Integrand size = 22, antiderivative size = 82 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=\frac {242}{625} \sqrt {1-2 x}+\frac {22}{375} (1-2 x)^{3/2}+\frac {2}{125} (1-2 x)^{5/2}-\frac {3}{35} (1-2 x)^{7/2}-\frac {242}{625} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
22/375*(1-2*x)^(3/2)+2/125*(1-2*x)^(5/2)-3/35*(1-2*x)^(7/2)-242/3125*arcta nh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+242/625*(1-2*x)^(1/2)
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=\frac {5 \sqrt {1-2 x} \left (4937+4370 x-12660 x^2+9000 x^3\right )-5082 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{65625} \]
(5*Sqrt[1 - 2*x]*(4937 + 4370*x - 12660*x^2 + 9000*x^3) - 5082*Sqrt[55]*Ar cTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/65625
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {90, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)}{5 x+3} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{5} \int \frac {(1-2 x)^{5/2}}{5 x+3}dx-\frac {3}{35} (1-2 x)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {3}{35} (1-2 x)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {3}{35} (1-2 x)^{7/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {3}{35} (1-2 x)^{7/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {3}{35} (1-2 x)^{7/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )+\frac {2}{25} (1-2 x)^{5/2}\right )-\frac {3}{35} (1-2 x)^{7/2}\) |
(-3*(1 - 2*x)^(7/2))/35 + ((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)/5
3.20.71.3.1 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 = 1.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}+\frac {\sqrt {1-2 x}\, \left (9000 x^{3}-12660 x^{2}+4370 x +4937\right )}{13125}\) | \(44\) |
risch | \(-\frac {\left (9000 x^{3}-12660 x^{2}+4370 x +4937\right ) \left (-1+2 x \right )}{13125 \sqrt {1-2 x}}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(49\) |
derivativedivides | \(\frac {22 \left (1-2 x \right )^{\frac {3}{2}}}{375}+\frac {2 \left (1-2 x \right )^{\frac {5}{2}}}{125}-\frac {3 \left (1-2 x \right )^{\frac {7}{2}}}{35}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}+\frac {242 \sqrt {1-2 x}}{625}\) | \(56\) |
default | \(\frac {22 \left (1-2 x \right )^{\frac {3}{2}}}{375}+\frac {2 \left (1-2 x \right )^{\frac {5}{2}}}{125}-\frac {3 \left (1-2 x \right )^{\frac {7}{2}}}{35}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}+\frac {242 \sqrt {1-2 x}}{625}\) | \(56\) |
trager | \(\left (\frac {24}{35} x^{3}-\frac {844}{875} x^{2}+\frac {874}{2625} x +\frac {4937}{13125}\right ) \sqrt {1-2 x}-\frac {121 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3125}\) | \(69\) |
-242/3125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+1/13125*(1-2*x)^(1 /2)*(9000*x^3-12660*x^2+4370*x+4937)
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=\frac {121}{3125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {1}{13125} \, {\left (9000 \, x^{3} - 12660 \, x^{2} + 4370 \, x + 4937\right )} \sqrt {-2 \, x + 1} \]
121/3125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/ (5*x + 3)) + 1/13125*(9000*x^3 - 12660*x^2 + 4370*x + 4937)*sqrt(-2*x + 1)
Time = 1.44 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=- \frac {3 \left (1 - 2 x\right )^{\frac {7}{2}}}{35} + \frac {2 \left (1 - 2 x\right )^{\frac {5}{2}}}{125} + \frac {22 \left (1 - 2 x\right )^{\frac {3}{2}}}{375} + \frac {242 \sqrt {1 - 2 x}}{625} + \frac {121 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} \]
-3*(1 - 2*x)**(7/2)/35 + 2*(1 - 2*x)**(5/2)/125 + 22*(1 - 2*x)**(3/2)/375 + 242*sqrt(1 - 2*x)/625 + 121*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/3125
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=-\frac {3}{35} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {2}{125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {22}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{3125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {242}{625} \, \sqrt {-2 \, x + 1} \]
-3/35*(-2*x + 1)^(7/2) + 2/125*(-2*x + 1)^(5/2) + 22/375*(-2*x + 1)^(3/2) + 121/3125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( -2*x + 1))) + 242/625*sqrt(-2*x + 1)
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=\frac {3}{35} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {2}{125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {22}{375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{3125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {242}{625} \, \sqrt {-2 \, x + 1} \]
3/35*(2*x - 1)^3*sqrt(-2*x + 1) + 2/125*(2*x - 1)^2*sqrt(-2*x + 1) + 22/37 5*(-2*x + 1)^(3/2) + 121/3125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(- 2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 242/625*sqrt(-2*x + 1)
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{3+5 x} \, dx=\frac {242\,\sqrt {1-2\,x}}{625}+\frac {22\,{\left (1-2\,x\right )}^{3/2}}{375}+\frac {2\,{\left (1-2\,x\right )}^{5/2}}{125}-\frac {3\,{\left (1-2\,x\right )}^{7/2}}{35}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{3125} \]