Integrand size = 24, antiderivative size = 95 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=\frac {22 \sqrt {1-2 x}}{3125}+\frac {2 (1-2 x)^{3/2}}{1875}-\frac {3897 (1-2 x)^{5/2}}{2500}+\frac {162}{175} (1-2 x)^{7/2}-\frac {3}{20} (1-2 x)^{9/2}-\frac {22 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125} \]
2/1875*(1-2*x)^(3/2)-3897/2500*(1-2*x)^(5/2)+162/175*(1-2*x)^(7/2)-3/20*(1 -2*x)^(9/2)-22/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+22/3125 *(1-2*x)^(1/2)
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=\frac {-5 \sqrt {1-2 x} \left (50858-123295 x-83565 x^2+171000 x^3+157500 x^4\right )-462 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{328125} \]
(-5*Sqrt[1 - 2*x]*(50858 - 123295*x - 83565*x^2 + 171000*x^3 + 157500*x^4) - 462*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/328125
Time = 0.21 (sec) , antiderivative size = 95, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^3}{5 x+3} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {27}{20} (1-2 x)^{7/2}-\frac {162}{25} (1-2 x)^{5/2}+\frac {(1-2 x)^{3/2}}{125 (5 x+3)}+\frac {3897}{500} (1-2 x)^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {22 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125}-\frac {3}{20} (1-2 x)^{9/2}+\frac {162}{175} (1-2 x)^{7/2}-\frac {3897 (1-2 x)^{5/2}}{2500}+\frac {2 (1-2 x)^{3/2}}{1875}+\frac {22 \sqrt {1-2 x}}{3125}\) |
(22*Sqrt[1 - 2*x])/3125 + (2*(1 - 2*x)^(3/2))/1875 - (3897*(1 - 2*x)^(5/2) )/2500 + (162*(1 - 2*x)^(7/2))/175 - (3*(1 - 2*x)^(9/2))/20 - (22*Sqrt[11/ 5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3125
3.19.98.3.1 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 = 0.99 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.52
method | result | size |
pseudoelliptic | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}-\frac {\sqrt {1-2 x}\, \left (157500 x^{4}+171000 x^{3}-83565 x^{2}-123295 x +50858\right )}{65625}\) | \(49\) |
risch | \(\frac {\left (157500 x^{4}+171000 x^{3}-83565 x^{2}-123295 x +50858\right ) \left (-1+2 x \right )}{65625 \sqrt {1-2 x}}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) | \(54\) |
derivativedivides | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{1875}-\frac {3897 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {162 \left (1-2 x \right )^{\frac {7}{2}}}{175}-\frac {3 \left (1-2 x \right )^{\frac {9}{2}}}{20}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}+\frac {22 \sqrt {1-2 x}}{3125}\) | \(65\) |
default | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{1875}-\frac {3897 \left (1-2 x \right )^{\frac {5}{2}}}{2500}+\frac {162 \left (1-2 x \right )^{\frac {7}{2}}}{175}-\frac {3 \left (1-2 x \right )^{\frac {9}{2}}}{20}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}+\frac {22 \sqrt {1-2 x}}{3125}\) | \(65\) |
trager | \(\left (-\frac {12}{5} x^{4}-\frac {456}{175} x^{3}+\frac {5571}{4375} x^{2}+\frac {24659}{13125} x -\frac {50858}{65625}\right ) \sqrt {1-2 x}+\frac {11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{15625}\) | \(74\) |
-22/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/65625*(1-2*x)^(1 /2)*(157500*x^4+171000*x^3-83565*x^2-123295*x+50858)
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=\frac {11}{15625} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - \frac {1}{65625} \, {\left (157500 \, x^{4} + 171000 \, x^{3} - 83565 \, x^{2} - 123295 \, x + 50858\right )} \sqrt {-2 \, x + 1} \]
11/15625*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/ (5*x + 3)) - 1/65625*(157500*x^4 + 171000*x^3 - 83565*x^2 - 123295*x + 508 58)*sqrt(-2*x + 1)
Time = 1.84 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=- \frac {3 \left (1 - 2 x\right )^{\frac {9}{2}}}{20} + \frac {162 \left (1 - 2 x\right )^{\frac {7}{2}}}{175} - \frac {3897 \left (1 - 2 x\right )^{\frac {5}{2}}}{2500} + \frac {2 \left (1 - 2 x\right )^{\frac {3}{2}}}{1875} + \frac {22 \sqrt {1 - 2 x}}{3125} + \frac {11 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{15625} \]
-3*(1 - 2*x)**(9/2)/20 + 162*(1 - 2*x)**(7/2)/175 - 3897*(1 - 2*x)**(5/2)/ 2500 + 2*(1 - 2*x)**(3/2)/1875 + 22*sqrt(1 - 2*x)/3125 + 11*sqrt(55)*(log( sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/15625
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=-\frac {3}{20} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} + \frac {162}{175} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {3897}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {2}{1875} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{15625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {22}{3125} \, \sqrt {-2 \, x + 1} \]
-3/20*(-2*x + 1)^(9/2) + 162/175*(-2*x + 1)^(7/2) - 3897/2500*(-2*x + 1)^( 5/2) + 2/1875*(-2*x + 1)^(3/2) + 11/15625*sqrt(55)*log(-(sqrt(55) - 5*sqrt (-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 22/3125*sqrt(-2*x + 1)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=-\frac {3}{20} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} - \frac {162}{175} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {3897}{2500} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {2}{1875} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{15625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {22}{3125} \, \sqrt {-2 \, x + 1} \]
-3/20*(2*x - 1)^4*sqrt(-2*x + 1) - 162/175*(2*x - 1)^3*sqrt(-2*x + 1) - 38 97/2500*(2*x - 1)^2*sqrt(-2*x + 1) + 2/1875*(-2*x + 1)^(3/2) + 11/15625*sq rt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2* x + 1))) + 22/3125*sqrt(-2*x + 1)
Time = 1.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx=\frac {22\,\sqrt {1-2\,x}}{3125}+\frac {2\,{\left (1-2\,x\right )}^{3/2}}{1875}-\frac {3897\,{\left (1-2\,x\right )}^{5/2}}{2500}+\frac {162\,{\left (1-2\,x\right )}^{7/2}}{175}-\frac {3\,{\left (1-2\,x\right )}^{9/2}}{20}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22{}\mathrm {i}}{15625} \]