Integrand size = 24, antiderivative size = 93 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {4774713 \sqrt {1-2 x}}{50000}+\frac {268707 (1-2 x)^{3/2}}{5000}-\frac {51057 (1-2 x)^{5/2}}{2500}+\frac {5751 (1-2 x)^{7/2}}{1400}-\frac {27}{80} (1-2 x)^{9/2}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]
268707/5000*(1-2*x)^(3/2)-51057/2500*(1-2*x)^(5/2)+5751/1400*(1-2*x)^(7/2) -27/80*(1-2*x)^(9/2)-2/171875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2 )-4774713/50000*(1-2*x)^(1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {3 \sqrt {1-2 x} \left (425872+348095 x+295290 x^2+160875 x^3+39375 x^4\right )}{21875}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]
(-3*Sqrt[1 - 2*x]*(425872 + 348095*x + 295290*x^2 + 160875*x^3 + 39375*x^4 ))/21875 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(3 x+2)^5}{\sqrt {1-2 x} (5 x+3)} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {243}{80} (1-2 x)^{7/2}-\frac {5751}{200} (1-2 x)^{5/2}+\frac {51057}{500} (1-2 x)^{3/2}-\frac {806121 \sqrt {1-2 x}}{5000}+\frac {1}{3125 (5 x+3) \sqrt {1-2 x}}+\frac {4774713}{50000 \sqrt {1-2 x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}}-\frac {27}{80} (1-2 x)^{9/2}+\frac {5751 (1-2 x)^{7/2}}{1400}-\frac {51057 (1-2 x)^{5/2}}{2500}+\frac {268707 (1-2 x)^{3/2}}{5000}-\frac {4774713 \sqrt {1-2 x}}{50000}\) |
(-4774713*Sqrt[1 - 2*x])/50000 + (268707*(1 - 2*x)^(3/2))/5000 - (51057*(1 - 2*x)^(5/2))/2500 + (5751*(1 - 2*x)^(7/2))/1400 - (27*(1 - 2*x)^(9/2))/8 0 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])
3.21.36.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 1.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53
method | result | size |
pseudoelliptic | \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}-\frac {3 \sqrt {1-2 x}\, \left (39375 x^{4}+160875 x^{3}+295290 x^{2}+348095 x +425872\right )}{21875}\) | \(49\) |
risch | \(\frac {3 \left (39375 x^{4}+160875 x^{3}+295290 x^{2}+348095 x +425872\right ) \left (-1+2 x \right )}{21875 \sqrt {1-2 x}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}\) | \(54\) |
derivativedivides | \(\frac {268707 \left (1-2 x \right )^{\frac {3}{2}}}{5000}-\frac {51057 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {5751 \left (1-2 x \right )^{\frac {7}{2}}}{1400}-\frac {27 \left (1-2 x \right )^{\frac {9}{2}}}{80}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}-\frac {4774713 \sqrt {1-2 x}}{50000}\) | \(65\) |
default | \(\frac {268707 \left (1-2 x \right )^{\frac {3}{2}}}{5000}-\frac {51057 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {5751 \left (1-2 x \right )^{\frac {7}{2}}}{1400}-\frac {27 \left (1-2 x \right )^{\frac {9}{2}}}{80}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}-\frac {4774713 \sqrt {1-2 x}}{50000}\) | \(65\) |
trager | \(\left (-\frac {27}{5} x^{4}-\frac {3861}{175} x^{3}-\frac {177174}{4375} x^{2}-\frac {208857}{4375} x -\frac {1277616}{21875}\right ) \sqrt {1-2 x}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{171875}\) | \(74\) |
-2/171875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-3/21875*(1-2*x)^(1 /2)*(39375*x^4+160875*x^3+295290*x^2+348095*x+425872)
Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {3}{21875} \, {\left (39375 \, x^{4} + 160875 \, x^{3} + 295290 \, x^{2} + 348095 \, x + 425872\right )} \sqrt {-2 \, x + 1} + \frac {1}{171875} \, \sqrt {55} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) \]
-3/21875*(39375*x^4 + 160875*x^3 + 295290*x^2 + 348095*x + 425872)*sqrt(-2 *x + 1) + 1/171875*sqrt(55)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3))
Time = 2.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=- \frac {27 \left (1 - 2 x\right )^{\frac {9}{2}}}{80} + \frac {5751 \left (1 - 2 x\right )^{\frac {7}{2}}}{1400} - \frac {51057 \left (1 - 2 x\right )^{\frac {5}{2}}}{2500} + \frac {268707 \left (1 - 2 x\right )^{\frac {3}{2}}}{5000} - \frac {4774713 \sqrt {1 - 2 x}}{50000} + \frac {\sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{171875} \]
-27*(1 - 2*x)**(9/2)/80 + 5751*(1 - 2*x)**(7/2)/1400 - 51057*(1 - 2*x)**(5 /2)/2500 + 268707*(1 - 2*x)**(3/2)/5000 - 4774713*sqrt(1 - 2*x)/50000 + sq rt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5)) /171875
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {27}{80} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} + \frac {5751}{1400} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {51057}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {268707}{5000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{171875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {4774713}{50000} \, \sqrt {-2 \, x + 1} \]
-27/80*(-2*x + 1)^(9/2) + 5751/1400*(-2*x + 1)^(7/2) - 51057/2500*(-2*x + 1)^(5/2) + 268707/5000*(-2*x + 1)^(3/2) + 1/171875*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 4774713/50000*sqrt(- 2*x + 1)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=-\frac {27}{80} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} - \frac {5751}{1400} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {51057}{2500} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {268707}{5000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{171875} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {4774713}{50000} \, \sqrt {-2 \, x + 1} \]
-27/80*(2*x - 1)^4*sqrt(-2*x + 1) - 5751/1400*(2*x - 1)^3*sqrt(-2*x + 1) - 51057/2500*(2*x - 1)^2*sqrt(-2*x + 1) + 268707/5000*(-2*x + 1)^(3/2) + 1/ 171875*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) - 4774713/50000*sqrt(-2*x + 1)
Time = 1.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)} \, dx=\frac {268707\,{\left (1-2\,x\right )}^{3/2}}{5000}-\frac {4774713\,\sqrt {1-2\,x}}{50000}-\frac {51057\,{\left (1-2\,x\right )}^{5/2}}{2500}+\frac {5751\,{\left (1-2\,x\right )}^{7/2}}{1400}-\frac {27\,{\left (1-2\,x\right )}^{9/2}}{80}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{171875} \]