Integrand size = 24, antiderivative size = 121 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=\frac {22 \sqrt {1-2 x}}{78125}+\frac {2 (1-2 x)^{3/2}}{46875}-\frac {4774713 (1-2 x)^{5/2}}{250000}+\frac {806121 (1-2 x)^{7/2}}{35000}-\frac {5673}{500} (1-2 x)^{9/2}+\frac {5751 (1-2 x)^{11/2}}{2200}-\frac {243 (1-2 x)^{13/2}}{1040}-\frac {22 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \]
2/46875*(1-2*x)^(3/2)-4774713/250000*(1-2*x)^(5/2)+806121/35000*(1-2*x)^(7 /2)-5673/500*(1-2*x)^(9/2)+5751/2200*(1-2*x)^(11/2)-243/1040*(1-2*x)^(13/2 )-22/390625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+22/78125*(1-2*x) ^(1/2)
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.59 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=\frac {-5 \sqrt {1-2 x} \left (1180568944-1321809935 x-4276774170 x^2-1659418875 x^3+6683000625 x^4+9100350000 x^5+3508312500 x^6\right )-66066 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1173046875} \]
(-5*Sqrt[1 - 2*x]*(1180568944 - 1321809935*x - 4276774170*x^2 - 1659418875 *x^3 + 6683000625*x^4 + 9100350000*x^5 + 3508312500*x^6) - 66066*Sqrt[55]* ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1173046875
Time = 0.22 (sec) , antiderivative size = 121, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^5}{5 x+3} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {243}{80} (1-2 x)^{11/2}-\frac {5751}{200} (1-2 x)^{9/2}+\frac {51057}{500} (1-2 x)^{7/2}-\frac {806121 (1-2 x)^{5/2}}{5000}+\frac {(1-2 x)^{3/2}}{3125 (5 x+3)}+\frac {4774713 (1-2 x)^{3/2}}{50000}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {22 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125}-\frac {243 (1-2 x)^{13/2}}{1040}+\frac {5751 (1-2 x)^{11/2}}{2200}-\frac {5673}{500} (1-2 x)^{9/2}+\frac {806121 (1-2 x)^{7/2}}{35000}-\frac {4774713 (1-2 x)^{5/2}}{250000}+\frac {2 (1-2 x)^{3/2}}{46875}+\frac {22 \sqrt {1-2 x}}{78125}\) |
(22*Sqrt[1 - 2*x])/78125 + (2*(1 - 2*x)^(3/2))/46875 - (4774713*(1 - 2*x)^ (5/2))/250000 + (806121*(1 - 2*x)^(7/2))/35000 - (5673*(1 - 2*x)^(9/2))/50 0 + (5751*(1 - 2*x)^(11/2))/2200 - (243*(1 - 2*x)^(13/2))/1040 - (22*Sqrt[ 11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/78125
3.19.96.3.1 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 = 59, normalized size of antiderivative = 0.49
method | result | size |
pseudoelliptic | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}-\frac {\sqrt {1-2 x}\, \left (3508312500 x^{6}+9100350000 x^{5}+6683000625 x^{4}-1659418875 x^{3}-4276774170 x^{2}-1321809935 x +1180568944\right )}{234609375}\) | \(59\) |
risch | \(\frac {\left (3508312500 x^{6}+9100350000 x^{5}+6683000625 x^{4}-1659418875 x^{3}-4276774170 x^{2}-1321809935 x +1180568944\right ) \left (-1+2 x \right )}{234609375 \sqrt {1-2 x}}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}\) | \(64\) |
derivativedivides | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{46875}-\frac {4774713 \left (1-2 x \right )^{\frac {5}{2}}}{250000}+\frac {806121 \left (1-2 x \right )^{\frac {7}{2}}}{35000}-\frac {5673 \left (1-2 x \right )^{\frac {9}{2}}}{500}+\frac {5751 \left (1-2 x \right )^{\frac {11}{2}}}{2200}-\frac {243 \left (1-2 x \right )^{\frac {13}{2}}}{1040}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}+\frac {22 \sqrt {1-2 x}}{78125}\) | \(83\) |
default | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{46875}-\frac {4774713 \left (1-2 x \right )^{\frac {5}{2}}}{250000}+\frac {806121 \left (1-2 x \right )^{\frac {7}{2}}}{35000}-\frac {5673 \left (1-2 x \right )^{\frac {9}{2}}}{500}+\frac {5751 \left (1-2 x \right )^{\frac {11}{2}}}{2200}-\frac {243 \left (1-2 x \right )^{\frac {13}{2}}}{1040}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}+\frac {22 \sqrt {1-2 x}}{78125}\) | \(83\) |
trager | \(\left (-\frac {972}{65} x^{6}-\frac {138672}{3575} x^{5}-\frac {509181}{17875} x^{4}+\frac {4425117}{625625} x^{3}+\frac {285118278}{15640625} x^{2}+\frac {264361987}{46921875} x -\frac {1180568944}{234609375}\right ) \sqrt {1-2 x}+\frac {11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{390625}\) | \(84\) |
