Integrand size = 19, antiderivative size = 96 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{5/2}}{15 (3+5 x)^{3/2}}+\frac {4 (1-2 x)^{3/2}}{15 \sqrt {3+5 x}}+\frac {4}{25} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {22}{25} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]
-2/15*(1-2*x)^(5/2)/(3+5*x)^(3/2)+22/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2 ))*10^(1/2)+4/15*(1-2*x)^(3/2)/(3+5*x)^(1/2)+4/25*(1-2*x)^(1/2)*(3+5*x)^(1 /2)
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=\frac {2}{375} \left (\frac {5 \sqrt {1-2 x} \left (79+190 x+30 x^2\right )}{(3+5 x)^{3/2}}+33 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {6}{5}+2 x}}{\sqrt {1-2 x}}\right )\right ) \]
(2*((5*Sqrt[1 - 2*x]*(79 + 190*x + 30*x^2))/(3 + 5*x)^(3/2) + 33*Sqrt[10]* ArcTan[Sqrt[6/5 + 2*x]/Sqrt[1 - 2*x]]))/375
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 57, 60, 64, 223}
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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle -\frac {2}{3} \int \frac {(1-2 x)^{3/2}}{(5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle -\frac {2}{3} \left (-\frac {6}{5} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {2}{3} \left (-\frac {6}{5} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle -\frac {2}{3} \left (-\frac {6}{5} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {2}{3} \left (-\frac {6}{5} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{15 (5 x+3)^{3/2}}\) |
(-2*(1 - 2*x)^(5/2))/(15*(3 + 5*x)^(3/2)) - (2*((-2*(1 - 2*x)^(3/2))/(5*Sq rt[3 + 5*x]) - (6*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11] *Sqrt[3 + 5*x]])/(5*Sqrt[10])))/5))/3
3.25.51.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
\[\int \frac {\left (1-2 x \right )^{\frac {5}{2}}}{\left (3+5 x \right )^{\frac {5}{2}}}d x\]
Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=-\frac {33 \, \sqrt {5} \sqrt {2} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 10 \, {\left (30 \, x^{2} + 190 \, x + 79\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{375 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/375*(33*sqrt(5)*sqrt(2)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(5)*sqrt(2) *(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 10*(30*x^2 + 190*x + 79)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Result contains complex when optimal does not.
Time = 3.83 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.68 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=\begin {cases} \frac {4 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )}{125} + \frac {308 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{1875} - \frac {242 \sqrt {10} \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{9375 \left (x + \frac {3}{5}\right )} + \frac {11 \sqrt {10} i \log {\left (\frac {1}{x + \frac {3}{5}} \right )}}{125} + \frac {11 \sqrt {10} i \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {22 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{125} & \text {for}\: \frac {1}{\left |{x + \frac {3}{5}}\right |} > \frac {10}{11} \\\frac {4 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )}{125} + \frac {308 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{1875} - \frac {242 \sqrt {10} i \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}}{9375 \left (x + \frac {3}{5}\right )} + \frac {11 \sqrt {10} i \log {\left (\frac {1}{x + \frac {3}{5}} \right )}}{125} - \frac {22 \sqrt {10} i \log {\left (\sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} + 1 \right )}}{125} & \text {otherwise} \end {cases} \]
Piecewise((4*sqrt(10)*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)/125 + 308*sqr t(10)*sqrt(-1 + 11/(10*(x + 3/5)))/1875 - 242*sqrt(10)*sqrt(-1 + 11/(10*(x + 3/5)))/(9375*(x + 3/5)) + 11*sqrt(10)*I*log(1/(x + 3/5))/125 + 11*sqrt( 10)*I*log(x + 3/5)/125 + 22*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/125, 1/Abs(x + 3/5) > 10/11), (4*sqrt(10)*I*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3 /5)/125 + 308*sqrt(10)*I*sqrt(1 - 11/(10*(x + 3/5)))/1875 - 242*sqrt(10)*I *sqrt(1 - 11/(10*(x + 3/5)))/(9375*(x + 3/5)) + 11*sqrt(10)*I*log(1/(x + 3 /5))/125 - 22*sqrt(10)*I*log(sqrt(1 - 11/(10*(x + 3/5))) + 1)/125, True))
Time = 0.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.34 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=\frac {11}{125} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{5 \, {\left (625 \, x^{4} + 1500 \, x^{3} + 1350 \, x^{2} + 540 \, x + 81\right )}} - \frac {11 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{30 \, {\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} - \frac {121 \, \sqrt {-10 \, x^{2} - x + 3}}{150 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {77 \, \sqrt {-10 \, x^{2} - x + 3}}{75 \, {\left (5 \, x + 3\right )}} \]
11/125*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 1/5*(-10*x^2 - x + 3)^(5/2 )/(625*x^4 + 1500*x^3 + 1350*x^2 + 540*x + 81) - 11/30*(-10*x^2 - x + 3)^( 3/2)/(125*x^3 + 225*x^2 + 135*x + 27) - 121/150*sqrt(-10*x^2 - x + 3)/(25* x^2 + 30*x + 9) + 77/75*sqrt(-10*x^2 - x + 3)/(5*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (67) = 134\).
Time = 0.37 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.66 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=-\frac {11 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{30000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {4}{625} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {22}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {99 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{2500 \, \sqrt {5 \, x + 3}} - \frac {11 \, \sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {27 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{1875 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-11/30000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 4/625*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 22/125*sqrt(10)*arcsin(1/1 1*sqrt(22)*sqrt(5*x + 3)) + 99/2500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sq rt(22))/sqrt(5*x + 3) - 11/1875*sqrt(10)*(5*x + 3)^(3/2)*(27*(sqrt(2)*sqrt (-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(2 2))^3
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{5/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (5\,x+3\right )}^{5/2}} \,d x \]