Optimal. Leaf size=567 \[ \frac {d (e x)^{m+1} \left (A \left (-2 a^2 d^2 n+a b c d (m-6 n+1)-b^2 c^2 (m-2 n+1)\right )+a B c (b c (m+1)-a d (m-6 n+1))\right )}{2 a^2 c e n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (A b (a d (m-5 n+1)-b c (m-2 n+1))+a B (b c (m+1)-a d (m-3 n+1)))}{2 a^2 e n^2 (b c-a d)^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+m (2-7 n)+12 n^2-7 n+1\right )-2 a b c d \left (m^2+m (2-5 n)+4 n^2-5 n+1\right )+b^2 c^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )+a B \left (-a^2 d^2 \left (m^2+m (2-5 n)+6 n^2-5 n+1\right )+2 a b c d (m+1) (m-3 n+1)-b^2 c^2 (m+1) (m-n+1)\right )\right )}{2 a^3 e (m+1) n^2 (b c-a d)^4}+\frac {d^2 (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m-4 n+1)-B c (m-3 n+1)))}{c^2 e (m+1) n (b c-a d)^4}+\frac {(e x)^{m+1} (A b-a B)}{2 a e n (b c-a d) \left (a+b x^n\right )^2 \left (c+d x^n\right )} \]
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Rubi [A] time = 2.34, antiderivative size = 567, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {595, 597, 364} \[ \frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+m (2-7 n)+12 n^2-7 n+1\right )-2 a b c d \left (m^2+m (2-5 n)+4 n^2-5 n+1\right )+b^2 c^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )+a B \left (-a^2 d^2 \left (m^2+m (2-5 n)+6 n^2-5 n+1\right )+2 a b c d (m+1) (m-3 n+1)-b^2 c^2 (m+1) (m-n+1)\right )\right )}{2 a^3 e (m+1) n^2 (b c-a d)^4}+\frac {d (e x)^{m+1} \left (A \left (-2 a^2 d^2 n+a b c d (m-6 n+1)-b^2 c^2 (m-2 n+1)\right )+a B c (b c (m+1)-a d (m-6 n+1))\right )}{2 a^2 c e n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (A b (a d (m-5 n+1)-b c (m-2 n+1))+a B (b c (m+1)-a d (m-3 n+1)))}{2 a^2 e n^2 (b c-a d)^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {d^2 (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m-4 n+1)-B c (m-3 n+1)))}{c^2 e (m+1) n (b c-a d)^4}+\frac {(e x)^{m+1} (A b-a B)}{2 a e n (b c-a d) \left (a+b x^n\right )^2 \left (c+d x^n\right )} \]
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 )^3 \left (c+d x^n\right )^2} \, dx &=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}-\frac {\int \frac {(e x)^m \left (-a B c (1+m)+A b c (1+m-2 n)+2 a A d n+(A b-a B) d (1+m-3 n) x^n\right )}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^2} \, dx}{2 a (b c-a d) n}\\ &=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}+\frac {(a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {\int \frac {(e x)^m \left (-c (1+m) (a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n)))+(b c-a d) n (a B c (1+m)-A b c (1+m-2 n)-2 a A d n)-d (a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (1+m-2 n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{2 a^2 (b c-a d)^2 n^2}\\ &=\frac {d \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right ) (e x)^{1+m}}{2 a^2 c (b c-a d)^3 e n^2 \left (c+d x^n\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}+\frac {(a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {\int \frac {(e x)^m \left (-n \left (a d (1+m) \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right )+(b c-a d) (c (1+m) (a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n)))-(b c-a d) n (a B c (1+m)-A (b c (1+m-2 n)+2 a d n)))\right )-b d (1+m-n) n \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right ) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 a^2 c (b c-a d)^3 n^3}\\ &=\frac {d \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right ) (e x)^{1+m}}{2 a^2 c (b c-a d)^3 e n^2 \left (c+d x^n\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}+\frac {(a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {\int \left (\frac {b c n \left (a B \left (2 a b c d (1+m) (1+m-3 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-5 n)-5 n+4 n^2\right )+a^2 d^2 \left (1+m^2+m (2-7 n)-7 n+12 n^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (a+b x^n\right )}+\frac {2 a^2 d^2 (b c (A d (1+m-4 n)-B c (1+m-3 n))+a d (B c (1+m)-A d (1+m-n))) n^2 (e x)^m}{(b c-a d) \left (c+d x^n\right )}\right ) \, dx}{2 a^2 c (b c-a d)^3 n^3}\\ &=\frac {d \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right ) (e x)^{1+m}}{2 a^2 c (b c-a d)^3 e n^2 \left (c+d x^n\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}+\frac {(a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {\left (d^2 (b c (A d (1+m-4 n)-B c (1+m-3 n))+a d (B c (1+m)-A d (1+m-n)))\right ) \int \frac {(e x)^m}{c+d x^n} \, dx}{c (b c-a d)^4 n}+\frac {\left (b \left (a B \left (2 a b c d (1+m) (1+m-3 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-5 n)-5 n+4 n^2\right )+a^2 d^2 \left (1+m^2+m (2-7 n)-7 n+12 n^2\right )\right )\right )\right ) \int \frac {(e x)^m}{a+b x^n} \, dx}{2 a^2 (b c-a d)^4 n^2}\\ &=\frac {d \left (a B c (b c (1+m)-a d (1+m-6 n))+A \left (a b c d (1+m-6 n)-b^2 c^2 (1+m-2 n)-2 a^2 d^2 n\right )\right ) (e x)^{1+m}}{2 a^2 c (b c-a d)^3 e n^2 \left (c+d x^n\right )}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2 \left (c+d x^n\right )}+\frac {(a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-5 n)-b c (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {b \left (a B \left (2 a b c d (1+m) (1+m-3 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-5 n)-5 n+4 n^2\right )+a^2 d^2 \left (1+m^2+m (2-7 n)-7 n+12 n^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{2 a^3 (b c-a d)^4 e (1+m) n^2}+\frac {d^2 (b c (A d (1+m-4 n)-B c (1+m-3 n))+a d (B c (1+m)-A d (1+m-n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c^2 (b c-a d)^4 e (1+m) n}\\ \end {align*}
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Mathematica [A] time = 0.53, size = 270, normalized size = 0.48 \[ \frac {x (e x)^m \left (\frac {b (A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^3}+\frac {b (b c-a d) (a B d-2 A b d+b B c) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^2}+\frac {d^2 (b c-a d) (B c-A d) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c^2}+\frac {d^2 (a B d-3 A b d+2 b B c) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c}-\frac {b d (a B d-3 A b d+2 b B c) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a}\right )}{(m+1) (b c-a d)^4} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{b^{3} d^{2} x^{5 \, n} + a^{3} c^{2} + {\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{4 \, n} + {\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{3 \, n} + {\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2 \, n} + {\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{n}}, x\right ) \]
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 \[ \int \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (b x^{n} + a\right )}^{3} {\left (d x^{n} + c\right )}^{2}}\,{d x} \]
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 \[ \int \frac {\left (B \,x^{n}+A \right ) \left (e x \right )^{m}}{\left (b \,x^{n}+a \right )^{3} \left (d \,x^{n}+c \right )^{2}}\, dx \]
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 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{{\left (a+b\,x^n\right )}^3\,{\left (c+d\,x^n\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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