Integrand size = 26, antiderivative size = 164 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {1626211523 \sqrt {1-2 x} \sqrt {3+5 x}}{1126400}-\frac {3315}{352} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{3/2}-\frac {123 (2+3 x)^3 (3+5 x)^{3/2}}{22 \sqrt {1-2 x}}+\frac {(2+3 x)^4 (3+5 x)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {3 \sqrt {1-2 x} (3+5 x)^{3/2} (22868329+10798680 x)}{281600}+\frac {1626211523 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{102400 \sqrt {10}} \]
1/3*(2+3*x)^4*(3+5*x)^(3/2)/(1-2*x)^(3/2)+1626211523/1024000*arcsin(1/11*2 2^(1/2)*(3+5*x)^(1/2))*10^(1/2)-123/22*(2+3*x)^3*(3+5*x)^(3/2)/(1-2*x)^(1/ 2)-3315/352*(2+3*x)^2*(3+5*x)^(3/2)*(1-2*x)^(1/2)-3/281600*(3+5*x)^(3/2)*( 22868329+10798680*x)*(1-2*x)^(1/2)-1626211523/1126400*(1-2*x)^(1/2)*(3+5*x )^(1/2)
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (739060191-2034703904 x+633940524 x^2+236669040 x^3+83548800 x^4+15552000 x^5\right )+4878634569 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{3072000 (1-2 x)^{3/2}} \]
(-10*Sqrt[3 + 5*x]*(739060191 - 2034703904*x + 633940524*x^2 + 236669040*x ^3 + 83548800*x^4 + 15552000*x^5) + 4878634569*Sqrt[10 - 20*x]*(-1 + 2*x)* ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(3072000*(1 - 2*x)^(3/2))
Time = 0.25 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {108, 27, 167, 27, 170, 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.
| \(\displaystyle \int \frac {(3 x+2)^4 (5 x+3)^{3/2}}{(1-2 x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {3 (3 x+2)^3 \sqrt {5 x+3} (55 x+34)}{2 (1-2 x)^{3/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(3 x+2)^4 (5 x+3)^{3/2}}{3 (1-2 x)^{3/2}}-\frac {1}{2} \int \frac {(3 x+2)^3 \sqrt {5 x+3} (55 x+34)}{(1-2 x)^{3/2}}dx\) |
| \(\Big \downarrow \) 167 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{11} \int -\frac {(3 x+2)^2 \sqrt {5 x+3} (16575 x+10312)}{2 \sqrt {1-2 x}}dx-\frac {123 (5 x+3)^{3/2} (3 x+2)^3}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \int \frac {(3 x+2)^2 \sqrt {5 x+3} (16575 x+10312)}{\sqrt {1-2 x}}dx-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (-\frac {1}{40} \int -\frac {5 (3 x+2) \sqrt {5 x+3} (809901 x+508994)}{2 \sqrt {1-2 x}}dx-\frac {3315}{8} \sqrt {1-2 x} (5 x+3)^{3/2} (3 x+2)^2\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {1}{16} \int \frac {(3 x+2) \sqrt {5 x+3} (809901 x+508994)}{\sqrt {1-2 x}}dx-\frac {3315}{8} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {1}{16} \left (\frac {1626211523}{800} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {3}{400} \sqrt {1-2 x} (5 x+3)^{3/2} (10798680 x+22868329)\right )-\frac {3315}{8} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {1}{16} \left (\frac {1626211523}{800} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{400} \sqrt {1-2 x} (5 x+3)^{3/2} (10798680 x+22868329)\right )-\frac {3315}{8} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 64 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {1}{16} \left (\frac {1626211523}{800} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{400} \sqrt {1-2 x} (5 x+3)^{3/2} (10798680 x+22868329)\right )-\frac {3315}{8} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{22} \left (\frac {1}{16} \left (\frac {1626211523}{800} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{400} \sqrt {1-2 x} (5 x+3)^{3/2} (10798680 x+22868329)\right )-\frac {3315}{8} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {123 (3 x+2)^3 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{3/2} (3 x+2)^4}{3 (1-2 x)^{3/2}}\) |
((2 + 3*x)^4*(3 + 5*x)^(3/2))/(3*(1 - 2*x)^(3/2)) + ((-123*(2 + 3*x)^3*(3 + 5*x)^(3/2))/(11*Sqrt[1 - 2*x]) + ((-3315*Sqrt[1 - 2*x]*(2 + 3*x)^2*(3 + 5*x)^(3/2))/8 + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)*(22868329 + 10798680*x) )/400 + (1626211523*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[ 2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/800)/16)/22)/2
3.26.89.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
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 = 171, normalized size of antiderivative = 1.04
| method | result | size |
