Integrand size = 19, antiderivative size = 402 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x \left (a+b x^n\right )^{1+p}}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}-\frac {d (a d (1+2 n)-b c (1+n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+3 n+n p)}-\frac {\left (a^3 d^3 \left (1+3 n+2 n^2\right )-3 a^2 b c d^2 (1+n) (1+n (3+p))+3 a b^2 c^2 d \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )-b^3 c^3 \left (1+3 n (2+p)+n^2 \left (11+12 p+3 p^2\right )+n^3 \left (6+11 p+6 p^2+p^3\right )\right )\right ) x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))} \]
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Time = 0.39 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {427, 542, 396, 252, 251} \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}-\frac {x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (a^3 d^3 \left (2 n^2+3 n+1\right )-3 a^2 b c d^2 (n+1) (n (p+3)+1)+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}-\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b^2 (n (p+2)+1) (n (p+3)+1)}+\frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)} \]
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Rule 251
Rule 252
Rule 396
Rule 427
Rule 542
Rubi steps \begin{align*} \text {integral}& = \frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}+\frac {\int \left (a+b x^n\right )^p \left (c+d x^n\right ) \left (-c (a d-b (c+c n (3+p)))-d (a d (1+2 n)-b (c+c n (5+p))) x^n\right ) \, dx}{b (1+n (3+p))} \\ & = -\frac {d (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}+\frac {\int \left (a+b x^n\right )^p \left (c \left (a^2 d^2 (1+2 n)-a b c d (2+n (7+2 p))+b^2 c^2 \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )\right )+d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x^n\right ) \, dx}{b^2 (1+n (2+p)) (1+n (3+p))} \\ & = \frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x \left (a+b x^n\right )^{1+p}}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}-\frac {d (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}-\frac {\left (a^3 d^3 \left (1+3 n+2 n^2\right )-3 a^2 b c d^2 (1+n) (1+n (3+p))+3 a b^2 c^2 d \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )-b^3 c^3 \left (1+3 n (2+p)+n^2 \left (11+12 p+3 p^2\right )+n^3 \left (6+11 p+6 p^2+p^3\right )\right )\right ) \int \left (a+b x^n\right )^p \, dx}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))} \\ & = \frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x \left (a+b x^n\right )^{1+p}}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}-\frac {d (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}-\frac {\left (\left (a^3 d^3 \left (1+3 n+2 n^2\right )-3 a^2 b c d^2 (1+n) (1+n (3+p))+3 a b^2 c^2 d \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )-b^3 c^3 \left (1+3 n (2+p)+n^2 \left (11+12 p+3 p^2\right )+n^3 \left (6+11 p+6 p^2+p^3\right )\right )\right ) \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^n}{a}\right )^p \, dx}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))} \\ & = \frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x \left (a+b x^n\right )^{1+p}}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}-\frac {d (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}-\frac {\left (a^3 d^3 \left (1+3 n+2 n^2\right )-3 a^2 b c d^2 (1+n) (1+n (3+p))+3 a b^2 c^2 d \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )-b^3 c^3 \left (1+3 n (2+p)+n^2 \left (11+12 p+3 p^2\right )+n^3 \left (6+11 p+6 p^2+p^3\right )\right )\right ) x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b x^n}{a}\right )}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))} \\ \end{align*}
Time = 5.31 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.42 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (\frac {3 c^2 d x^n \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},-p,2+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+n}+\frac {3 c d^2 x^{2 n} \operatorname {Hypergeometric2F1}\left (2+\frac {1}{n},-p,3+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+2 n}+\frac {d^3 x^{3 n} \operatorname {Hypergeometric2F1}\left (3+\frac {1}{n},-p,4+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+3 n}+c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )\right ) \]
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\[\int \left (a +b \,x^{n}\right )^{p} \left (c +d \,x^{n}\right )^{3}d x\]
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\[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\int { {\left (d x^{n} + c\right )}^{3} {\left (b x^{n} + a\right )}^{p} \,d x } \]
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Result contains complex when optimal does not.
Time = 42.88 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.60 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\frac {a^{\frac {1}{n}} a^{p - \frac {1}{n}} c^{3} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{n}, - p \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 a^{1 + \frac {1}{n}} a^{p - 1 - \frac {1}{n}} c^{2} d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} + \frac {3 a^{2 + \frac {1}{n}} a^{p - 2 - \frac {1}{n}} c d^{2} x^{2 n + 1} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {a^{3 + \frac {1}{n}} a^{p - 3 - \frac {1}{n}} d^{3} x^{3 n + 1} \Gamma \left (3 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 3 + \frac {1}{n} \\ 4 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (4 + \frac {1}{n}\right )} \]
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\[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\int { {\left (d x^{n} + c\right )}^{3} {\left (b x^{n} + a\right )}^{p} \,d x } \]
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Exception generated. \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\int {\left (a+b\,x^n\right )}^p\,{\left (c+d\,x^n\right )}^3 \,d x \]
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