Optimal. Leaf size=230 \[ \frac{2 (a+b x) (d+e x)^{3/2} (A b-a B)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) \sqrt{d+e x} (A b-a B) (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.144906, antiderivative size = 230, 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 = {770, 80, 50, 63, 208} \[ \frac{2 (a+b x) (d+e x)^{3/2} (A b-a B)}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) \sqrt{d+e x} (A b-a B) (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 770
Rule 80
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(A+B x) (d+e x)^{3/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(A+B x) (d+e x)^{3/2}}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (\frac{5}{2} A b^2 e-\frac{5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{5 b^2 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right ) \left (\frac{5}{2} A b^2 e-\frac{5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{5 b^4 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (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{2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right )^2 \left (\frac{5}{2} A b^2 e-\frac{5}{2} a b 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}{5 b^6 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (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{2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (4 \left (b^2 d-a b e\right )^2 \left (\frac{5}{2} A b^2 e-\frac{5}{2} a b 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 )}{5 b^6 e^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (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{2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (A b-a B) (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 [A] time = 0.212623, size = 127, normalized size = 0.55 \[ \frac{2 (a+b x) \left (\frac{5 e (A b-a B) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{3 b^{5/2}}+B (d+e x)^{5/2}\right )}{5 b e \sqrt{(a+b x)^2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.01, size = 414, normalized size = 1.8 \begin{align*}{\frac{2\,bx+2\,a}{15\,{b}^{3}e} \left ( 3\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{2}+5\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}e+15\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}b{e}^{3}-30\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{2}d{e}^{2}+15\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){b}^{3}{d}^{2}e-5\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe-15\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}+30\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}-15\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{2}{d}^{2}e-15\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ab{e}^{2}+15\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}de+15\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}-15\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \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{3}{2}}}{\sqrt{{\left (b x + a\right )}^{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.4507, size = 795, normalized size = 3.46 \begin{align*} \left [-\frac{15 \,{\left ({\left (B a b - A b^{2}\right )} d e -{\left (B a^{2} - A a b\right )} 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 (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \,{\left (B a b - A b^{2}\right )} d e + 15 \,{\left (B a^{2} - A a b\right )} e^{2} +{\left (6 \, B b^{2} d e - 5 \,{\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, b^{3} e}, \frac{2 \,{\left (15 \,{\left ({\left (B a b - A b^{2}\right )} d e -{\left (B a^{2} - A a b\right )} 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 (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \,{\left (B a b - A b^{2}\right )} d e + 15 \,{\left (B a^{2} - A a b\right )} e^{2} +{\left (6 \, B b^{2} d e - 5 \,{\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}\right )}}{15 \, b^{3} e}\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 = 1.14437, size = 413, normalized size = 1.8 \begin{align*} -\frac{2 \,{\left (B a b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm{sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm{sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{3}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{4} e^{4} \mathrm{sgn}\left (b x + a\right ) - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} e^{5} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} e^{5} \mathrm{sgn}\left (b x + a\right ) - 15 \, \sqrt{x e + d} B a b^{3} d e^{5} \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} A b^{4} d e^{5} \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} B a^{2} b^{2} e^{6} \mathrm{sgn}\left (b x + a\right ) - 15 \, \sqrt{x e + d} A a b^{3} e^{6} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{15 \, b^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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