Integrand size = 26, antiderivative size = 84 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {2 \sqrt {1-2 x} (2+3 x)^2}{55 \sqrt {3+5 x}}-\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (979+300 x)}{4400}+\frac {2493 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{400 \sqrt {10}} \]
2493/4000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/55*(2+3*x)^2*(1-2 *x)^(1/2)/(3+5*x)^(1/2)-3/4400*(979+300*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\frac {-10 \sqrt {1-2 x} \left (9451+19305 x+5940 x^2\right )-27423 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{44000 \sqrt {3+5 x}} \]
(-10*Sqrt[1 - 2*x]*(9451 + 19305*x + 5940*x^2) - 27423*Sqrt[30 + 50*x]*Arc Tan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(44000*Sqrt[3 + 5*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.
| \(\displaystyle \int \frac {(3 x+2)^3}{\sqrt {1-2 x} (5 x+3)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {2}{55} \int -\frac {3 (3 x+2) (25 x+26)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{55} \int \frac {(3 x+2) (25 x+26)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {3}{55} \left (\frac {9141}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (300 x+979)\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {3}{55} \left (\frac {9141}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (300 x+979)\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {3}{55} \left (\frac {9141 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (300 x+979)\right )-\frac {2 \sqrt {1-2 x} (3 x+2)^2}{55 \sqrt {5 x+3}}\) |
(-2*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*Sqrt[3 + 5*x]) + (3*(-1/80*(Sqrt[1 - 2* x]*Sqrt[3 + 5*x]*(979 + 300*x)) + (9141*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/ (80*Sqrt[10])))/55
3.25.99.3.1 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 = 1.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18
| method | result | size |
| default | \(\frac {\left (137115 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -118800 x^{2} \sqrt {-10 x^{2}-x +3}+82269 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-386100 x \sqrt {-10 x^{2}-x +3}-189020 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{88000 \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) | \(99\) |
1/88000*(137115*10^(1/2)*arcsin(20/11*x+1/11)*x-118800*x^2*(-10*x^2-x+3)^( 1/2)+82269*10^(1/2)*arcsin(20/11*x+1/11)-386100*x*(-10*x^2-x+3)^(1/2)-1890 20*(-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 = 81, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {27423 \, \sqrt {10} {\left (5 \, x + 3\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (5940 \, x^{2} + 19305 \, x + 9451\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{88000 \, {\left (5 \, x + 3\right )}} \]
-1/88000*(27423*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*(5940*x^2 + 19305*x + 9451)*s qrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)
\[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\frac {2493}{8000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {27}{100} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {1431}{2000} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{1375 \, {\left (5 \, x + 3\right )}} \]
2493/8000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 27/100*sqrt(-10*x^2 - x + 3)*x - 1431/2000*sqrt(-10*x^2 - x + 3) - 2/1375*sqrt(-10*x^2 - x + 3)/( 5*x + 3)
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.32 \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=-\frac {27}{10000} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 41 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {2493}{4000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{13750 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{6875 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \]
-27/10000*(4*sqrt(5)*(5*x + 3) + 41*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 2493/4000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/13750*sqrt(1 0)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/6875*sqrt(10)*sq rt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
Timed out. \[ \int \frac {(2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^3}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{3/2}} \,d x \]