Integrand size = 26, antiderivative size = 94 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {17951 \sqrt {1-2 x} \sqrt {3+5 x}}{1760}+\frac {49 (3+5 x)^{3/2}}{22 \sqrt {1-2 x}}+\frac {9}{40} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {17951 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{160 \sqrt {10}} \]
-17951/1600*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+49/22*(3+5*x)^(3/ 2)/(1-2*x)^(1/2)+9/40*(3+5*x)^(3/2)*(1-2*x)^(1/2)+17951/1760*(1-2*x)^(1/2) *(3+5*x)^(1/2)
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-2809+1518 x+360 x^2\right )+17951 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{1600 \sqrt {1-2 x}} \]
(-10*Sqrt[3 + 5*x]*(-2809 + 1518*x + 360*x^2) + 17951*Sqrt[10 - 20*x]*ArcT an[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(1600*Sqrt[1 - 2*x])
Time = 0.19 (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, 90, 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.
\(\displaystyle \int \frac {(3 x+2)^2 \sqrt {5 x+3}}{(1-2 x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}-\frac {1}{22} \int \frac {\sqrt {5 x+3} (198 x+853)}{2 \sqrt {1-2 x}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}-\frac {1}{44} \int \frac {\sqrt {5 x+3} (198 x+853)}{\sqrt {1-2 x}}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{44} \left (\frac {99}{10} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {17951}{20} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx\right )+\frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{44} \left (\frac {99}{10} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {17951}{20} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )+\frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{44} \left (\frac {99}{10} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {17951}{20} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )+\frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{44} \left (\frac {99}{10} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {17951}{20} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )\right )+\frac {49 (5 x+3)^{3/2}}{22 \sqrt {1-2 x}}\) |
(49*(3 + 5*x)^(3/2))/(22*Sqrt[1 - 2*x]) + ((99*Sqrt[1 - 2*x]*(3 + 5*x)^(3/ 2))/10 - (17951*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11 ]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/20)/44
3.26.21.3.1 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 = 3.98 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13
method | result | size |
default | \(-\frac {\left (35902 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -7200 x^{2} \sqrt {-10 x^{2}-x +3}-17951 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-30360 x \sqrt {-10 x^{2}-x +3}+56180 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{3200 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(106\) |
-1/3200*(35902*10^(1/2)*arcsin(20/11*x+1/11)*x-7200*x^2*(-10*x^2-x+3)^(1/2 )-17951*10^(1/2)*arcsin(20/11*x+1/11)-30360*x*(-10*x^2-x+3)^(1/2)+56180*(- 10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)/(-10*x^2-x+3)^(1/2 )
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\frac {17951 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (360 \, x^{2} + 1518 \, x - 2809\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3200 \, {\left (2 \, x - 1\right )}} \]
1/3200*(17951*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(360*x^2 + 1518*x - 2809)*sqrt( 5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{2} \sqrt {5 x + 3}}{\left (1 - 2 x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=-\frac {17951}{3200} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {9}{8} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {849}{160} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{4 \, {\left (2 \, x - 1\right )}} \]
-17951/3200*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 9/8*sqrt(-10*x^2 - x + 3)*x + 849/160*sqrt(-10*x^2 - x + 3) - 49/4*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=-\frac {17951}{1600} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6 \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 181 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 17951 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4000 \, {\left (2 \, x - 1\right )}} \]
-17951/1600*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/4000*(6*(12*s qrt(5)*(5*x + 3) + 181*sqrt(5))*(5*x + 3) - 17951*sqrt(5))*sqrt(5*x + 3)*s qrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^2\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]