Optimal. Leaf size=366 \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{2 c^2 e n^2 (b c-a d)^2 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2} \]
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Rubi [A] time = 1.22, antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {595, 597, 364} \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{2 c^2 e n^2 (b c-a d)^2 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2} \]
Antiderivative was successfully verified.
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Rule 364
Rule 595
Rule 597
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx &=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {\int \frac {(e x)^m \left (-a B c (1+m)+a A d (1+m-2 n)+2 A b c n-b (B c-A d) (1+m-2 n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{2 c (b c-a d) n}\\ &=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\int \frac {(e x)^m \left (-a (1+m) (b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n)))-(b c-a d) n (a B c (1+m)-a A d (1+m-2 n)-2 A b c n)-b (b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (1+m-n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 c^2 (b c-a d)^2 n^2}\\ &=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\int \left (\frac {2 b^2 (A b-a B) c^2 n^2 (e x)^m}{(b c-a d) \left (a+b x^n\right )}+\frac {\left (-b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)+a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)-2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^n\right )}\right ) \, dx}{2 c^2 (b c-a d)^2 n^2}\\ &=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {\left (b^2 (A b-a B)\right ) \int \frac {(e x)^m}{a+b x^n} \, dx}{(b c-a d)^3}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{2 c^2 (b c-a d)^3 n^2}\\ &=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a (b c-a d)^3 e (1+m)}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^3 e (1+m) n^2}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 201, normalized size = 0.55 \[ \frac {x (e x)^m \left (\frac {b^2 (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a}+\frac {(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c^3}-\frac {d (A b-a B) (b c-a d) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c^2}-\frac {b d (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c}\right )}{(m+1) (b c-a d)^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{b d^{3} x^{4 \, n} + a c^{3} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{3 \, n} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} x^{2 \, n} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{n}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )} {\left (d x^{n} + c\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{n}+A \right ) \left (e x \right )^{m}}{\left (b \,x^{n}+a \right ) \left (d \,x^{n}+c \right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ {\left ({\left ({\left (m^{2} - m {\left (5 \, n - 2\right )} + 6 \, n^{2} - 5 \, n + 1\right )} b^{2} c^{2} d e^{m} - 2 \, {\left (m^{2} - 2 \, m {\left (2 \, n - 1\right )} + 3 \, n^{2} - 4 \, n + 1\right )} a b c d^{2} e^{m} + {\left (m^{2} - m {\left (3 \, n - 2\right )} + 2 \, n^{2} - 3 \, n + 1\right )} a^{2} d^{3} e^{m}\right )} A - {\left ({\left (m^{2} - m {\left (3 \, n - 2\right )} + 2 \, n^{2} - 3 \, n + 1\right )} b^{2} c^{3} e^{m} - 2 \, {\left (m^{2} - 2 \, m {\left (n - 1\right )} - 2 \, n + 1\right )} a b c^{2} d e^{m} + {\left (m^{2} - m {\left (n - 2\right )} - n + 1\right )} a^{2} c d^{2} e^{m}\right )} B\right )} \int -\frac {x^{m}}{2 \, {\left (b^{3} c^{6} n^{2} - 3 \, a b^{2} c^{5} d n^{2} + 3 \, a^{2} b c^{4} d^{2} n^{2} - a^{3} c^{3} d^{3} n^{2} + {\left (b^{3} c^{5} d n^{2} - 3 \, a b^{2} c^{4} d^{2} n^{2} + 3 \, a^{2} b c^{3} d^{3} n^{2} - a^{3} c^{2} d^{4} n^{2}\right )} x^{n}\right )}}\,{d x} + {\left (B a b^{2} e^{m} - A b^{3} e^{m}\right )} \int -\frac {x^{m}}{a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3} + {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{n}}\,{d x} - \frac {{\left ({\left (a c d^{2} e^{m} {\left (m - 3 \, n + 1\right )} - b c^{2} d e^{m} {\left (m - 5 \, n + 1\right )}\right )} A - {\left (a c^{2} d e^{m} {\left (m - n + 1\right )} - b c^{3} e^{m} {\left (m - 3 \, n + 1\right )}\right )} B\right )} x x^{m} + {\left ({\left (a d^{3} e^{m} {\left (m - 2 \, n + 1\right )} - b c d^{2} e^{m} {\left (m - 4 \, n + 1\right )}\right )} A + {\left (b c^{2} d e^{m} {\left (m - 2 \, n + 1\right )} - a c d^{2} e^{m} {\left (m + 1\right )}\right )} B\right )} x e^{\left (m \log \relax (x) + n \log \relax (x)\right )}}{2 \, {\left (b^{2} c^{6} n^{2} - 2 \, a b c^{5} d n^{2} + a^{2} c^{4} d^{2} n^{2} + {\left (b^{2} c^{4} d^{2} n^{2} - 2 \, a b c^{3} d^{3} n^{2} + a^{2} c^{2} d^{4} n^{2}\right )} x^{2 \, n} + 2 \, {\left (b^{2} c^{5} d n^{2} - 2 \, a b c^{4} d^{2} n^{2} + a^{2} c^{3} d^{3} n^{2}\right )} x^{n}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{\left (a+b\,x^n\right )\,{\left (c+d\,x^n\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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