Integrand size = 19, antiderivative size = 341 \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=-\frac {d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x}{a b^4 n (1+n) (1+2 n)}-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {(b c-a d)^3 (b c (1-n)-a d (1+3 n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 b^4 n} \]
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Time = 0.42 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {424, 542, 396, 251} \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=-\frac {x (b c-a d)^3 (b c (1-n)-a d (3 n+1)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 b^4 n}-\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{a b^3 n (n+1) (2 n+1)}-\frac {d x \left (-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a^2 b c d^2 \left (16 n^3+26 n^2+15 n+3\right )-a b^2 c^2 d \left (12 n^3+17 n^2+12 n+3\right )+b^3 c^3 \left (2 n^2+3 n+1\right )\right )}{a b^4 n (n+1) (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{a b^2 n (2 n+1)}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )} \]
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Rule 251
Rule 396
Rule 424
Rule 542
Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {\left (c+d x^n\right )^2 \left (c (a d-b c (1-n))+d (a d (1+3 n)-b (c+2 c n)) x^n\right )}{a+b x^n} \, dx}{a b n} \\ & = \frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {\left (c+d x^n\right ) \left (c \left (2 a b c d (1+2 n)-a^2 d^2 (1+3 n)-b^2 c^2 \left (1+n-2 n^2\right )\right )-d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x^n\right )}{a+b x^n} \, dx}{a b^2 n (1+2 n)} \\ & = -\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {c \left (3 a b^2 c^2 d \left (1+3 n+2 n^2\right )+a^3 d^3 \left (1+5 n+6 n^2\right )-a^2 b c d^2 \left (3+12 n+13 n^2\right )-b^3 c^3 \left (1+2 n-n^2-2 n^3\right )\right )-d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x^n}{a+b x^n} \, dx}{a b^3 n (1+n) (1+2 n)} \\ & = -\frac {d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x}{a b^4 n (1+n) (1+2 n)}-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {\left ((b c-a d)^3 (b c (1-n)-a d (1+3 n))\right ) \int \frac {1}{a+b x^n} \, dx}{a b^4 n} \\ & = -\frac {d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x}{a b^4 n (1+n) (1+2 n)}-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {(b c-a d)^3 (b c (1-n)-a d (1+3 n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b^4 n} \\ \end{align*}
Time = 6.39 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\frac {x \left (\frac {4 a b^3 c^3 d-6 a^2 b^2 c^2 d^2+4 a^3 b c d^3-a^4 d^4+b^4 c^4 (-1+n)}{a^2 n}+\frac {(-b c+a d)^3 (b c (-1+n)+a d (1+3 n))}{a^2 n}+\frac {2 b d^3 (2 b c-a d) x^n}{1+n}+\frac {b^2 d^4 x^{2 n}}{1+2 n}+\frac {(b c-a d)^4}{a n \left (a+b x^n\right )}+\frac {(b c-a d)^3 (b c (-1+n)+a d (1+3 n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 n}\right )}{b^4} \]
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\[\int \frac {\left (c +d \,x^{n}\right )^{4}}{\left (a +b \,x^{n}\right )^{2}}d x\]
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\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int \frac {\left (c + d x^{n}\right )^{4}}{\left (a + b x^{n}\right )^{2}}\, dx \]
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\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int \frac {{\left (c+d\,x^n\right )}^4}{{\left (a+b\,x^n\right )}^2} \,d x \]
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