Optimal. Leaf size=128 \[ \frac{1}{9} (1-2 x)^{3/2} (5 x+3)^{3/2}+\frac{37}{180} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{1781 \sqrt{1-2 x} \sqrt{5 x+3}}{2160}+\frac{19573 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{6480 \sqrt{10}}-\frac{14}{81} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]
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Rubi [A] time = 0.0522678, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {101, 154, 157, 54, 216, 93, 204} \[ \frac{1}{9} (1-2 x)^{3/2} (5 x+3)^{3/2}+\frac{37}{180} \sqrt{1-2 x} (5 x+3)^{3/2}-\frac{1781 \sqrt{1-2 x} \sqrt{5 x+3}}{2160}+\frac{19573 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{6480 \sqrt{10}}-\frac{14}{81} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right ) \]
Antiderivative was successfully verified.
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Rule 101
Rule 154
Rule 157
Rule 54
Rule 216
Rule 93
Rule 204
Rubi steps
\begin{align*} \int \frac{(1-2 x)^{3/2} (3+5 x)^{3/2}}{2+3 x} \, dx &=\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}-\frac{1}{9} \int \frac{\left (-30-\frac{111 x}{2}\right ) \sqrt{1-2 x} \sqrt{3+5 x}}{2+3 x} \, dx\\ &=\frac{37}{180} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}-\frac{1}{270} \int \frac{\left (-\frac{801}{2}-\frac{5343 x}{4}\right ) \sqrt{3+5 x}}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=-\frac{1781 \sqrt{1-2 x} \sqrt{3+5 x}}{2160}+\frac{37}{180} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}+\frac{\int \frac{\frac{23493}{4}+\frac{58719 x}{8}}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{1620}\\ &=-\frac{1781 \sqrt{1-2 x} \sqrt{3+5 x}}{2160}+\frac{37}{180} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}+\frac{49}{81} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx+\frac{19573 \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx}{12960}\\ &=-\frac{1781 \sqrt{1-2 x} \sqrt{3+5 x}}{2160}+\frac{37}{180} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}+\frac{98}{81} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )+\frac{19573 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{6480 \sqrt{5}}\\ &=-\frac{1781 \sqrt{1-2 x} \sqrt{3+5 x}}{2160}+\frac{37}{180} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{1}{9} (1-2 x)^{3/2} (3+5 x)^{3/2}+\frac{19573 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{6480 \sqrt{10}}-\frac{14}{81} \sqrt{7} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )\\ \end{align*}
Mathematica [A] time = 0.0718475, size = 105, normalized size = 0.82 \[ \frac{30 \sqrt{5 x+3} \left (4800 x^3-6360 x^2+1438 x+271\right )-19573 \sqrt{10-20 x} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-11200 \sqrt{7-14 x} \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{64800 \sqrt{1-2 x}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.009, size = 115, normalized size = 0.9 \begin{align*}{\frac{1}{129600}\sqrt{1-2\,x}\sqrt{3+5\,x} \left ( -144000\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+19573\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +11200\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +118800\,x\sqrt{-10\,{x}^{2}-x+3}+16260\,\sqrt{-10\,{x}^{2}-x+3} \right ){\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.75738, size = 112, normalized size = 0.88 \begin{align*} \frac{1}{9} \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}} + \frac{37}{36} \, \sqrt{-10 \, x^{2} - x + 3} x + \frac{19573}{129600} \, \sqrt{10} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) + \frac{7}{81} \, \sqrt{7} \arcsin \left (\frac{37 \, x}{11 \,{\left | 3 \, x + 2 \right |}} + \frac{20}{11 \,{\left | 3 \, x + 2 \right |}}\right ) - \frac{449}{2160} \, \sqrt{-10 \, x^{2} - x + 3} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.60132, size = 352, normalized size = 2.75 \begin{align*} -\frac{1}{2160} \,{\left (2400 \, x^{2} - 1980 \, x - 271\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1} - \frac{7}{81} \, \sqrt{7} \arctan \left (\frac{\sqrt{7}{\left (37 \, x + 20\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{14 \,{\left (10 \, x^{2} + x - 3\right )}}\right ) - \frac{19573}{129600} \, \sqrt{10} \arctan \left (\frac{\sqrt{10}{\left (20 \, x + 1\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{20 \,{\left (10 \, x^{2} + x - 3\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{\left (1 - 2 x\right )^{\frac{3}{2}} \left (5 x + 3\right )^{\frac{3}{2}}}{3 x + 2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.29895, size = 251, normalized size = 1.96 \begin{align*} \frac{7}{810} \, \sqrt{70} \sqrt{10}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{70} \sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} - \frac{1}{10800} \,{\left (12 \,{\left (8 \, \sqrt{5}{\left (5 \, x + 3\right )} - 81 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 1781 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5} + \frac{19573}{129600} \, \sqrt{10}{\left (\pi + 2 \, \arctan \left (-\frac{\sqrt{5 \, x + 3}{\left (\frac{{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}\right )\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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