Optimal. Leaf size=342 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (-m^2+2 m+3\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) (b c-a d)^2}+\frac{d^2 (e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 (b c-a d)} \]
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Rubi [A] time = 0.774935, antiderivative size = 342, normalized size of antiderivative = 1., 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 = {579, 584, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (-m^2+2 m+3\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) (b c-a d)^2}+\frac{d^2 (e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 (b c-a d)} \]
Antiderivative was successfully verified.
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Rule 579
Rule 584
Rule 364
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 )} \, dx &=\frac{(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2}-\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 (3-m) x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, 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}+\frac{(A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right )}+\frac{\int \frac{(e x)^m \left (-a B c (1+m) (a d (5-m)-b (c-c m))+A \left (8 a^2 d^2-a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+d (1-m) (A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac{(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2}+\frac{(A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right )}+\frac{\int \left (\frac{\left (A b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (3+2 m-m^2\right )-a^2 d^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+A d) (e x)^m}{(-b c+a d) \left (c+d x^2\right )}\right ) \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac{(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2}+\frac{(A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right )}+\frac{\left (d^2 (B c-A d)\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{(b c-a d)^3}+\frac{\left (A b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (3+2 m-m^2\right )-a^2 d^2 \left (3-4 m+m^2\right )\right )\right ) \int \frac{(e x)^m}{a+b x^2} \, dx}{8 a^2 (b c-a d)^3}\\ &=\frac{(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2}+\frac{(A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right )}+\frac{\left (A b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (3+2 m-m^2\right )-a^2 d^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)^3 e (1+m)}+\frac{d^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{d x^2}{c}\right )}{c (b c-a d)^3 e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.213849, size = 195, normalized size = 0.57 \[ \frac{x (e x)^m \left (\frac{b (b c-a d) (B c-A d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^2}+\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^3}+\frac{b d (A d-B c) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a}+\frac{d^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c}\right )}{(m+1) (b c-a d)^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( ex \right ) ^{m}}{ \left ( b{x}^{2}+a \right ) ^{3} \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{b^{3} d x^{8} +{\left (b^{3} c + 3 \, a b^{2} d\right )} x^{6} + 3 \,{\left (a b^{2} c + a^{2} b d\right )} x^{4} + a^{3} c +{\left (3 \, a^{2} b c + a^{3} d\right )} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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