Integrand size = 26, antiderivative size = 138 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {3+5 x}}+\frac {21483 \sqrt {1-2 x} \sqrt {3+5 x}}{80000}+\frac {651 (1-2 x)^{3/2} \sqrt {3+5 x}}{8000}+\frac {651 (1-2 x)^{5/2} \sqrt {3+5 x}}{22000}-\frac {9}{200} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {236313 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{80000 \sqrt {10}} \]
236313/800000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/275*(1-2*x)^( 7/2)/(3+5*x)^(1/2)+651/8000*(1-2*x)^(3/2)*(3+5*x)^(1/2)+651/22000*(1-2*x)^ (5/2)*(3+5*x)^(1/2)-9/200*(1-2*x)^(7/2)*(3+5*x)^(1/2)+21483/80000*(1-2*x)^ (1/2)*(3+5*x)^(1/2)
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {10 \sqrt {1-2 x} \left (79699+134625 x-112620 x^2-77600 x^3+144000 x^4\right )-236313 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{800000 \sqrt {3+5 x}} \]
(10*Sqrt[1 - 2*x]*(79699 + 134625*x - 112620*x^2 - 77600*x^3 + 144000*x^4) - 236313*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(800000*S qrt[3 + 5*x])
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {100, 27, 90, 60, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)^2}{(5 x+3)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {2}{275} \int \frac {9 (1-2 x)^{5/2} (55 x+39)}{2 \sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {9}{275} \int \frac {(1-2 x)^{5/2} (55 x+39)}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \int \frac {(1-2 x)^{5/2}}{\sqrt {5 x+3}}dx-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \left (\frac {11}{6} \int \frac {(1-2 x)^{3/2}}{\sqrt {5 x+3}}dx+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \left (\frac {11}{6} \left (\frac {33}{20} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {9}{275} \left (\frac {217}{16} \left (\frac {11}{6} \left (\frac {33}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )+\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{3/2}\right )+\frac {1}{15} \sqrt {5 x+3} (1-2 x)^{5/2}\right )-\frac {11}{8} (1-2 x)^{7/2} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{7/2}}{275 \sqrt {5 x+3}}\) |
(-2*(1 - 2*x)^(7/2))/(275*Sqrt[3 + 5*x]) + (9*((-11*(1 - 2*x)^(7/2)*Sqrt[3 + 5*x])/8 + (217*(((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/15 + (11*(((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))/6))/16))/275
3.25.38.3.1 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*(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*((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 = 1.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {\left (2880000 x^{4} \sqrt {-10 x^{2}-x +3}-1552000 x^{3} \sqrt {-10 x^{2}-x +3}+1181565 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -2252400 x^{2} \sqrt {-10 x^{2}-x +3}+708939 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+2692500 x \sqrt {-10 x^{2}-x +3}+1593980 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{1600000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(133\) |
1/1600000*(2880000*x^4*(-10*x^2-x+3)^(1/2)-1552000*x^3*(-10*x^2-x+3)^(1/2) +1181565*10^(1/2)*arcsin(20/11*x+1/11)*x-2252400*x^2*(-10*x^2-x+3)^(1/2)+7 08939*10^(1/2)*arcsin(20/11*x+1/11)+2692500*x*(-10*x^2-x+3)^(1/2)+1593980* (-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {236313 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (144000 \, x^{4} - 77600 \, x^{3} - 112620 \, x^{2} + 134625 \, x + 79699\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1600000 \, {\left (5 \, x + 3\right )}} \]
-1/1600000*(236313*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*(144000*x^4 - 77600*x^3 - 112620*x^2 + 134625*x + 79699)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
\[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \left (3 x + 2\right )^{2}}{\left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=-\frac {18 \, x^{5}}{5 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {187 \, x^{4}}{50 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3691 \, x^{3}}{2000 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {38187 \, x^{2}}{8000 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {236313}{1600000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {24773 \, x}{80000 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {79699}{80000 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-18/5*x^5/sqrt(-10*x^2 - x + 3) + 187/50*x^4/sqrt(-10*x^2 - x + 3) + 3691/ 2000*x^3/sqrt(-10*x^2 - x + 3) - 38187/8000*x^2/sqrt(-10*x^2 - x + 3) - 23 6313/1600000*sqrt(10)*arcsin(-20/11*x - 1/11) - 24773/80000*x/sqrt(-10*x^2 - x + 3) + 79699/80000/sqrt(-10*x^2 - x + 3)
Time = 0.37 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\frac {1}{2000000} \, {\left (4 \, {\left (8 \, {\left (36 \, \sqrt {5} {\left (5 \, x + 3\right )} - 529 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 16905 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 61545 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {236313}{800000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {121 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{31250 \, \sqrt {5 \, x + 3}} + \frac {242 \, \sqrt {10} \sqrt {5 \, x + 3}}{15625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
1/2000000*(4*(8*(36*sqrt(5)*(5*x + 3) - 529*sqrt(5))*(5*x + 3) + 16905*sqr t(5))*(5*x + 3) + 61545*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 236313/80 0000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 121/31250*sqrt(10)*(sq rt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 242/15625*sqrt(10)*sqrt( 5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^{3/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,{\left (3\,x+2\right )}^2}{{\left (5\,x+3\right )}^{3/2}} \,d x \]