Optimal. Leaf size=81 \[ -\frac{2}{15} (2 x+3)^{5/2}+\frac{62}{27} (2 x+3)^{3/2}+\frac{526}{27} \sqrt{2 x+3}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{850}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]
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Rubi [A] time = 0.0789454, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {824, 826, 1166, 207} \[ -\frac{2}{15} (2 x+3)^{5/2}+\frac{62}{27} (2 x+3)^{3/2}+\frac{526}{27} \sqrt{2 x+3}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{850}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]
Antiderivative was successfully verified.
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Rule 824
Rule 826
Rule 1166
Rule 207
Rubi steps
\begin{align*} \int \frac{(5-x) (3+2 x)^{5/2}}{2+5 x+3 x^2} \, dx &=-\frac{2}{15} (3+2 x)^{5/2}+\frac{1}{3} \int \frac{(3+2 x)^{3/2} (49+31 x)}{2+5 x+3 x^2} \, dx\\ &=\frac{62}{27} (3+2 x)^{3/2}-\frac{2}{15} (3+2 x)^{5/2}+\frac{1}{9} \int \frac{\sqrt{3+2 x} (317+263 x)}{2+5 x+3 x^2} \, dx\\ &=\frac{526}{27} \sqrt{3+2 x}+\frac{62}{27} (3+2 x)^{3/2}-\frac{2}{15} (3+2 x)^{5/2}+\frac{1}{27} \int \frac{1801+1639 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac{526}{27} \sqrt{3+2 x}+\frac{62}{27} (3+2 x)^{3/2}-\frac{2}{15} (3+2 x)^{5/2}+\frac{2}{27} \operatorname{Subst}\left (\int \frac{-1315+1639 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{526}{27} \sqrt{3+2 x}+\frac{62}{27} (3+2 x)^{3/2}-\frac{2}{15} (3+2 x)^{5/2}-36 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )+\frac{4250}{27} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{526}{27} \sqrt{3+2 x}+\frac{62}{27} (3+2 x)^{3/2}-\frac{2}{15} (3+2 x)^{5/2}+12 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{850}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}
Mathematica [A] time = 0.0699868, size = 64, normalized size = 0.79 \[ 12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{2}{405} \left (3 \sqrt{2 x+3} \left (36 x^2-202 x-1699\right )+2125 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.01, size = 71, normalized size = 0.9 \begin{align*} -{\frac{2}{15} \left ( 3+2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{62}{27} \left ( 3+2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{526}{27}\sqrt{3+2\,x}}-{\frac{850\,\sqrt{15}}{81}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }+6\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\,\ln \left ( -1+\sqrt{3+2\,x} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.48779, size = 119, normalized size = 1.47 \begin{align*} -\frac{2}{15} \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + \frac{62}{27} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + \frac{425}{81} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{526}{27} \, \sqrt{2 \, x + 3} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.14703, size = 246, normalized size = 3.04 \begin{align*} \frac{425}{81} \, \sqrt{5} \sqrt{3} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) - \frac{2}{135} \,{\left (36 \, x^{2} - 202 \, x - 1699\right )} \sqrt{2 \, x + 3} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 73.6479, size = 126, normalized size = 1.56 \begin{align*} - \frac{2 \left (2 x + 3\right )^{\frac{5}{2}}}{15} + \frac{62 \left (2 x + 3\right )^{\frac{3}{2}}}{27} + \frac{526 \sqrt{2 x + 3}}{27} + \frac{4250 \left (\begin{cases} - \frac{\sqrt{15} \operatorname{acoth}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 > \frac{5}{3} \\- \frac{\sqrt{15} \operatorname{atanh}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 < \frac{5}{3} \end{cases}\right )}{27} - 6 \log{\left (\sqrt{2 x + 3} - 1 \right )} + 6 \log{\left (\sqrt{2 x + 3} + 1 \right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.09626, size = 124, normalized size = 1.53 \begin{align*} -\frac{2}{15} \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + \frac{62}{27} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + \frac{425}{81} \, \sqrt{15} \log \left (\frac{{\left | -2 \, \sqrt{15} + 6 \, \sqrt{2 \, x + 3} \right |}}{2 \,{\left (\sqrt{15} + 3 \, \sqrt{2 \, x + 3}\right )}}\right ) + \frac{526}{27} \, \sqrt{2 \, x + 3} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt{2 \, x + 3} - 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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