Integrand size = 24, antiderivative size = 92 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {140}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 92, 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, 159, 162, 65, 212} \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {140}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x} \]
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Rule 65
Rule 100
Rule 159
Rule 162
Rule 212
Rubi steps \begin{align*} \text {integral}& = \frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {1}{3} \int \frac {\sqrt {1-2 x} (96+39 x)}{(2+3 x) (3+5 x)} \, dx \\ & = \frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {2}{45} \int \frac {954-\frac {813 x}{2}}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx \\ & = \frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}-\frac {490}{3} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {1331}{5} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = \frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {490}{3} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {1331}{5} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = \frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {140}{3} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1}{225} \left (\frac {15 \sqrt {1-2 x} (87+8 x)}{2+3 x}+3500 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2178 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
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Time = 1.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {8 \sqrt {1-2 x}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {98 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
default | \(\frac {8 \sqrt {1-2 x}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {98 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
risch | \(-\frac {16 x^{2}+166 x -87}{15 \left (2+3 x \right ) \sqrt {1-2 x}}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}\) | \(64\) |
pseudoelliptic | \(\frac {3500 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}-2178 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right ) \sqrt {55}+15 \sqrt {1-2 x}\, \left (87+8 x \right )}{450+675 x}\) | \(70\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (87+8 x \right )}{30+45 x}+\frac {70 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{9}-\frac {121 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{25}\) | \(111\) |
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Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1089 \, \sqrt {11} \sqrt {5} {\left (3 \, x + 2\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 1750 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 15 \, {\left (8 \, x + 87\right )} \sqrt {-2 \, x + 1}}{225 \, {\left (3 \, x + 2\right )}} \]
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Time = 14.26 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.35 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {8 \sqrt {1 - 2 x}}{45} - \frac {203 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{27} + \frac {121 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{25} + \frac {1372 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{9} \]
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Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {98\,\sqrt {1-2\,x}}{27\,\left (2\,x+\frac {4}{3}\right )}+\frac {8\,\sqrt {1-2\,x}}{45}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,140{}\mathrm {i}}{9}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{25} \]
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