Optimal. Leaf size=275 \[ \frac{35 b e^2 (a+b x)}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^4}+\frac{35 e^2 (a+b x)}{12 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}-\frac{35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}}+\frac{7 e}{4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]
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Rubi [A] time = 0.124641, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \[ \frac{35 b e^2 (a+b x)}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^4}+\frac{35 e^2 (a+b x)}{12 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}-\frac{35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}}+\frac{7 e}{4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]
Antiderivative was successfully verified.
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Rule 646
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(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{1}{\left (a b+b^2 x\right )^3 (d+e x)^{5/2}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^2 (d+e x)^{5/2}} \, dx}{4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 b e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 b^2 e^2 \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 d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 b^2 e \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 d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0240417, size = 67, normalized size = 0.24 \[ \frac{2 e^2 (a+b x) \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 \sqrt{(a+b x)^2} (d+e x)^{3/2} (b d-a e)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.279, size = 388, normalized size = 1.4 \begin{align*}{\frac{bx+a}{12\, \left ( ae-bd \right ) ^{4}} \left ( 105\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}{x}^{2}{b}^{4}{e}^{2}+210\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}xa{b}^{3}{e}^{2}+105\,\sqrt{ \left ( ae-bd \right ) b}{x}^{3}{b}^{3}{e}^{3}+105\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}{a}^{2}{b}^{2}{e}^{2}+175\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}a{b}^{2}{e}^{3}+140\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{3}d{e}^{2}+56\,\sqrt{ \left ( ae-bd \right ) b}x{a}^{2}b{e}^{3}+238\,\sqrt{ \left ( ae-bd \right ) b}xa{b}^{2}d{e}^{2}+21\,\sqrt{ \left ( ae-bd \right ) b}x{b}^{3}{d}^{2}e-8\,\sqrt{ \left ( ae-bd \right ) b}{a}^{3}{e}^{3}+80\,\sqrt{ \left ( ae-bd \right ) b}{a}^{2}bd{e}^{2}+39\,\sqrt{ \left ( ae-bd \right ) b}a{b}^{2}{d}^{2}e-6\,\sqrt{ \left ( ae-bd \right ) b}{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}} \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{1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.80953, size = 2472, normalized size = 8.99 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right )^{\frac{5}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.27909, size = 844, normalized size = 3.07 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{4} d^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} + \frac{2 \,{\left (9 \,{\left (x e + d\right )} b e^{2} + b d e^{2} - a e^{3}\right )}}{3 \,{\left (b^{4} d^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left (x e + d\right )}^{\frac{3}{2}}} + \frac{11 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} e^{2} - 13 \, \sqrt{x e + d} b^{3} d e^{2} + 13 \, \sqrt{x e + d} a b^{2} e^{3}}{4 \,{\left (b^{4} d^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 4 \, a^{3} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{4} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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