Integrand size = 22, antiderivative size = 89 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {616}{625} \sqrt {1-2 x}+\frac {56}{375} (1-2 x)^{3/2}+\frac {56 (1-2 x)^{5/2}}{1375}-\frac {(1-2 x)^{7/2}}{55 (3+5 x)}-\frac {616}{625} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
56/375*(1-2*x)^(3/2)+56/1375*(1-2*x)^(5/2)-1/55*(1-2*x)^(7/2)/(3+5*x)-616/ 3125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+616/625*(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (6579+8630 x-3820 x^2+1800 x^3\right )}{3+5 x}-1848 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{9375} \]
((5*Sqrt[1 - 2*x]*(6579 + 8630*x - 3820*x^2 + 1800*x^3))/(3 + 5*x) - 1848* Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/9375
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 60, 60, 60, 73, 219}
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.
\(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {28}{55} \int \frac {(1-2 x)^{5/2}}{5 x+3}dx-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {28}{55} \left (\frac {11}{5} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {28}{55} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {28}{55} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {28}{55} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {28}{55} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{7/2}}{55 (5 x+3)}\) |
-1/55*(1 - 2*x)^(7/2)/(3 + 5*x) + (28*((2*(1 - 2*x)^(5/2))/25 + (11*((2*(1 - 2*x)^(3/2))/15 + (11*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[ 5/11]*Sqrt[1 - 2*x]])/5))/5))/5))/55
3.20.83.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 1.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {3600 x^{4}-9440 x^{3}+21080 x^{2}+4528 x -6579}{1875 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {616 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(56\) |
pseudoelliptic | \(\frac {-1848 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+5 \sqrt {1-2 x}\, \left (1800 x^{3}-3820 x^{2}+8630 x +6579\right )}{28125+46875 x}\) | \(57\) |
derivativedivides | \(\frac {6 \left (1-2 x \right )^{\frac {5}{2}}}{125}+\frac {62 \left (1-2 x \right )^{\frac {3}{2}}}{375}+\frac {638 \sqrt {1-2 x}}{625}+\frac {242 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {616 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(63\) |
default | \(\frac {6 \left (1-2 x \right )^{\frac {5}{2}}}{125}+\frac {62 \left (1-2 x \right )^{\frac {3}{2}}}{375}+\frac {638 \sqrt {1-2 x}}{625}+\frac {242 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {616 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(63\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (1800 x^{3}-3820 x^{2}+8630 x +6579\right )}{5625+9375 x}+\frac {308 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3125}\) | \(77\) |
-1/1875*(3600*x^4-9440*x^3+21080*x^2+4528*x-6579)/(3+5*x)/(1-2*x)^(1/2)-61 6/3125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {924 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (1800 \, x^{3} - 3820 \, x^{2} + 8630 \, x + 6579\right )} \sqrt {-2 \, x + 1}}{9375 \, {\left (5 \, x + 3\right )}} \]
1/9375*(924*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1 ) + 5*x - 8)/(5*x + 3)) + 5*(1800*x^3 - 3820*x^2 + 8630*x + 6579)*sqrt(-2* x + 1))/(5*x + 3)
Time = 27.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.21 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {6 \left (1 - 2 x\right )^{\frac {5}{2}}}{125} + \frac {62 \left (1 - 2 x\right )^{\frac {3}{2}}}{375} + \frac {638 \sqrt {1 - 2 x}}{625} + \frac {297 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} - \frac {5324 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
6*(1 - 2*x)**(5/2)/125 + 62*(1 - 2*x)**(3/2)/375 + 638*sqrt(1 - 2*x)/625 + 297*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt( 55)/5))/3125 - 5324*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x) /11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > - sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {6}{125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {62}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {308}{3125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {638}{625} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
6/125*(-2*x + 1)^(5/2) + 62/375*(-2*x + 1)^(3/2) + 308/3125*sqrt(55)*log(- (sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 638/625*sqr t(-2*x + 1) - 121/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {6}{125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {62}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {308}{3125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {638}{625} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
6/125*(2*x - 1)^2*sqrt(-2*x + 1) + 62/375*(-2*x + 1)^(3/2) + 308/3125*sqrt (55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 638/625*sqrt(-2*x + 1) - 121/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.69 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx=\frac {638\,\sqrt {1-2\,x}}{625}-\frac {242\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}+\frac {62\,{\left (1-2\,x\right )}^{3/2}}{375}+\frac {6\,{\left (1-2\,x\right )}^{5/2}}{125}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,616{}\mathrm {i}}{3125} \]