n)-b^2 c^2 (1+n)-a^2 d^2 \left (1-3 n+2 n^2\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{2 c^3 d^2 n^2} \]
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Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {424, 393, 251} \[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=-\frac {x \left (-a^2 d^2 \left (2 n^2-3 n+1\right )+2 a b c d (1-n)-b^2 c^2 (n+1)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{2 c^3 d^2 n^2}+\frac {x (b c-a d) (a d (1-2 n)-b c (n+1))}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac {x (b c-a d) \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2} \]
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Rule 251
Rule 393
Rule 424
Rubi steps \begin{align*} \text {integral}& = -\frac {(b c-a d) x \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2}+\frac {\int \frac {a (b c-a d (1-2 n))-b (a d (1-n)-b c (1+n)) x^n}{\left (c+d x^n\right )^2} \, dx}{2 c d n} \\ & = -\frac {(b c-a d) x \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2}+\frac {(b c-a d) (a d (1-2 n)-b c (1+n)) x}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac {\left (2 a b c d (1-n)-b^2 c^2 (1+n)-a^2 d^2 \left (1-3 n+2 n^2\right )\right ) \int \frac {1}{c+d x^n} \, dx}{2 c^2 d^2 n^2} \\ & = -\frac {(b c-a d) x \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2}+\frac {(b c-a d) (a d (1-2 n)-b c (1+n)) x}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac {\left (2 a b c d (1-n)-b^2 c^2 (1+n)-a^2 d^2 \left (1-3 n+2 n^2\right )\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{2 c^3 d^2 n^2} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\frac {x \left (\frac {c^2 (b c-a d)^2 n}{\left (c+d x^n\right )^2}-\frac {c (b c-a d) (a d (-1+2 n)+b (c+2 c n))}{c+d x^n}+\left (2 a b c d (-1+n)+b^2 c^2 (1+n)+a^2 d^2 \left (1-3 n+2 n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )\right )}{2 c^3 d^2 n^2} \]
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\[\int \frac {\left (a +b \,x^{n}\right )^{2}}{\left (c +d \,x^{n}\right )^{3}}d x\]
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\[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{2}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{2}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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\[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{2}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^n\right )^2}{\left (c+d x^n\right )^3} \, dx=\int \frac {{\left (a+b\,x^n\right )}^2}{{\left (c+d\,x^n\right )}^3} \,d x \]
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