Integrand size = 26, antiderivative size = 94 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{3/2}}{825 (3+5 x)^{3/2}}-\frac {12 (1-2 x)^{3/2}}{275 \sqrt {3+5 x}}+\frac {3}{55} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {3 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{5 \sqrt {10}} \]
-2/825*(1-2*x)^(3/2)/(3+5*x)^(3/2)+3/50*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2) )*10^(1/2)-12/275*(1-2*x)^(3/2)/(3+5*x)^(1/2)+3/55*(1-2*x)^(1/2)*(3+5*x)^( 1/2)
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=\frac {\sqrt {1-2 x} \left (59+278 x+297 x^2\right )}{165 (3+5 x)^{3/2}}-\frac {3 \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{5 \sqrt {10}} \]
(Sqrt[1 - 2*x]*(59 + 278*x + 297*x^2))/(165*(3 + 5*x)^(3/2)) - (3*ArcTan[S qrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(5*Sqrt[10])
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {100, 27, 87, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^2}{(5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{825} \int \frac {99 \sqrt {1-2 x} (15 x+11)}{2 (5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{25} \int \frac {\sqrt {1-2 x} (15 x+11)}{(5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {3}{25} \left (\frac {25}{11} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {4 (1-2 x)^{3/2}}{11 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3}{25} \left (\frac {25}{11} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {4 (1-2 x)^{3/2}}{11 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {3}{25} \left (\frac {25}{11} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {4 (1-2 x)^{3/2}}{11 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {3}{25} \left (\frac {25}{11} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {4 (1-2 x)^{3/2}}{11 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{825 (5 x+3)^{3/2}}\) |
(-2*(1 - 2*x)^(3/2))/(825*(3 + 5*x)^(3/2)) + (3*((-4*(1 - 2*x)^(3/2))/(11* Sqrt[3 + 5*x]) + (25*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/ 11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/11))/25
3.24.20.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((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*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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]
Time = 1.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.20
method | result | size |
default | \(\frac {\left (2475 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+2970 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +5940 x^{2} \sqrt {-10 x^{2}-x +3}+891 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+5560 x \sqrt {-10 x^{2}-x +3}+1180 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{3300 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(113\) |
1/3300*(2475*10^(1/2)*arcsin(20/11*x+1/11)*x^2+2970*10^(1/2)*arcsin(20/11* x+1/11)*x+5940*x^2*(-10*x^2-x+3)^(1/2)+891*10^(1/2)*arcsin(20/11*x+1/11)+5 560*x*(-10*x^2-x+3)^(1/2)+1180*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2 -x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=-\frac {99 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (297 \, x^{2} + 278 \, x + 59\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3300 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/3300*(99*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*s qrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 20*(297*x^2 + 278*x + 59)* sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1 - 2 x} \left (3 x + 2\right )^{2}}{\left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (67) = 134\).
Time = 0.40 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{330000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {9}{625} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {3}{50} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {131 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{27500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {393 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{20625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
-1/330000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 9/625*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 3/50*sqrt(10)*arcsin(1/11* sqrt(22)*sqrt(5*x + 3)) - 131/27500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sq rt(22))/sqrt(5*x + 3) + 1/20625*sqrt(10)*(5*x + 3)^(3/2)*(393*(sqrt(2)*sqr t(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt( 22))^3
Timed out. \[ \int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^2}{{\left (5\,x+3\right )}^{5/2}} \,d x \]