Optimal. Leaf size=491 \[ -\frac {d \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (5+4 m-m^2\right )-a^2 d^2 \left (15-8 m+m^2\right )\right )+A b \left (a^2 d^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^4 e (1+m)}+\frac {d^2 (b c (B c (5-m)-A d (7-m))+a d (A d (1-m)+B c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^4 e (1+m)} \]
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Rubi [A]
time = 0.98, antiderivative size = 491, 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 = {593, 598, 371}
\begin {gather*} -\frac {d (e x)^{m+1} \left (A \left (4 a^2 d^2+a b c d (11-m)-b^2 c^2 (3-m)\right )-a B c (a d (11-m)+b c (m+1))\right )}{8 a^2 c e \left (c+d x^2\right ) (b c-a d)^3}+\frac {(e x)^{m+1} (A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)^2}+\frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-12 m+35\right )-2 a b c d \left (m^2-8 m+7\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (-m^2+4 m+5\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^4}+\frac {d^2 (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (5-m)-A d (7-m)))}{2 c^2 e (m+1) (b c-a d)^4}+\frac {(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 \left (c+d x^2\right ) (b c-a d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 593
Rule 598
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^3 \left (c+d x^2\right )^2} \, dx &=\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (4 a A d-A b c (3-m)-a B c (1+m)-(A b-a B) d (5-m) x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx}{4 a (b c-a d)}\\ &=\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (-a B c (1+m) (a d (7-m)-b (c-c m))+A \left (8 a^2 d^2-a b c d \left (5-10 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+d (3-m) (A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a^2 (b c-a d)^2}\\ &=-\frac {d \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (-2 \left (a B c \left (4 a^2 d^2-b^2 c^2 (1-m)+a b c d (9-m)\right ) (1+m)-A \left (24 a^2 b c d^2-4 a^3 d^3 (1-m)-a b^2 c^2 d \left (11-12 m+m^2\right )+b^3 c^3 \left (3-4 m+m^2\right )\right )\right )-2 b d (1-m) \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a^2 c (b c-a d)^3}\\ &=-\frac {d \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\int \left (\frac {2 b c \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (5+4 m-m^2\right )-a^2 d^2 \left (15-8 m+m^2\right )\right )+A b \left (a^2 d^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {8 a^2 d^2 (b c (B c (5-m)-A d (7-m))+a d (A d (1-m)+B c (1+m))) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{16 a^2 c (b c-a d)^3}\\ &=-\frac {d \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\left (d^2 (b c (B c (5-m)-A d (7-m))+a d (A d (1-m)+B c (1+m)))\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{2 c (b c-a d)^4}+\frac {\left (b \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (5+4 m-m^2\right )-a^2 d^2 \left (15-8 m+m^2\right )\right )+A b \left (a^2 d^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{8 a^2 (b c-a d)^4}\\ &=-\frac {d \left (A \left (4 a^2 d^2-b^2 c^2 (3-m)+a b c d (11-m)\right )-a B c (a d (11-m)+b c (1+m))\right ) (e x)^{1+m}}{8 a^2 c (b c-a d)^3 e \left (c+d x^2\right )}+\frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2 \left (c+d x^2\right )}+\frac {(A b (b c (3-m)-a d (9-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b \left (a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (5+4 m-m^2\right )-a^2 d^2 \left (15-8 m+m^2\right )\right )+A b \left (a^2 d^2 \left (35-12 m+m^2\right )-2 a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^4 e (1+m)}+\frac {d^2 (b c (B c (5-m)-A d (7-m))+a d (A d (1-m)+B c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^4 e (1+m)}\\ \end {align*}
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Mathematica [A]
time = 1.62, size = 268, normalized size = 0.55 \begin {gather*} \frac {x (e x)^m \left (-a^2 b c^2 d (2 b B c-3 A b d+a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^3 c d^2 (2 b B c-3 A b d+a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d) \left (a b c^2 (b B c-2 A b d+a B d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^3 d^2 (B c-A d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+b (A b-a B) c^2 (b c-a d) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )\right )}{a^3 c^2 (b c-a d)^4 (1+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right )}{\left (b \,x^{2}+a \right )^{3} \left (d \,x^{2}+c \right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^2+a\right )}^3\,{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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