Integrand size = 19, antiderivative size = 94 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {3+5 x}}-\frac {33}{50} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {1}{5} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {363 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{50 \sqrt {10}} \]
-363/500*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/5*(1-2*x)^(5/2)/(3 +5*x)^(1/2)-1/5*(1-2*x)^(3/2)*(3+5*x)^(1/2)-33/50*(1-2*x)^(1/2)*(3+5*x)^(1 /2)
Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=\frac {10 \sqrt {1-2 x} \left (-149-75 x+20 x^2\right )-363 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {6}{5}+2 x}}{\sqrt {1-2 x}}\right )}{500 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(-149 - 75*x + 20*x^2) - 363*Sqrt[30 + 50*x]*ArcTan[Sqrt [6/5 + 2*x]/Sqrt[1 - 2*x]])/(500*Sqrt[3 + 5*x])
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 60, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(5 x+3)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle -2 \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -2 \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -2 \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle -2 \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -2 \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )-\frac {2 (1-2 x)^{5/2}}{5 \sqrt {5 x+3}}\) |
(-2*(1 - 2*x)^(5/2))/(5*Sqrt[3 + 5*x]) - 2*(((1 - 2*x)^(3/2)*Sqrt[3 + 5*x] )/10 + (33*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/20)
3.25.40.3.1 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]
\[\int \frac {\left (1-2 x \right )^{\frac {5}{2}}}{\left (3+5 x \right )^{\frac {3}{2}}}d x\]
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=\frac {363 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (20 \, x^{2} - 75 \, x - 149\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1000 \, {\left (5 \, x + 3\right )}} \]
1/1000*(363*sqrt(10)*(5*x + 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(20*x^2 - 75*x - 149)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
Result contains complex when optimal does not.
Time = 3.85 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.43 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=\begin {cases} \frac {4 i \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{5 \sqrt {10 x - 5}} - \frac {121 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{25 \sqrt {10 x - 5}} + \frac {121 i \sqrt {x + \frac {3}{5}}}{250 \sqrt {10 x - 5}} + \frac {363 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{500} + \frac {2662 i}{625 \sqrt {x + \frac {3}{5}} \sqrt {10 x - 5}} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\- \frac {363 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{500} - \frac {4 \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{5 \sqrt {5 - 10 x}} + \frac {121 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{25 \sqrt {5 - 10 x}} - \frac {121 \sqrt {x + \frac {3}{5}}}{250 \sqrt {5 - 10 x}} - \frac {2662}{625 \sqrt {5 - 10 x} \sqrt {x + \frac {3}{5}}} & \text {otherwise} \end {cases} \]
Piecewise((4*I*(x + 3/5)**(5/2)/(5*sqrt(10*x - 5)) - 121*I*(x + 3/5)**(3/2 )/(25*sqrt(10*x - 5)) + 121*I*sqrt(x + 3/5)/(250*sqrt(10*x - 5)) + 363*sqr t(10)*I*acosh(sqrt(110)*sqrt(x + 3/5)/11)/500 + 2662*I/(625*sqrt(x + 3/5)* sqrt(10*x - 5)), Abs(x + 3/5) > 11/10), (-363*sqrt(10)*asin(sqrt(110)*sqrt (x + 3/5)/11)/500 - 4*(x + 3/5)**(5/2)/(5*sqrt(5 - 10*x)) + 121*(x + 3/5)* *(3/2)/(25*sqrt(5 - 10*x)) - 121*sqrt(x + 3/5)/(250*sqrt(5 - 10*x)) - 2662 /(625*sqrt(5 - 10*x)*sqrt(x + 3/5)), True))
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=-\frac {4 \, x^{3}}{5 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {17 \, x^{2}}{5 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {363}{1000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {223 \, x}{50 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {149}{50 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-4/5*x^3/sqrt(-10*x^2 - x + 3) + 17/5*x^2/sqrt(-10*x^2 - x + 3) + 363/1000 *sqrt(10)*arcsin(-20/11*x - 1/11) + 223/50*x/sqrt(-10*x^2 - x + 3) - 149/5 0/sqrt(-10*x^2 - x + 3)
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=\frac {1}{1250} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 99 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {363}{500} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {121 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{1250 \, \sqrt {5 \, x + 3}} + \frac {242 \, \sqrt {10} \sqrt {5 \, x + 3}}{625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
1/1250*(4*sqrt(5)*(5*x + 3) - 99*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 363/500*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 121/1250*sqrt(10)*( sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 242/625*sqrt(10)*sqrt( 5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(1-2 x)^{5/2}}{(3+5 x)^{3/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}}{{\left (5\,x+3\right )}^{3/2}} \,d x \]