Integrand size = 24, antiderivative size = 121 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}+624 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 121, 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} \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=624 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )+\frac {995 \sqrt {1-2 x}}{22 (5 x+3)}-\frac {15 \sqrt {1-2 x}}{2 (5 x+3)^2}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^2} \]
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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+3 x) (3+5 x)^2}-\int \frac {-18+25 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^3} \, dx \\ & = -\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {1}{22} \int \frac {-1298+1485 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx \\ & = -\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}-\frac {1}{242} \int \frac {-53614+32835 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx \\ & = -\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}-936 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {33325}{22} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}+936 \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {33325}{22} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {\sqrt {1-2 x} \left (5662+18410 x+14925 x^2\right )}{22 (2+3 x) (3+5 x)^2}+624 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
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Time = 3.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {29850 x^{3}+21895 x^{2}-7086 x -5662}{22 \left (3+5 x \right )^{2} \sqrt {1-2 x}\, \left (2+3 x \right )}+\frac {624 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}-\frac {6665 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}\) | \(76\) |
derivativedivides | \(\frac {-\frac {3325 \left (1-2 x \right )^{\frac {3}{2}}}{11}+655 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {6665 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}-\frac {6 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}+\frac {624 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}\) | \(82\) |
default | \(\frac {-\frac {3325 \left (1-2 x \right )^{\frac {3}{2}}}{11}+655 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {6665 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}-\frac {6 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}+\frac {624 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}\) | \(82\) |
pseudoelliptic | \(\frac {151008 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \left (3+5 x \right )^{2} \sqrt {21}-93310 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right ) \left (3+5 x \right )^{2} \sqrt {55}+77 \sqrt {1-2 x}\, \left (14925 x^{2}+18410 x +5662\right )}{1694 \left (2+3 x \right ) \left (3+5 x \right )^{2}}\) | \(97\) |
trager | \(\frac {\left (14925 x^{2}+18410 x +5662\right ) \sqrt {1-2 x}}{22 \left (3+5 x \right )^{2} \left (2+3 x \right )}-\frac {6665 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{242}-\frac {312 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{7}\) | \(123\) |
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Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {46655 \, \sqrt {11} \sqrt {5} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 75504 \, \sqrt {7} \sqrt {3} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \, {\left (14925 \, x^{2} + 18410 \, x + 5662\right )} \sqrt {-2 \, x + 1}}{1694 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \]
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Time = 52.50 (sec) , antiderivative size = 488, normalized size of antiderivative = 4.03 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=- \frac {309 \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 {309 \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} + 252 \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 ) + 1360 \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 ) + 440 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \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 = 1.06 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {6665}{242} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {312}{7} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {14925 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 66670 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 74393 \, \sqrt {-2 \, x + 1}}{11 \, {\left (75 \, {\left (2 \, x - 1\right )}^{3} + 505 \, {\left (2 \, x - 1\right )}^{2} + 2266 \, x - 286\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {6665}{242} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {312}{7} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {9 \, \sqrt {-2 \, x + 1}}{3 \, x + 2} - \frac {5 \, {\left (665 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1441 \, \sqrt {-2 \, x + 1}\right )}}{44 \, {\left (5 \, x + 3\right )}^{2}} \]
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Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {624\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{7}-\frac {6665\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{121}+\frac {\frac {6763\,\sqrt {1-2\,x}}{75}-\frac {13334\,{\left (1-2\,x\right )}^{3/2}}{165}+\frac {199\,{\left (1-2\,x\right )}^{5/2}}{11}}{\frac {2266\,x}{75}+\frac {101\,{\left (2\,x-1\right )}^2}{15}+{\left (2\,x-1\right )}^3-\frac {286}{75}} \]
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