Optimal. Leaf size=433 \[ -\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (-m^2-2 m+3\right )+B c \left (m^2+6 m+5\right )\right )+b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^3 d^4 e (m+1)}-\frac{b (e x)^{m+1} \left (2 a^2 d^2 (m+1) (A d (3-m)+B c (m+1))+3 a b c d (m+3) (A d (m+1)-B c (m+5))-b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^2 d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (a d (m+3) (A d (3-m)+B c (m+1))+b c (m+5) (A d (m+3)-B c (m+7)))}{8 c^2 d^3 e^3 (m+3)}+\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (A d (m+3)-B c (m+7)))}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]
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Rubi [A] time = 1.13765, antiderivative size = 433, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {577, 570, 364} \[ -\frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (-m^2-2 m+3\right )+B c \left (m^2+6 m+5\right )\right )+b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^3 d^4 e (m+1)}-\frac{b (e x)^{m+1} \left (2 a^2 d^2 (m+1) (A d (3-m)+B c (m+1))+3 a b c d (m+3) (A d (m+1)-B c (m+5))-b^2 c^2 (m+5) (A d (m+3)-B c (m+7))\right )}{8 c^2 d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (a d (m+3) (A d (3-m)+B c (m+1))+b c (m+5) (A d (m+3)-B c (m+7)))}{8 c^2 d^3 e^3 (m+3)}+\frac{\left (a+b x^2\right )^2 (e x)^{m+1} (a d (A d (3-m)+B c (m+1))+b c (A d (m+3)-B c (m+7)))}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{4 c d e \left (c+d x^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 577
Rule 570
Rule 364
Rubi steps
\begin{align*} \int \frac{(e x)^m \left (a+b x^2\right )^3 \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}-\frac{\int \frac{(e x)^m \left (a+b x^2\right )^2 \left (-a (A d (3-m)+B c (1+m))+b (A d (3+m)-B c (7+m)) x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac{(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{\int \frac{(e x)^m \left (a+b x^2\right ) \left (a (a d (1-m) (A d (3-m)+B c (1+m))-b c (1+m) (A d (3+m)-B c (7+m)))-b (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) x^2\right )}{c+d x^2} \, dx}{8 c^2 d^2}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac{(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}+\frac{\int \left (-\frac{b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^m}{d^2}-\frac{b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{2+m}}{d e^2}+\frac{\left (35 b^3 B c^4-15 A b^3 c^3 d-45 a b^2 B c^3 d+9 a A b^2 c^2 d^2+9 a^2 b B c^2 d^2+3 a^2 A b c d^3+a^3 B c d^3+3 a^3 A d^4+12 b^3 B c^4 m-8 A b^3 c^3 d m-24 a b^2 B c^3 d m+12 a A b^2 c^2 d^2 m+12 a^2 b B c^2 d^2 m-4 a^3 A d^4 m+b^3 B c^4 m^2-A b^3 c^3 d m^2-3 a b^2 B c^3 d m^2+3 a A b^2 c^2 d^2 m^2+3 a^2 b B c^2 d^2 m^2-3 a^2 A b c d^3 m^2-a^3 B c d^3 m^2+a^3 A d^4 m^2\right ) (e x)^m}{d^2 \left (c+d x^2\right )}\right ) \, dx}{8 c^2 d^2}\\ &=-\frac{b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^{1+m}}{8 c^2 d^4 e (1+m)}-\frac{b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{3+m}}{8 c^2 d^3 e^3 (3+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac{(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{\left ((b c-a d) \left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+b^2 c^2 (5+m) (A d (3+m)-B c (7+m))+2 a b c d \left (A d \left (3-2 m-m^2\right )+B c \left (5+6 m+m^2\right )\right )\right )\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{8 c^2 d^4}\\ &=-\frac{b \left (2 a^2 d^2 (1+m) (A d (3-m)+B c (1+m))+3 a b c d (3+m) (A d (1+m)-B c (5+m))-b^2 c^2 (5+m) (A d (3+m)-B c (7+m))\right ) (e x)^{1+m}}{8 c^2 d^4 e (1+m)}-\frac{b^2 (a d (3+m) (A d (3-m)+B c (1+m))+b c (5+m) (A d (3+m)-B c (7+m))) (e x)^{3+m}}{8 c^2 d^3 e^3 (3+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{4 c d e \left (c+d x^2\right )^2}+\frac{(a d (A d (3-m)+B c (1+m))+b c (A d (3+m)-B c (7+m))) (e x)^{1+m} \left (a+b x^2\right )^2}{8 c^2 d^2 e \left (c+d x^2\right )}-\frac{(b c-a d) \left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+b^2 c^2 (5+m) (A d (3+m)-B c (7+m))+2 a b c d \left (A d \left (3-2 m-m^2\right )+B c \left (5+6 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{d x^2}{c}\right )}{8 c^3 d^4 e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.356362, size = 222, normalized size = 0.51 \[ \frac{x (e x)^m \left (\frac{b^2 (3 a B d+A b d-3 b B c)}{m+1}-\frac{(b c-a d)^2 \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (-a B d-3 A b d+4 b B c)}{c^2 (m+1)}+\frac{(b c-a d)^3 (B c-A d) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c^3 (m+1)}+\frac{3 b (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (-a B d-A b d+2 b B c)}{c (m+1)}+\frac{b^3 B d x^2}{m+3}\right )}{d^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( b{x}^{2}+a \right ) ^{3} \left ( ex \right ) ^{m}}{ \left ( d{x}^{2}+c \right ) ^{3}}}\, 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 (b x^{2} + a\right )}^{3} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}}\,{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 b^{3} x^{8} +{\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} x^{4} + A a^{3} +{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )} \left (e x\right )^{m}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, 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 (b x^{2} + a\right )}^{3} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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