Integrand size = 26, antiderivative size = 113 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{15 (3+5 x)^{3/2}}-\frac {392 \sqrt {1-2 x} (2+3 x)^2}{825 \sqrt {3+5 x}}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (1243+1740 x)}{11000}+\frac {1071 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1000 \sqrt {10}} \]
1071/10000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/15*(2+3*x)^3*(1- 2*x)^(1/2)/(3+5*x)^(3/2)-392/825*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)+7/1 1000*(1243+1740*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=\frac {\frac {10 \sqrt {1-2 x} \left (11567+75470 x+147015 x^2+89100 x^3\right )}{(3+5 x)^{3/2}}-35343 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{330000} \]
((10*Sqrt[1 - 2*x]*(11567 + 75470*x + 147015*x^2 + 89100*x^3))/(3 + 5*x)^( 3/2) - 35343*Sqrt[10]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/330000
Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {108, 27, 167, 27, 164, 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.
| \(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^3}{(5 x+3)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {2}{15} \int \frac {7 (1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {14}{15} \int \frac {(1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {14}{15} \left (\frac {2}{55} \int \frac {3 (34-145 x) (3 x+2)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {28 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {14}{15} \left (\frac {3}{55} \int \frac {(34-145 x) (3 x+2)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {28 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {14}{15} \left (\frac {3}{55} \left (\frac {1683}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1740 x+1243)\right )-\frac {28 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {14}{15} \left (\frac {3}{55} \left (\frac {1683}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1740 x+1243)\right )-\frac {28 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {14}{15} \left (\frac {3}{55} \left (\frac {1683 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}+\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (1740 x+1243)\right )-\frac {28 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{15 (5 x+3)^{3/2}}\) |
(-2*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(15*(3 + 5*x)^(3/2)) + (14*((-28*Sqrt[1 - 2 *x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) + (3*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(12 43 + 1740*x))/80 + (1683*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(80*Sqrt[10]))) /55))/15
3.24.19.3.1 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[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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
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 = 130, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {\left (883575 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+1782000 x^{3} \sqrt {-10 x^{2}-x +3}+1060290 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +2940300 x^{2} \sqrt {-10 x^{2}-x +3}+318087 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1509400 x \sqrt {-10 x^{2}-x +3}+231340 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{660000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(130\) |
1/660000*(883575*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1782000*x^3*(-10*x^2-x+ 3)^(1/2)+1060290*10^(1/2)*arcsin(20/11*x+1/11)*x+2940300*x^2*(-10*x^2-x+3) ^(1/2)+318087*10^(1/2)*arcsin(20/11*x+1/11)+1509400*x*(-10*x^2-x+3)^(1/2)+ 231340*(-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 = 96, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=-\frac {35343 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 20 \, {\left (89100 \, x^{3} + 147015 \, x^{2} + 75470 \, x + 11567\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{660000 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
-1/660000*(35343*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 20*(89100*x^3 + 14701 5*x^2 + 75470*x + 11567)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1 - 2 x} \left (3 x + 2\right )^{3}}{\left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.52 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=\frac {27}{25000} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{1650000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {1071}{10000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {197 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{137500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {591 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{103125 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]
27/25000*(4*sqrt(5)*(5*x + 3) - 3*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1/1650000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 1071/10000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 197/137500*sq rt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/103125*sqrt( 10)*(5*x + 3)^(3/2)*(591*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
Timed out. \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^{5/2}} \, dx=\int \frac {\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^3}{{\left (5\,x+3\right )}^{5/2}} \,d x \]