Integrand size = 26, antiderivative size = 84 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {7 (2+3 x)^2 \sqrt {3+5 x}}{11 \sqrt {1-2 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (25003+10380 x)}{8800}-\frac {56421 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{800 \sqrt {10}} \]
-56421/8000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^2*(3 +5*x)^(1/2)/(1-2*x)^(1/2)+3/8800*(25003+10380*x)*(1-2*x)^(1/2)*(3+5*x)^(1/ 2)
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-97409+51678 x+11880 x^2\right )+620631 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{88000 \sqrt {1-2 x}} \]
(-10*Sqrt[3 + 5*x]*(-97409 + 51678*x + 11880*x^2) + 620631*Sqrt[10 - 20*x] *ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(88000*Sqrt[1 - 2*x])
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, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{3/2} \sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {1}{11} \int \frac {3 (3 x+2) (173 x+106)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \int \frac {(3 x+2) (173 x+106)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {206877}{800} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{400} \sqrt {1-2 x} \sqrt {5 x+3} (10380 x+25003)\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {206877 \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}}{2000}-\frac {1}{400} \sqrt {1-2 x} \sqrt {5 x+3} (10380 x+25003)\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {206877 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{400 \sqrt {10}}-\frac {1}{400} \sqrt {1-2 x} \sqrt {5 x+3} (10380 x+25003)\right )\) |
(7*(2 + 3*x)^2*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) - (3*(-1/400*(Sqrt[1 - 2* x]*Sqrt[3 + 5*x]*(25003 + 10380*x)) + (206877*ArcSin[Sqrt[2/11]*Sqrt[3 + 5 *x]])/(400*Sqrt[10])))/22
3.26.52.3.1 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[(-(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[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 = 1.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.26
method | result | size |
default | \(-\frac {\left (1241262 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -237600 x^{2} \sqrt {-10 x^{2}-x +3}-620631 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1033560 x \sqrt {-10 x^{2}-x +3}+1948180 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{176000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\) | \(106\) |
-1/176000*(1241262*10^(1/2)*arcsin(20/11*x+1/11)*x-237600*x^2*(-10*x^2-x+3 )^(1/2)-620631*10^(1/2)*arcsin(20/11*x+1/11)-1033560*x*(-10*x^2-x+3)^(1/2) +1948180*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(-1+2*x)/(-10*x^ 2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {620631 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (11880 \, x^{2} + 51678 \, x - 97409\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{176000 \, {\left (2 \, x - 1\right )}} \]
1/176000*(620631*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*(11880*x^2 + 51678*x - 97409 )*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\left (1 - 2 x\right )^{\frac {3}{2}} \sqrt {5 x + 3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {56421}{16000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {27}{40} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {2619}{800} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {343 \, \sqrt {-10 \, x^{2} - x + 3}}{44 \, {\left (2 \, x - 1\right )}} \]
-56421/16000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 27/40*sqrt(-10*x^2 - x + 3)*x + 2619/800*sqrt(-10*x^2 - x + 3) - 343/44*sqrt(-10*x^2 - x + 3)/ (2*x - 1)
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {56421}{8000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (594 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 63 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 620695 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{220000 \, {\left (2 \, x - 1\right )}} \]
-56421/8000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/220000*(594*( 4*sqrt(5)*(5*x + 3) + 63*sqrt(5))*(5*x + 3) - 620695*sqrt(5))*sqrt(5*x + 3 )*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^3}{{\left (1-2\,x\right )}^{3/2}\,\sqrt {5\,x+3}} \,d x \]