Optimal. Leaf size=198 \[ -\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) \sqrt{d+e x} (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.121972, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \[ -\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) \sqrt{d+e x} (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 768
Rule 646
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(5 e) \int \frac{(d+e x)^{3/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx}{2 b}\\ &=-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{2 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e (b d-a e) (a+b x) \sqrt{d+e x}}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (b d-a e)^{3/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0325365, size = 66, normalized size = 0.33 \[ \frac{2 e (a+b x) (d+e x)^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b (d+e x)}{a e-b d}\right )}{7 \sqrt{(a+b x)^2} (a e-b d)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.015, size = 409, normalized size = 2.1 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{3\,{b}^{3}} \left ( 2\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}x{b}^{2}e+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}-30\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}d{e}^{2}+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{b}^{3}{d}^{2}e+2\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe-12\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xab{e}^{2}+12\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}x{b}^{2}de+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}-30\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{2}{d}^{2}e-15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}+18\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde-3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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 + a\right )}{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.01768, size = 707, normalized size = 3.57 \begin{align*} \left [-\frac{15 \,{\left (a b d e - a^{2} e^{2} +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) - 2 \,{\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \,{\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, -\frac{15 \,{\left (a b d e - a^{2} e^{2} +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (2 \, b^{2} e^{2} x^{2} - 3 \, b^{2} d^{2} + 20 \, a b d e - 15 \, a^{2} e^{2} + 2 \,{\left (7 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (b^{4} x + a b^{3}\right )}}\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 [A] time = 1.21331, size = 363, normalized size = 1.83 \begin{align*} \frac{5 \,{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\sqrt{-b^{2} d + a b e} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{{\left (\sqrt{x e + d} b^{2} d^{2} e^{2} - 2 \, \sqrt{x e + d} a b d e^{3} + \sqrt{x e + d} a^{2} e^{4}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{4} e^{4} + 6 \, \sqrt{x e + d} b^{4} d e^{4} - 6 \, \sqrt{x e + d} a b^{3} e^{5}\right )} e^{\left (-3\right )}}{3 \, b^{6} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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