Integrand size = 26, antiderivative size = 84 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=-\frac {(95621-33462 x) \sqrt {3+5 x}}{14520 \sqrt {1-2 x}}+\frac {7 (2+3 x)^2 \sqrt {3+5 x}}{33 (1-2 x)^{3/2}}+\frac {1593 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{40 \sqrt {10}} \]
1593/400*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/33*(2+3*x)^2*(3+5* x)^(1/2)/(1-2*x)^(3/2)-1/14520*(95621-33462*x)*(3+5*x)^(1/2)/(1-2*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {-10 \sqrt {3+5 x} \left (83301-261664 x+39204 x^2\right )+578259 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{145200 (1-2 x)^{3/2}} \]
(-10*Sqrt[3 + 5*x]*(83301 - 261664*x + 39204*x^2) + 578259*Sqrt[10 - 20*x] *(-1 + 2*x)*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(145200*(1 - 2*x)^(3/2) )
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {109, 27, 160, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{5/2} \sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{33 (1-2 x)^{3/2}}-\frac {1}{33} \int \frac {(3 x+2) (507 x+310)}{2 (1-2 x)^{3/2} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{33 (1-2 x)^{3/2}}-\frac {1}{66} \int \frac {(3 x+2) (507 x+310)}{(1-2 x)^{3/2} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {1}{66} \left (\frac {52569}{40} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {(95621-33462 x) \sqrt {5 x+3}}{220 \sqrt {1-2 x}}\right )+\frac {7 \sqrt {5 x+3} (3 x+2)^2}{33 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{66} \left (\frac {52569}{100} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {(95621-33462 x) \sqrt {5 x+3}}{220 \sqrt {1-2 x}}\right )+\frac {7 \sqrt {5 x+3} (3 x+2)^2}{33 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{66} \left (\frac {52569 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{20 \sqrt {10}}-\frac {(95621-33462 x) \sqrt {5 x+3}}{220 \sqrt {1-2 x}}\right )+\frac {7 \sqrt {5 x+3} (3 x+2)^2}{33 (1-2 x)^{3/2}}\) |
(7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(33*(1 - 2*x)^(3/2)) + (-1/220*((95621 - 334 62*x)*Sqrt[3 + 5*x])/Sqrt[1 - 2*x] + (52569*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x ]])/(20*Sqrt[10]))/66
3.27.10.3.1 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[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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[(b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Simp[(a*d*f*h*m + b*(d* (f*g + e*h) - c*f*h*(m + 2)))/(b^2*d) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 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 = 1.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {\left (2313036 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-2313036 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -784080 x^{2} \sqrt {-10 x^{2}-x +3}+578259 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+5233280 x \sqrt {-10 x^{2}-x +3}-1666020 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{290400 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(120\) |
1/290400*(2313036*10^(1/2)*arcsin(20/11*x+1/11)*x^2-2313036*10^(1/2)*arcsi n(20/11*x+1/11)*x-784080*x^2*(-10*x^2-x+3)^(1/2)+578259*10^(1/2)*arcsin(20 /11*x+1/11)+5233280*x*(-10*x^2-x+3)^(1/2)-1666020*(-10*x^2-x+3)^(1/2))*(3+ 5*x)^(1/2)*(1-2*x)^(1/2)/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=-\frac {578259 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (39204 \, x^{2} - 261664 \, x + 83301\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{290400 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
-1/290400*(578259*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(39204*x^2 - 261664 *x + 83301)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\left (1 - 2 x\right )^{\frac {5}{2}} \sqrt {5 x + 3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {1593}{800} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {27}{40} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {343 \, \sqrt {-10 \, x^{2} - x + 3}}{132 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {11123 \, \sqrt {-10 \, x^{2} - x + 3}}{1452 \, {\left (2 \, x - 1\right )}} \]
1593/800*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/40*sqrt(-10*x^2 - x + 3) + 343/132*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 11123/1452*sqrt(-1 0*x^2 - x + 3)/(2*x - 1)
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\frac {1593}{400} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (9801 \, \sqrt {5} {\left (5 \, x + 3\right )} - 385886 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 6360321 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1815000 \, {\left (2 \, x - 1\right )}^{2}} \]
1593/400*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/1815000*(4*(9801 *sqrt(5)*(5*x + 3) - 385886*sqrt(5))*(5*x + 3) + 6360321*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^3}{{\left (1-2\,x\right )}^{5/2}\,\sqrt {5\,x+3}} \,d x \]