Optimal. Leaf size=168 \[ \frac{e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.0875864, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {646, 47, 51, 63, 208} \[ \frac{e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 646
Rule 47
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\sqrt{d+e x}}{\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{\sqrt{d+e x}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (e \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^2 \sqrt{d+e x}} \, dx}{4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e \sqrt{d+e x}}{4 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (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 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e \sqrt{d+e x}}{4 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (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 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e \sqrt{d+e x}}{4 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0253277, size = 67, normalized size = 0.4 \[ -\frac{2 e^2 (a+b x) (d+e x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 \sqrt{(a+b x)^2} (b d-a e)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.271, size = 200, normalized size = 1.2 \begin{align*}{\frac{bx+a}{4\, \left ( ae-bd \right ) b} \left ( \arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){x}^{2}{b}^{2}{e}^{2}+2\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xab{e}^{2}+\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{{\frac{3}{2}}}b+\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){a}^{2}{e}^{2}-\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ae+\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}bd \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{\sqrt{e x + d}}{{\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 = 1.70585, size = 949, normalized size = 5.65 \begin{align*} \left [-\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) + 2 \,{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}, -\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) +{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}\right ] \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{\sqrt{d + e x}}{\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 [A] time = 1.21077, size = 297, normalized size = 1.77 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{2} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b e \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{{\left (x e + d\right )}^{\frac{3}{2}} b e^{2} + \sqrt{x e + d} b d e^{2} - \sqrt{x e + d} a e^{3}}{4 \,{\left (b^{2} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b e \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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