Integrand size = 24, antiderivative size = 106 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {748 \sqrt {1-2 x}}{15 (3+5 x)}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}-\frac {910}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1562}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 154, 162, 65, 212} \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {910}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1562}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2) (5 x+3)}-\frac {748 \sqrt {1-2 x}}{15 (5 x+3)} \]
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Rule 65
Rule 100
Rule 154
Rule 162
Rule 212
Rubi steps \begin{align*} \text {integral}& = \frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}+\frac {1}{3} \int \frac {(131-31 x) \sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx \\ & = -\frac {748 \sqrt {1-2 x}}{15 (3+5 x)}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}+\frac {1}{15} \int \frac {-3771+2306 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx \\ & = -\frac {748 \sqrt {1-2 x}}{15 (3+5 x)}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}+\frac {3185}{3} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-\frac {8591}{5} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {748 \sqrt {1-2 x}}{15 (3+5 x)}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}-\frac {3185}{3} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+\frac {8591}{5} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {748 \sqrt {1-2 x}}{15 (3+5 x)}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x) (3+5 x)}-\frac {910}{3} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1562}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {2}{225} \left (-\frac {15 \sqrt {1-2 x} (1461+2314 x)}{12+38 x+30 x^2}-11375 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+7029 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
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Time = 3.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.65
method | result | size |
risch | \(\frac {\left (1461+2314 x \right ) \left (-1+2 x \right )}{15 \left (15 x^{2}+19 x +6\right ) \sqrt {1-2 x}}-\frac {910 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}+\frac {1562 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}\) | \(69\) |
derivativedivides | \(\frac {242 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {1562 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}+\frac {98 \sqrt {1-2 x}}{9 \left (-\frac {4}{3}-2 x \right )}-\frac {910 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(70\) |
default | \(\frac {242 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {1562 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}+\frac {98 \sqrt {1-2 x}}{9 \left (-\frac {4}{3}-2 x \right )}-\frac {910 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(70\) |
pseudoelliptic | \(\frac {-22750 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {21}+14058 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {55}-15 \sqrt {1-2 x}\, \left (1461+2314 x \right )}{3375 x^{2}+4275 x +1350}\) | \(85\) |
trager | \(-\frac {\left (1461+2314 x \right ) \sqrt {1-2 x}}{15 \left (15 x^{2}+19 x +6\right )}-\frac {455 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{9}-\frac {781 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{25}\) | \(116\) |
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Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {7029 \, \sqrt {11} \sqrt {5} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 11375 \, \sqrt {7} \sqrt {3} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 15 \, {\left (2314 \, x + 1461\right )} \sqrt {-2 \, x + 1}}{225 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]
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Time = 26.76 (sec) , antiderivative size = 325, normalized size of antiderivative = 3.07 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {448 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{9} - \frac {792 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{25} - \frac {1372 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{3} - \frac {5324 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{5} \]
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Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {781}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {455}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {4 \, {\left (1157 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2618 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {781}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {455}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (1157 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2618 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {1562\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{25}-\frac {910\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{9}-\frac {\frac {10472\,\sqrt {1-2\,x}}{225}-\frac {4628\,{\left (1-2\,x\right )}^{3/2}}{225}}{\frac {136\,x}{15}+{\left (2\,x-1\right )}^2+\frac {3}{5}} \]
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