Integrand size = 24, antiderivative size = 120 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {1512 \sqrt {1-2 x} (2+3 x)^2}{75625}-\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}-\frac {189 \sqrt {1-2 x} (8976+2875 x)}{756250}-\frac {22113 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{378125 \sqrt {55}} \]
[Out]
Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 154, 158, 152, 65, 212} \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {22113 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{378125 \sqrt {55}}-\frac {\sqrt {1-2 x} (3 x+2)^4}{110 (5 x+3)^2}-\frac {201 \sqrt {1-2 x} (3 x+2)^3}{6050 (5 x+3)}-\frac {1512 \sqrt {1-2 x} (3 x+2)^2}{75625}-\frac {189 \sqrt {1-2 x} (2875 x+8976)}{756250} \]
[In]
[Out]
Rule 65
Rule 100
Rule 152
Rule 154
Rule 158
Rule 212
Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {1}{110} \int \frac {(-150-183 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^2} \, dx \\ & = -\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}-\frac {\int \frac {(-6237-3024 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx}{6050} \\ & = -\frac {1512 \sqrt {1-2 x} (2+3 x)^2}{75625}-\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}+\frac {\int \frac {(2+3 x) (348138+543375 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{151250} \\ & = -\frac {1512 \sqrt {1-2 x} (2+3 x)^2}{75625}-\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}-\frac {189 \sqrt {1-2 x} (8976+2875 x)}{756250}+\frac {22113 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{756250} \\ & = -\frac {1512 \sqrt {1-2 x} (2+3 x)^2}{75625}-\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}-\frac {189 \sqrt {1-2 x} (8976+2875 x)}{756250}-\frac {22113 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{756250} \\ & = -\frac {1512 \sqrt {1-2 x} (2+3 x)^2}{75625}-\frac {\sqrt {1-2 x} (2+3 x)^4}{110 (3+5 x)^2}-\frac {201 \sqrt {1-2 x} (2+3 x)^3}{6050 (3+5 x)}-\frac {189 \sqrt {1-2 x} (8976+2875 x)}{756250}-\frac {22113 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{378125 \sqrt {55}} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {-\frac {55 \sqrt {1-2 x} \left (16525496+63610155 x+76970520 x^2+32506650 x^3+7350750 x^4\right )}{(3+5 x)^2}-44226 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{41593750} \]
[In]
[Out]
Time = 3.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51
method | result | size |
risch | \(\frac {14701500 x^{5}+57662550 x^{4}+121434390 x^{3}+50249790 x^{2}-30559163 x -16525496}{756250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {22113 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{20796875}\) | \(61\) |
pseudoelliptic | \(\frac {-44226 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (7350750 x^{4}+32506650 x^{3}+76970520 x^{2}+63610155 x +16525496\right )}{41593750 \left (3+5 x \right )^{2}}\) | \(65\) |
derivativedivides | \(-\frac {243 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {513 \left (1-2 x \right )^{\frac {3}{2}}}{625}-\frac {39393 \sqrt {1-2 x}}{12500}+\frac {\frac {333 \left (1-2 x \right )^{\frac {3}{2}}}{75625}-\frac {67 \sqrt {1-2 x}}{6875}}{\left (-6-10 x \right )^{2}}-\frac {22113 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{20796875}\) | \(75\) |
default | \(-\frac {243 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {513 \left (1-2 x \right )^{\frac {3}{2}}}{625}-\frac {39393 \sqrt {1-2 x}}{12500}+\frac {\frac {333 \left (1-2 x \right )^{\frac {3}{2}}}{75625}-\frac {67 \sqrt {1-2 x}}{6875}}{\left (-6-10 x \right )^{2}}-\frac {22113 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{20796875}\) | \(75\) |
trager | \(-\frac {\left (7350750 x^{4}+32506650 x^{3}+76970520 x^{2}+63610155 x +16525496\right ) \sqrt {1-2 x}}{756250 \left (3+5 x \right )^{2}}-\frac {22113 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{41593750}\) | \(82\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {22113 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (7350750 \, x^{4} + 32506650 \, x^{3} + 76970520 \, x^{2} + 63610155 \, x + 16525496\right )} \sqrt {-2 \, x + 1}}{41593750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
[In]
[Out]
Timed out. \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {243}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {513}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {22113}{41593750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {39393}{12500} \, \sqrt {-2 \, x + 1} + \frac {333 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{75625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
[In]
[Out]
none
Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {243}{2500} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {513}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {22113}{41593750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {39393}{12500} \, \sqrt {-2 \, x + 1} + \frac {333 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{302500 \, {\left (5 \, x + 3\right )}^{2}} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {513\,{\left (1-2\,x\right )}^{3/2}}{625}-\frac {39393\,\sqrt {1-2\,x}}{12500}-\frac {243\,{\left (1-2\,x\right )}^{5/2}}{2500}-\frac {\frac {67\,\sqrt {1-2\,x}}{171875}-\frac {333\,{\left (1-2\,x\right )}^{3/2}}{1890625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22113{}\mathrm {i}}{20796875} \]
[In]
[Out]