Optimal. Leaf size=55 \[ -\frac{26}{5 \sqrt{2 x+3}}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]
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Rubi [A] time = 0.0450021, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {828, 826, 1166, 207} \[ -\frac{26}{5 \sqrt{2 x+3}}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]
Antiderivative was successfully verified.
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Rule 828
Rule 826
Rule 1166
Rule 207
Rubi steps
\begin{align*} \int \frac{5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx &=-\frac{26}{5 \sqrt{3+2 x}}+\frac{1}{5} \int \frac{-9-39 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{26}{5 \sqrt{3+2 x}}+\frac{2}{5} \operatorname{Subst}\left (\int \frac{99-39 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{26}{5 \sqrt{3+2 x}}+\frac{102}{5} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-36 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{26}{5 \sqrt{3+2 x}}+12 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{34}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}
Mathematica [A] time = 0.0393925, size = 55, normalized size = 1. \[ -\frac{26}{5 \sqrt{2 x+3}}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{34}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.011, size = 53, normalized size = 1. \begin{align*} -{\frac{34\,\sqrt{15}}{25}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-{\frac{26}{5}{\frac{1}{\sqrt{3+2\,x}}}}+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 = 1.44915, size = 95, normalized size = 1.73 \begin{align*} \frac{17}{25} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{26}{5 \, \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 [B] time = 1.59019, size = 274, normalized size = 4.98 \begin{align*} \frac{17 \, \sqrt{5} \sqrt{3}{\left (2 \, x + 3\right )} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 150 \,{\left (2 \, x + 3\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 150 \,{\left (2 \, x + 3\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 130 \, \sqrt{2 \, x + 3}}{25 \,{\left (2 \, x + 3\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 36.4474, size = 102, normalized size = 1.85 \begin{align*} \frac{102 \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 )}{5} - 6 \log{\left (\sqrt{2 x + 3} - 1 \right )} + 6 \log{\left (\sqrt{2 x + 3} + 1 \right )} - \frac{26}{5 \sqrt{2 x + 3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.10559, size = 100, normalized size = 1.82 \begin{align*} \frac{17}{25} \, \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{26}{5 \, \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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