| default | \(\frac {\left (-311040000 x^{5} \sqrt {-10 x^{2}-x +3}-1670976000 x^{4} \sqrt {-10 x^{2}-x +3}+19514538276 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-4733380800 x^{3} \sqrt {-10 x^{2}-x +3}-19514538276 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -12678810480 x^{2} \sqrt {-10 x^{2}-x +3}+4878634569 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+40694078080 x \sqrt {-10 x^{2}-x +3}-14781203820 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{6144000 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) | \(171\) |
1/6144000*(-311040000*x^5*(-10*x^2-x+3)^(1/2)-1670976000*x^4*(-10*x^2-x+3) ^(1/2)+19514538276*10^(1/2)*arcsin(20/11*x+1/11)*x^2-4733380800*x^3*(-10*x ^2-x+3)^(1/2)-19514538276*10^(1/2)*arcsin(20/11*x+1/11)*x-12678810480*x^2* (-10*x^2-x+3)^(1/2)+4878634569*10^(1/2)*arcsin(20/11*x+1/11)+40694078080*x *(-10*x^2-x+3)^(1/2)-14781203820*(-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 = 106, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=-\frac {4878634569 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (15552000 \, x^{5} + 83548800 \, x^{4} + 236669040 \, x^{3} + 633940524 \, x^{2} - 2034703904 \, x + 739060191\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{6144000 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]
-1/6144000*(4878634569*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*(15552000*x^5 + 83548800*x^4 + 236669040*x^3 + 633940524*x^2 - 2034703904*x + 739060191) *sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{4} \left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \]
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.47 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {81}{64} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {1666460963}{2048000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {251559}{12800} i \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x - \frac {21}{11}\right ) + \frac {10161}{1280} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {2079}{32} \, \sqrt {10 \, x^{2} - 21 \, x + 8} x + \frac {29403}{5120} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {43659}{640} \, \sqrt {10 \, x^{2} - 21 \, x + 8} - \frac {34897797}{102400} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2401 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{96 \, {\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} + \frac {1029 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {1323 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{32 \, {\left (2 \, x - 1\right )}} + \frac {26411 \, \sqrt {-10 \, x^{2} - x + 3}}{192 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {491519 \, \sqrt {-10 \, x^{2} - x + 3}}{192 \, {\left (2 \, x - 1\right )}} \]
81/64*(-10*x^2 - x + 3)^(3/2)*x + 1666460963/2048000*sqrt(5)*sqrt(2)*arcsi n(20/11*x + 1/11) + 251559/12800*I*sqrt(5)*sqrt(2)*arcsin(20/11*x - 21/11) + 10161/1280*(-10*x^2 - x + 3)^(3/2) - 2079/32*sqrt(10*x^2 - 21*x + 8)*x + 29403/5120*sqrt(-10*x^2 - x + 3)*x + 43659/640*sqrt(10*x^2 - 21*x + 8) - 34897797/102400*sqrt(-10*x^2 - x + 3) - 2401/96*(-10*x^2 - x + 3)^(3/2)/( 8*x^3 - 12*x^2 + 6*x - 1) + 1029/8*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) + 1323/32*(-10*x^2 - x + 3)^(3/2)/(2*x - 1) + 26411/192*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 491519/192*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\frac {1626211523}{1024000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (9 \, {\left (12 \, {\left (8 \, {\left (36 \, \sqrt {5} {\left (5 \, x + 3\right )} + 427 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 42657 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 9855815 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 3252423046 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 53664980259 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{38400000 \, {\left (2 \, x - 1\right )}^{2}} \]
1626211523/1024000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/384000 00*(4*(9*(12*(8*(36*sqrt(5)*(5*x + 3) + 427*sqrt(5))*(5*x + 3) + 42657*sqr t(5))*(5*x + 3) + 9855815*sqrt(5))*(5*x + 3) - 3252423046*sqrt(5))*(5*x + 3) + 53664980259*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x)^4 (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{5/2}} \,d x \]