Integrand size = 22, antiderivative size = 69 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=\frac {22}{125} \sqrt {1-2 x}+\frac {2}{75} (1-2 x)^{3/2}-\frac {3}{25} (1-2 x)^{5/2}-\frac {22}{125} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
2/75*(1-2*x)^(3/2)-3/25*(1-2*x)^(5/2)-22/625*arctanh(1/11*55^(1/2)*(1-2*x) ^(1/2))*55^(1/2)+22/125*(1-2*x)^(1/2)
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=\frac {5 \sqrt {1-2 x} \left (31+160 x-180 x^2\right )-66 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1875} \]
(5*Sqrt[1 - 2*x]*(31 + 160*x - 180*x^2) - 66*Sqrt[55]*ArcTanh[Sqrt[5/11]*S qrt[1 - 2*x]])/1875
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {90, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)}{5 x+3} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{5} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {3}{25} (1-2 x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {3}{25} (1-2 x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {3}{25} (1-2 x)^{5/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {3}{25} (1-2 x)^{5/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {3}{25} (1-2 x)^{5/2}\) |
(-3*(1 - 2*x)^(5/2))/25 + ((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
3.19.100.3.1 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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 0.98 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}-\frac {\sqrt {1-2 x}\, \left (180 x^{2}-160 x -31\right )}{375}\) | \(39\) |
risch | \(\frac {\left (180 x^{2}-160 x -31\right ) \left (-1+2 x \right )}{375 \sqrt {1-2 x}}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\) | \(44\) |
derivativedivides | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {3 \left (1-2 x \right )^{\frac {5}{2}}}{25}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}+\frac {22 \sqrt {1-2 x}}{125}\) | \(47\) |
default | \(\frac {2 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {3 \left (1-2 x \right )^{\frac {5}{2}}}{25}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}+\frac {22 \sqrt {1-2 x}}{125}\) | \(47\) |
trager | \(\left (-\frac {12}{25} x^{2}+\frac {32}{75} x +\frac {31}{375}\right ) \sqrt {1-2 x}+\frac {11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{625}\) | \(64\) |
-22/625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/375*(1-2*x)^(1/2)* (180*x^2-160*x-31)
Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=\frac {11}{625} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - \frac {1}{375} \, {\left (180 \, x^{2} - 160 \, x - 31\right )} \sqrt {-2 \, x + 1} \]
11/625*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5 *x + 3)) - 1/375*(180*x^2 - 160*x - 31)*sqrt(-2*x + 1)
Time = 1.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=- \frac {3 \left (1 - 2 x\right )^{\frac {5}{2}}}{25} + \frac {2 \left (1 - 2 x\right )^{\frac {3}{2}}}{75} + \frac {22 \sqrt {1 - 2 x}}{125} + \frac {11 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{625} \]
-3*(1 - 2*x)**(5/2)/25 + 2*(1 - 2*x)**(3/2)/75 + 22*sqrt(1 - 2*x)/125 + 11 *sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/ 5))/625
Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=-\frac {3}{25} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {2}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {22}{125} \, \sqrt {-2 \, x + 1} \]
-3/25*(-2*x + 1)^(5/2) + 2/75*(-2*x + 1)^(3/2) + 11/625*sqrt(55)*log(-(sqr t(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 22/125*sqrt(-2* x + 1)
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=-\frac {3}{25} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {2}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {11}{625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {22}{125} \, \sqrt {-2 \, x + 1} \]
-3/25*(2*x - 1)^2*sqrt(-2*x + 1) + 2/75*(-2*x + 1)^(3/2) + 11/625*sqrt(55) *log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1) )) + 22/125*sqrt(-2*x + 1)
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{3+5 x} \, dx=\frac {22\,\sqrt {1-2\,x}}{125}+\frac {2\,{\left (1-2\,x\right )}^{3/2}}{75}-\frac {3\,{\left (1-2\,x\right )}^{5/2}}{25}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22{}\mathrm {i}}{625} \]