-22/390625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/234609375*(1-2* x)^(1/2)*(3508312500*x^6+9100350000*x^5+6683000625*x^4-1659418875*x^3-4276 774170*x^2-1321809935*x+1180568944)
Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=\frac {11}{390625} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - \frac {1}{234609375} \, {\left (3508312500 \, x^{6} + 9100350000 \, x^{5} + 6683000625 \, x^{4} - 1659418875 \, x^{3} - 4276774170 \, x^{2} - 1321809935 \, x + 1180568944\right )} \sqrt {-2 \, x + 1} \]
11/390625*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8) /(5*x + 3)) - 1/234609375*(3508312500*x^6 + 9100350000*x^5 + 6683000625*x^ 4 - 1659418875*x^3 - 4276774170*x^2 - 1321809935*x + 1180568944)*sqrt(-2*x + 1)
Time = 2.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=- \frac {243 \left (1 - 2 x\right )^{\frac {13}{2}}}{1040} + \frac {5751 \left (1 - 2 x\right )^{\frac {11}{2}}}{2200} - \frac {5673 \left (1 - 2 x\right )^{\frac {9}{2}}}{500} + \frac {806121 \left (1 - 2 x\right )^{\frac {7}{2}}}{35000} - \frac {4774713 \left (1 - 2 x\right )^{\frac {5}{2}}}{250000} + \frac {2 \left (1 - 2 x\right )^{\frac {3}{2}}}{46875} + \frac {22 \sqrt {1 - 2 x}}{78125} + \frac {11 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{390625} \]
-243*(1 - 2*x)**(13/2)/1040 + 5751*(1 - 2*x)**(11/2)/2200 - 5673*(1 - 2*x) **(9/2)/500 + 806121*(1 - 2*x)**(7/2)/35000 - 4774713*(1 - 2*x)**(5/2)/250 000 + 2*(1 - 2*x)**(3/2)/46875 + 22*sqrt(1 - 2*x)/78125 + 11*sqrt(55)*(log (sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/390625
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=-\frac {243}{1040} \, {\left (-2 \, x + 1\right )}^{\frac {13}{2}} + \frac {5751}{2200} \, {\left (-2 \, x + 1\right )}^{\frac {11}{2}} - \frac {5673}{500} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} + \frac {806121}{35000} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {4774713}{250000} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {2}{46875} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{390625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {22}{78125} \, \sqrt {-2 \, x + 1} \]
-243/1040*(-2*x + 1)^(13/2) + 5751/2200*(-2*x + 1)^(11/2) - 5673/500*(-2*x + 1)^(9/2) + 806121/35000*(-2*x + 1)^(7/2) - 4774713/250000*(-2*x + 1)^(5 /2) + 2/46875*(-2*x + 1)^(3/2) + 11/390625*sqrt(55)*log(-(sqrt(55) - 5*sqr t(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 22/78125*sqrt(-2*x + 1)
Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=-\frac {243}{1040} \, {\left (2 \, x - 1\right )}^{6} \sqrt {-2 \, x + 1} - \frac {5751}{2200} \, {\left (2 \, x - 1\right )}^{5} \sqrt {-2 \, x + 1} - \frac {5673}{500} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} - \frac {806121}{35000} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {4774713}{250000} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {2}{46875} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{390625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {22}{78125} \, \sqrt {-2 \, x + 1} \]
-243/1040*(2*x - 1)^6*sqrt(-2*x + 1) - 5751/2200*(2*x - 1)^5*sqrt(-2*x + 1 ) - 5673/500*(2*x - 1)^4*sqrt(-2*x + 1) - 806121/35000*(2*x - 1)^3*sqrt(-2 *x + 1) - 4774713/250000*(2*x - 1)^2*sqrt(-2*x + 1) + 2/46875*(-2*x + 1)^( 3/2) + 11/390625*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sq rt(55) + 5*sqrt(-2*x + 1))) + 22/78125*sqrt(-2*x + 1)
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^5}{3+5 x} \, dx=\frac {22\,\sqrt {1-2\,x}}{78125}+\frac {2\,{\left (1-2\,x\right )}^{3/2}}{46875}-\frac {4774713\,{\left (1-2\,x\right )}^{5/2}}{250000}+\frac {806121\,{\left (1-2\,x\right )}^{7/2}}{35000}-\frac {5673\,{\left (1-2\,x\right )}^{9/2}}{500}+\frac {5751\,{\left (1-2\,x\right )}^{11/2}}{2200}-\frac {243\,{\left (1-2\,x\right )}^{13/2}}{1040}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22{}\mathrm {i}}{390625} \]