Optimal. Leaf size=295 \[ -\frac{(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} (a d (A d (1-m)+B c (m+1))-b c (A d (-m-2 p+1)+B c (m+2 p+1))) F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{2 c^2 d e (m+1) (b c-a d)}-\frac{b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} (B c-A d) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 c d e (m+1) (b c-a d)}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]
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Rubi [A] time = 0.413457, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {579, 584, 365, 364, 511, 510} \[ -\frac{(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} (a d (A d (1-m)+B c (m+1))-b c (A d (-m-2 p+1)+B c (m+2 p+1))) F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{2 c^2 d e (m+1) (b c-a d)}-\frac{b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} (B c-A d) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 c d e (m+1) (b c-a d)}+\frac{(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]
Antiderivative was successfully verified.
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Rule 579
Rule 584
Rule 365
Rule 364
Rule 511
Rule 510
Rubi steps
\begin{align*} \int \frac{(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx &=\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}+\frac{\int \frac{(e x)^m \left (a+b x^2\right )^p \left (2 A b c-a A d (1-m)-a B c (1+m)-b (B c-A d) (1+m+2 p) x^2\right )}{c+d x^2} \, dx}{2 c (b c-a d)}\\ &=\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}+\frac{\int \left (-\frac{b (B c-A d) (1+m+2 p) (e x)^m \left (a+b x^2\right )^p}{d}+\frac{(d (2 A b c-a A d (1-m)-a B c (1+m))+b c (B c-A d) (1+m+2 p)) (e x)^m \left (a+b x^2\right )^p}{d \left (c+d x^2\right )}\right ) \, dx}{2 c (b c-a d)}\\ &=\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac{(b (B c-A d) (1+m+2 p)) \int (e x)^m \left (a+b x^2\right )^p \, dx}{2 c d (b c-a d)}-\frac{(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) \int \frac{(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{2 c d (b c-a d)}\\ &=\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac{\left (b (B c-A d) (1+m+2 p) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int (e x)^m \left (1+\frac{b x^2}{a}\right )^p \, dx}{2 c d (b c-a d)}-\frac{\left ((a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{(e x)^m \left (1+\frac{b x^2}{a}\right )^p}{c+d x^2} \, dx}{2 c d (b c-a d)}\\ &=\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac{(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1+m}{2};-p,1;\frac{3+m}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{2 c^2 d (b c-a d) e (1+m)}-\frac{b (B c-A d) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},-p;\frac{3+m}{2};-\frac{b x^2}{a}\right )}{2 c d (b c-a d) e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.213634, size = 128, normalized size = 0.43 \[ \frac{x (e x)^m \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left ((A d-B c) F_1\left (\frac{m+1}{2};-p,2;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+B c F_1\left (\frac{m+1}{2};-p,1;\frac{m+3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )}{c^2 d (m+1)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( b{x}^{2}+a \right ) ^{p} \left ( B{x}^{2}+A \right ) }{ \left ( d{x}^{2}+c \right ) ^{2}}}\, 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 )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{2}}\,{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 (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{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 (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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