Optimal. Leaf size=250 \[ -\frac{d^2 e (4 m+11) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) (m+3) \sqrt{d^2-e^2 x^2}}+\frac{e \sqrt{d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}+\frac{d^3 (4 m+5) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) (m+2) \sqrt{d^2-e^2 x^2}}-\frac{3 d \sqrt{d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)} \]
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Rubi [A] time = 0.380661, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {852, 1809, 808, 365, 364} \[ -\frac{d^2 e (4 m+11) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2) (m+3) \sqrt{d^2-e^2 x^2}}+\frac{e \sqrt{d^2-e^2 x^2} (g x)^{m+2}}{g^2 (m+3)}+\frac{d^3 (4 m+5) \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1) (m+2) \sqrt{d^2-e^2 x^2}}-\frac{3 d \sqrt{d^2-e^2 x^2} (g x)^{m+1}}{g (m+2)} \]
Antiderivative was successfully verified.
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Rule 852
Rule 1809
Rule 808
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \frac{(g x)^m \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx &=\int \frac{(g x)^m (d-e x)^3}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{e (g x)^{2+m} \sqrt{d^2-e^2 x^2}}{g^2 (3+m)}-\frac{\int \frac{(g x)^m \left (-d^3 e^2 (3+m)+d^2 e^3 (11+4 m) x-3 d e^4 (3+m) x^2\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{e^2 (3+m)}\\ &=-\frac{3 d (g x)^{1+m} \sqrt{d^2-e^2 x^2}}{g (2+m)}+\frac{e (g x)^{2+m} \sqrt{d^2-e^2 x^2}}{g^2 (3+m)}+\frac{\int \frac{(g x)^m \left (d^3 e^4 (3+m) (5+4 m)-d^2 e^5 (2+m) (11+4 m) x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{e^4 (2+m) (3+m)}\\ &=-\frac{3 d (g x)^{1+m} \sqrt{d^2-e^2 x^2}}{g (2+m)}+\frac{e (g x)^{2+m} \sqrt{d^2-e^2 x^2}}{g^2 (3+m)}+\frac{\left (d^3 (5+4 m)\right ) \int \frac{(g x)^m}{\sqrt{d^2-e^2 x^2}} \, dx}{2+m}-\frac{\left (d^2 e (11+4 m)\right ) \int \frac{(g x)^{1+m}}{\sqrt{d^2-e^2 x^2}} \, dx}{g (3+m)}\\ &=-\frac{3 d (g x)^{1+m} \sqrt{d^2-e^2 x^2}}{g (2+m)}+\frac{e (g x)^{2+m} \sqrt{d^2-e^2 x^2}}{g^2 (3+m)}+\frac{\left (d^3 (5+4 m) \sqrt{1-\frac{e^2 x^2}{d^2}}\right ) \int \frac{(g x)^m}{\sqrt{1-\frac{e^2 x^2}{d^2}}} \, dx}{(2+m) \sqrt{d^2-e^2 x^2}}-\frac{\left (d^2 e (11+4 m) \sqrt{1-\frac{e^2 x^2}{d^2}}\right ) \int \frac{(g x)^{1+m}}{\sqrt{1-\frac{e^2 x^2}{d^2}}} \, dx}{g (3+m) \sqrt{d^2-e^2 x^2}}\\ &=-\frac{3 d (g x)^{1+m} \sqrt{d^2-e^2 x^2}}{g (2+m)}+\frac{e (g x)^{2+m} \sqrt{d^2-e^2 x^2}}{g^2 (3+m)}+\frac{d^3 (5+4 m) (g x)^{1+m} \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\frac{e^2 x^2}{d^2}\right )}{g (1+m) (2+m) \sqrt{d^2-e^2 x^2}}-\frac{d^2 e (11+4 m) (g x)^{2+m} \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (2+m) (3+m) \sqrt{d^2-e^2 x^2}}\\ \end{align*}
Mathematica [A] time = 0.203725, size = 245, normalized size = 0.98 \[ \frac{x \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}} (g x)^m \left (d^3 \left (m^3+9 m^2+26 m+24\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )-e (m+1) x \left (3 d^2 \left (m^2+7 m+12\right ) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )+e (m+2) x \left (e (m+3) x \, _2F_1\left (\frac{1}{2},\frac{m+4}{2};\frac{m+6}{2};\frac{e^2 x^2}{d^2}\right )-3 d (m+4) \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};\frac{e^2 x^2}{d^2}\right )\right )\right )\right )}{(m+1) (m+2) (m+3) (m+4) (d-e x) (d+e x)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.548, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m}}{ \left ( ex+d \right ) ^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{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 (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} \left (g x\right )^{m}}{{\left (e x + d\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 (e^{2} x^{2} - 2 \, d e x + d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}} \left (g x\right )^{m}}{e x + d}, 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 (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} \left (g x\right )^{m}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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