Optimal. Leaf size=202 \[ \frac{15 e^2 (a+b x) \sqrt{d+e x}}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{15 e^2 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.094029, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {646, 47, 50, 63, 208} \[ \frac{15 e^2 (a+b x) \sqrt{d+e x}}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{15 e^2 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 646
Rule 47
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{5/2}}{2 b (a+b x) \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}}{\left (a b+b^2 x\right )^2} \, dx}{4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 e^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{8 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{15 e^2 (a+b x) \sqrt{d+e x}}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 e^2 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{15 e^2 (a+b x) \sqrt{d+e x}}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 e \left (b^2 d-a b e\right ) \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 )}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{15 e^2 (a+b x) \sqrt{d+e x}}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{15 e^2 \sqrt{b d-a e} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.038997, size = 67, normalized size = 0.33 \[ -\frac{2 e^2 (a+b x) (d+e x)^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b (d+e x)}{b d-a e}\right )}{7 \sqrt{(a+b x)^2} (b d-a e)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.275, size = 413, normalized size = 2. \begin{align*}{\frac{bx+a}{4\,{b}^{3}} \left ( -15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}a{b}^{2}{e}^{3}+15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{3}d{e}^{2}+8\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{x}^{2}{b}^{2}{e}^{2}-30\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}+30\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}d{e}^{2}+9\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe-9\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}d+16\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xab{e}^{2}-15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}+15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}+15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}-14\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde+7\,\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 (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 = 1.64884, size = 728, normalized size = 3.6 \begin{align*} \left [\frac{15 \,{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\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 (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} -{\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{15 \,{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\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 (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} -{\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} 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 = 1.22455, size = 332, normalized size = 1.64 \begin{align*} \frac{15 \,{\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \, \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{2 \, \sqrt{x e + d} e^{2}}{b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{9 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} d e^{2} - 7 \, \sqrt{x e + d} b^{2} d^{2} e^{2} - 9 \,{\left (x e + d\right )}^{\frac{3}{2}} a b e^{3} + 14 \, \sqrt{x e + d} a b d e^{3} - 7 \, \sqrt{x e + d} a^{2} e^{4}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3} \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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