Integrand size = 26, antiderivative size = 135 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {4246733 \sqrt {1-2 x} \sqrt {3+5 x}}{14080}-\frac {101 (2+3 x)^2 (3+5 x)^{3/2}}{22 \sqrt {1-2 x}}+\frac {(2+3 x)^3 (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {3 \sqrt {1-2 x} (3+5 x)^{3/2} (59719+28200 x)}{3520}+\frac {4246733 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{1280 \sqrt {10}} \]
1/3*(2+3*x)^3*(3+5*x)^(3/2)/(1-2*x)^(3/2)+4246733/12800*arcsin(1/11*22^(1/ 2)*(3+5*x)^(1/2))*10^(1/2)-101/22*(2+3*x)^2*(3+5*x)^(3/2)/(1-2*x)^(1/2)-3/ 3520*(3+5*x)^(3/2)*(59719+28200*x)*(1-2*x)^(1/2)-4246733/14080*(1-2*x)^(1/ 2)*(3+5*x)^(1/2)
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.61 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (1925361-5349344 x+1544724 x^2+447120 x^3+86400 x^4\right )+12740199 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{38400 (1-2 x)^{3/2}} \]
(-10*Sqrt[3 + 5*x]*(1925361 - 5349344*x + 1544724*x^2 + 447120*x^3 + 86400 *x^4) + 12740199*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(38400*(1 - 2*x)^(3/2))
Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {108, 27, 167, 27, 164, 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.
| \(\displaystyle \int \frac {(3 x+2)^3 (5 x+3)^{3/2}}{(1-2 x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {(3 x+2)^3 (5 x+3)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {3 (3 x+2)^2 \sqrt {5 x+3} (45 x+28)}{2 (1-2 x)^{3/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^3 (5 x+3)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{2} \int \frac {(3 x+2)^2 \sqrt {5 x+3} (45 x+28)}{(1-2 x)^{3/2}}dx\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{11} \int -\frac {(3 x+2) \sqrt {5 x+3} (10575 x+6646)}{2 \sqrt {1-2 x}}dx-\frac {101 (5 x+3)^{3/2} (3 x+2)^2}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \int \frac {(3 x+2) \sqrt {5 x+3} (10575 x+6646)}{\sqrt {1-2 x}}dx-\frac {101 (3 x+2)^2 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {4246733}{160} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (28200 x+59719)\right )-\frac {101 (3 x+2)^2 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {4246733}{160} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (28200 x+59719)\right )-\frac {101 (3 x+2)^2 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {4246733}{160} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (28200 x+59719)\right )-\frac {101 (3 x+2)^2 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {4246733}{160} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (28200 x+59719)\right )-\frac {101 (3 x+2)^2 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\) |
((2 + 3*x)^3*(3 + 5*x)^(3/2))/(3*(1 - 2*x)^(3/2)) + ((-101*(2 + 3*x)^2*(3 + 5*x)^(3/2))/(11*Sqrt[1 - 2*x]) + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)*(597 19 + 28200*x))/80 + (4246733*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*Arc Sin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/160)/22)/2
3.26.90.3.1 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_))^(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 = 1.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(\frac {\left (-1728000 x^{4} \sqrt {-10 x^{2}-x +3}+50960796 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-8942400 x^{3} \sqrt {-10 x^{2}-x +3}-50960796 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -30894480 x^{2} \sqrt {-10 x^{2}-x +3}+12740199 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+106986880 x \sqrt {-10 x^{2}-x +3}-38507220 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{76800 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(154\) |
1/76800*(-1728000*x^4*(-10*x^2-x+3)^(1/2)+50960796*10^(1/2)*arcsin(20/11*x +1/11)*x^2-8942400*x^3*(-10*x^2-x+3)^(1/2)-50960796*10^(1/2)*arcsin(20/11* x+1/11)*x-30894480*x^2*(-10*x^2-x+3)^(1/2)+12740199*10^(1/2)*arcsin(20/11* x+1/11)+106986880*x*(-10*x^2-x+3)^(1/2)-38507220*(-10*x^2-x+3)^(1/2))*(1-2 *x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {12740199 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (86400 \, x^{4} + 447120 \, x^{3} + 1544724 \, x^{2} - 5349344 \, x + 1925361\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{76800 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
-1/76800*(12740199*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*(86400*x^4 + 44712 0*x^3 + 1544724*x^2 - 5349344*x + 1925361)*sqrt(5*x + 3)*sqrt(-2*x + 1))/( 4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \]
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.56 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {428267}{2560} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {35937}{25600} i \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x - \frac {21}{11}\right ) + \frac {9}{16} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {297}{64} \, \sqrt {10 \, x^{2} - 21 \, x + 8} x + \frac {6237}{1280} \, \sqrt {10 \, x^{2} - 21 \, x + 8} - \frac {6237}{128} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {343 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{48 \, {\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} + \frac {441 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {189 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{32 \, {\left (2 \, x - 1\right )}} + \frac {3773 \, \sqrt {-10 \, x^{2} - x + 3}}{96 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {3479 \, \sqrt {-10 \, x^{2} - x + 3}}{6 \, {\left (2 \, x - 1\right )}} \]
428267/2560*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 35937/25600*I*sqrt(5) *sqrt(2)*arcsin(20/11*x - 21/11) + 9/16*(-10*x^2 - x + 3)^(3/2) - 297/64*s qrt(10*x^2 - 21*x + 8)*x + 6237/1280*sqrt(10*x^2 - 21*x + 8) - 6237/128*sq rt(-10*x^2 - x + 3) - 343/48*(-10*x^2 - x + 3)^(3/2)/(8*x^3 - 12*x^2 + 6*x - 1) + 441/16*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) + 189/32*(-10*x^2 - x + 3)^(3/2)/(2*x - 1) + 3773/96*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1 ) + 3479/6*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {4246733}{12800} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (27 \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 111 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 8579 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 8493466 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 140142189 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{480000 \, {\left (2 \, x - 1\right )}^{2}} \]
4246733/12800*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/480000*(4*( 27*(4*(8*sqrt(5)*(5*x + 3) + 111*sqrt(5))*(5*x + 3) + 8579*sqrt(5))*(5*x + 3) - 8493466*sqrt(5))*(5*x + 3) + 140142189*sqrt(5))*sqrt(5*x + 3)*sqrt(- 10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x)^3 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^3\,{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{5/2}} \,d x \]