Optimal. Leaf size=61 \[ -\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (5 x+3)}-\frac {134 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
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Rubi [A] time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {89, 80, 63, 206} \begin {gather*} -\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (5 x+3)}-\frac {134 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 80
Rule 89
Rule 206
Rubi steps
\begin {align*} \int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x}}{275 (3+5 x)}+\frac {1}{275} \int \frac {364+495 x}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (3+5 x)}+\frac {67}{275} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (3+5 x)}-\frac {67}{275} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {9}{25} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{275 (3+5 x)}-\frac {134 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 53, normalized size = 0.87 \begin {gather*} -\frac {\sqrt {1-2 x} (495 x+298)}{275 (5 x+3)}-\frac {134 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.11, size = 61, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {1-2 x} (495 (1-2 x)-1091)}{275 (5 (1-2 x)-11)}-\frac {134 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.51, size = 59, normalized size = 0.97 \begin {gather*} \frac {67 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (495 \, x + 298\right )} \sqrt {-2 \, x + 1}}{15125 \, {\left (5 \, x + 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.24, size = 65, normalized size = 1.07 \begin {gather*} \frac {67}{15125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {9}{25} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 45, normalized size = 0.74 \begin {gather*} -\frac {134 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{15125}-\frac {9 \sqrt {-2 x +1}}{25}+\frac {2 \sqrt {-2 x +1}}{1375 \left (-2 x -\frac {6}{5}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.09, size = 62, normalized size = 1.02 \begin {gather*} \frac {67}{15125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {9}{25} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{275 \, {\left (5 \, x + 3\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.23, size = 44, normalized size = 0.72 \begin {gather*} -\frac {134\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {2\,\sqrt {1-2\,x}}{1375\,\left (2\,x+\frac {6}{5}\right )}-\frac {9\,\sqrt {1-2\,x}}{25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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