Integrand size = 24, antiderivative size = 93 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=\frac {16807}{528 (1-2 x)^{3/2}}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {51057}{500} \sqrt {1-2 x}+\frac {1917}{200} (1-2 x)^{3/2}-\frac {243}{400} (1-2 x)^{5/2}-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \]
16807/528/(1-2*x)^(3/2)+1917/200*(1-2*x)^(3/2)-243/400*(1-2*x)^(5/2)-2/831 875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-156065/968/(1-2*x)^(1/2) -51057/500*(1-2*x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=\frac {-\frac {55 \left (10097264-30775791 x+13976226 x^2+2597265 x^3+441045 x^4\right )}{(1-2 x)^{3/2}}-6 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{2495625} \]
((-55*(10097264 - 30775791*x + 13976226*x^2 + 2597265*x^3 + 441045*x^4))/( 1 - 2*x)^(3/2) - 6*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/2495625
Time = 0.22 (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 = {98, 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.
\(\displaystyle \int \frac {(3 x+2)^5}{(1-2 x)^{5/2} (5 x+3)} \, dx\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \int \left (\frac {243 x^2}{20 \sqrt {1-2 x}}+\frac {1134 x}{25 \sqrt {1-2 x}}+\frac {152793}{2000 \sqrt {1-2 x}}+\frac {1}{15125 \sqrt {1-2 x} (5 x+3)}-\frac {156065}{968 (1-2 x)^{3/2}}+\frac {16807}{176 (1-2 x)^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}}-\frac {243}{400} (1-2 x)^{5/2}+\frac {1917}{200} (1-2 x)^{3/2}-\frac {51057}{500} \sqrt {1-2 x}-\frac {156065}{968 \sqrt {1-2 x}}+\frac {16807}{528 (1-2 x)^{3/2}}\) |
16807/(528*(1 - 2*x)^(3/2)) - 156065/(968*Sqrt[1 - 2*x]) - (51057*Sqrt[1 - 2*x])/500 + (1917*(1 - 2*x)^(3/2))/200 - (243*(1 - 2*x)^(5/2))/400 - (2*A rcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15125*Sqrt[55])
3.22.70.3.1 Defintions of rubi rules used
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Time = 1.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.60
method | result | size |
pseudoelliptic | \(-\frac {243 \left (-\frac {4 \sqrt {1-2 x}\, \sqrt {55}\, \left (x -\frac {1}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{8085825}+x^{4}+\frac {53 x^{3}}{9}+\frac {1426 x^{2}}{45}-\frac {10258597 x}{147015}+\frac {10097264}{441045}\right )}{25 \left (1-2 x \right )^{\frac {3}{2}}}\) | \(56\) |
derivativedivides | \(\frac {16807}{528 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {1917 \left (1-2 x \right )^{\frac {3}{2}}}{200}-\frac {243 \left (1-2 x \right )^{\frac {5}{2}}}{400}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{831875}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {51057 \sqrt {1-2 x}}{500}\) | \(65\) |
default | \(\frac {16807}{528 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {1917 \left (1-2 x \right )^{\frac {3}{2}}}{200}-\frac {243 \left (1-2 x \right )^{\frac {5}{2}}}{400}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{831875}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {51057 \sqrt {1-2 x}}{500}\) | \(65\) |
trager | \(-\frac {\left (441045 x^{4}+2597265 x^{3}+13976226 x^{2}-30775791 x +10097264\right ) \sqrt {1-2 x}}{45375 \left (-1+2 x \right )^{2}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{831875}\) | \(82\) |
-243/25/(1-2*x)^(3/2)*(-4/8085825*(1-2*x)^(1/2)*55^(1/2)*(x-1/2)*arctanh(1 /11*55^(1/2)*(1-2*x)^(1/2))+x^4+53/9*x^3+1426/45*x^2-10258597/147015*x+100 97264/441045)
Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=\frac {3 \, \sqrt {55} {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (441045 \, x^{4} + 2597265 \, x^{3} + 13976226 \, x^{2} - 30775791 \, x + 10097264\right )} \sqrt {-2 \, x + 1}}{2495625 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
1/2495625*(3*sqrt(55)*(4*x^2 - 4*x + 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(441045*x^4 + 2597265*x^3 + 13976226*x^2 - 30775791* x + 10097264)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
Time = 3.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=- \frac {243 \left (1 - 2 x\right )^{\frac {5}{2}}}{400} + \frac {1917 \left (1 - 2 x\right )^{\frac {3}{2}}}{200} - \frac {51057 \sqrt {1 - 2 x}}{500} + \frac {\sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{831875} - \frac {156065}{968 \sqrt {1 - 2 x}} + \frac {16807}{528 \left (1 - 2 x\right )^{\frac {3}{2}}} \]
-243*(1 - 2*x)**(5/2)/400 + 1917*(1 - 2*x)**(3/2)/200 - 51057*sqrt(1 - 2*x )/500 + sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sq rt(55)/5))/831875 - 156065/(968*sqrt(1 - 2*x)) + 16807/(528*(1 - 2*x)**(3/ 2))
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=-\frac {243}{400} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {1917}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{831875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {51057}{500} \, \sqrt {-2 \, x + 1} + \frac {2401 \, {\left (780 \, x - 313\right )}}{5808 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \]
-243/400*(-2*x + 1)^(5/2) + 1917/200*(-2*x + 1)^(3/2) + 1/831875*sqrt(55)* log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 51057/ 500*sqrt(-2*x + 1) + 2401/5808*(780*x - 313)/(-2*x + 1)^(3/2)
Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=-\frac {243}{400} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {1917}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{831875} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {51057}{500} \, \sqrt {-2 \, x + 1} - \frac {2401 \, {\left (780 \, x - 313\right )}}{5808 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \]
-243/400*(2*x - 1)^2*sqrt(-2*x + 1) + 1917/200*(-2*x + 1)^(3/2) + 1/831875 *sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( -2*x + 1))) - 51057/500*sqrt(-2*x + 1) - 2401/5808*(780*x - 313)/((2*x - 1 )*sqrt(-2*x + 1))
Time = 1.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^5}{(1-2 x)^{5/2} (3+5 x)} \, dx=\frac {\frac {156065\,x}{484}-\frac {751513}{5808}}{{\left (1-2\,x\right )}^{3/2}}-\frac {51057\,\sqrt {1-2\,x}}{500}+\frac {1917\,{\left (1-2\,x\right )}^{3/2}}{200}-\frac {243\,{\left (1-2\,x\right )}^{5/2}}{400}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{831875} \]