Integrand size = 24, antiderivative size = 131 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}-\frac {7209}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {101, 156, 162, 65, 212} \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {7209}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {1045 \sqrt {1-2 x}}{14 (5 x+3)}+\frac {52 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)}+\frac {\sqrt {1-2 x}}{2 (3 x+2)^2 (5 x+3)} \]
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Rule 65
Rule 101
Rule 156
Rule 162
Rule 212
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}-\frac {1}{2} \int \frac {-18+25 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^2} \, dx \\ & = \frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}-\frac {1}{14} \int \frac {-1363+1560 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx \\ & = -\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}+\frac {1}{154} \int \frac {-56309+34485 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx \\ & = -\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}+\frac {21627}{14} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-2500 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}-\frac {21627}{14} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+2500 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {1045 \sqrt {1-2 x}}{14 (3+5 x)}+\frac {\sqrt {1-2 x}}{2 (2+3 x)^2 (3+5 x)}+\frac {52 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)}-\frac {7209}{7} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x} \left (3965+12228 x+9405 x^2\right )}{14 (2+3 x)^2 (3+5 x)}-\frac {7209}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+1000 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 1.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {18810 x^{3}+15051 x^{2}-4298 x -3965}{14 \left (3+5 x \right ) \sqrt {1-2 x}\, \left (2+3 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}\) | \(76\) |
derivativedivides | \(\frac {10 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {\frac {1251 \left (1-2 x \right )^{\frac {3}{2}}}{7}-423 \sqrt {1-2 x}}{\left (-4-6 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}\) | \(82\) |
default | \(\frac {10 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {1000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {\frac {1251 \left (1-2 x \right )^{\frac {3}{2}}}{7}-423 \sqrt {1-2 x}}{\left (-4-6 x \right )^{2}}-\frac {7209 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}\) | \(82\) |
pseudoelliptic | \(\frac {-158598 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{2} \left (3+5 x \right ) \sqrt {21}+98000 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right )^{2} \left (3+5 x \right ) \sqrt {55}-77 \sqrt {1-2 x}\, \left (9405 x^{2}+12228 x +3965\right )}{1078 \left (2+3 x \right )^{2} \left (3+5 x \right )}\) | \(97\) |
trager | \(-\frac {\left (9405 x^{2}+12228 x +3965\right ) \sqrt {1-2 x}}{14 \left (2+3 x \right )^{2} \left (3+5 x \right )}-\frac {500 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{11}+\frac {7209 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{98}\) | \(123\) |
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Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {49000 \, \sqrt {11} \sqrt {5} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 79299 \, \sqrt {7} \sqrt {3} {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 77 \, {\left (9405 \, x^{2} + 12228 \, x + 3965\right )} \sqrt {-2 \, x + 1}}{1078 \, {\left (45 \, x^{3} + 87 \, x^{2} + 56 \, x + 12\right )}} \]
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Time = 52.05 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.73 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {505 \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 )}{7} - \frac {505 \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 )}{11} - 816 \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 ) + 168 \left (\begin {cases} \frac {\sqrt {21} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )^{2}}\right )}{1029} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right ) - 1100 \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 ) \]
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Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {500}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {7209}{98} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {9405 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 43266 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 49721 \, \sqrt {-2 \, x + 1}}{7 \, {\left (45 \, {\left (2 \, x - 1\right )}^{3} + 309 \, {\left (2 \, x - 1\right )}^{2} + 1414 \, x - 168\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=-\frac {500}{11} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7209}{98} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {25 \, \sqrt {-2 \, x + 1}}{5 \, x + 3} + \frac {9 \, {\left (139 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 329 \, \sqrt {-2 \, x + 1}\right )}}{28 \, {\left (3 \, x + 2\right )}^{2}} \]
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Time = 1.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 (3+5 x)^2} \, dx=\frac {1000\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}-\frac {7209\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{49}-\frac {\frac {7103\,\sqrt {1-2\,x}}{45}-\frac {14422\,{\left (1-2\,x\right )}^{3/2}}{105}+\frac {209\,{\left (1-2\,x\right )}^{5/2}}{7}}{\frac {1414\,x}{45}+\frac {103\,{\left (2\,x-1\right )}^2}{15}+{\left (2\,x-1\right )}^3-\frac {56}{15}} \]
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