Integrand size = 24, antiderivative size = 120 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}+\frac {1428 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}-\frac {12803 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \]
-12803/1890625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-21/68750*(704 -375*x)*(1-2*x)^(1/2)+1428/6875*(2+3*x)^2*(1-2*x)^(1/2)-1/10*(2+3*x)^4*(1- 2*x)^(1/2)/(3+5*x)^2-131/550*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {\frac {55 \sqrt {1-2 x} \left (-121976-200305 x+506880 x^2+1103850 x^3+445500 x^4\right )}{(3+5 x)^2}-25606 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3781250} \]
((55*Sqrt[1 - 2*x]*(-121976 - 200305*x + 506880*x^2 + 1103850*x^3 + 445500 *x^4))/(3 + 5*x)^2 - 25606*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/378 1250
Time = 0.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {108, 166, 27, 170, 25, 164, 73, 219}
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.
| \(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^4}{(5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{10} \int \frac {(10-27 x) (3 x+2)^3}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{55} \int \frac {7 (121-408 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \int \frac {(121-408 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \left (\frac {408}{25} \sqrt {1-2 x} (3 x+2)^2-\frac {1}{25} \int -\frac {(1154-1125 x) (3 x+2)}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \left (\frac {1}{25} \int \frac {(1154-1125 x) (3 x+2)}{\sqrt {1-2 x} (5 x+3)}dx+\frac {408}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \left (\frac {1}{25} \left (\frac {1829}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3}{5} (704-375 x) \sqrt {1-2 x}\right )+\frac {408}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \left (\frac {1}{25} \left (-\frac {1829}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3}{5} \sqrt {1-2 x} (704-375 x)\right )+\frac {408}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{10} \left (\frac {7}{55} \left (\frac {1}{25} \left (-\frac {3658 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {3}{5} \sqrt {1-2 x} (704-375 x)\right )+\frac {408}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}\) |
-1/10*(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^2 + ((-131*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(55*(3 + 5*x)) + (7*((408*Sqrt[1 - 2*x]*(2 + 3*x)^2)/25 + ((-3*(70 4 - 375*x)*Sqrt[1 - 2*x])/5 - (3658*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5* Sqrt[55]))/25))/55)/10
3.19.48.3.1 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] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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] && ILtQ[m, -1] && GtQ[n, 0]
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 0.99 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51
| method | result | size |
| risch | \(-\frac {891000 x^{5}+1762200 x^{4}-90090 x^{3}-907490 x^{2}-43647 x +121976}{68750 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {12803 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1890625}\) | \(61\) |
| pseudoelliptic | \(\frac {-25606 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+55 \sqrt {1-2 x}\, \left (445500 x^{4}+1103850 x^{3}+506880 x^{2}-200305 x -121976\right )}{3781250 \left (3+5 x \right )^{2}}\) | \(65\) |
| derivativedivides | \(\frac {81 \left (1-2 x \right )^{\frac {5}{2}}}{1250}-\frac {369 \left (1-2 x \right )^{\frac {3}{2}}}{1250}+\frac {108 \sqrt {1-2 x}}{3125}+\frac {\frac {263 \left (1-2 x \right )^{\frac {3}{2}}}{6875}-\frac {53 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {12803 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1890625}\) | \(75\) |
| default | \(\frac {81 \left (1-2 x \right )^{\frac {5}{2}}}{1250}-\frac {369 \left (1-2 x \right )^{\frac {3}{2}}}{1250}+\frac {108 \sqrt {1-2 x}}{3125}+\frac {\frac {263 \left (1-2 x \right )^{\frac {3}{2}}}{6875}-\frac {53 \sqrt {1-2 x}}{625}}{\left (-6-10 x \right )^{2}}-\frac {12803 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1890625}\) | \(75\) |
| trager | \(\frac {\left (445500 x^{4}+1103850 x^{3}+506880 x^{2}-200305 x -121976\right ) \sqrt {1-2 x}}{68750 \left (3+5 x \right )^{2}}+\frac {12803 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3781250}\) | \(82\) |
-1/68750*(891000*x^5+1762200*x^4-90090*x^3-907490*x^2-43647*x+121976)/(3+5 *x)^2/(1-2*x)^(1/2)-12803/1890625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^ (1/2)
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {12803 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (445500 \, x^{4} + 1103850 \, x^{3} + 506880 \, x^{2} - 200305 \, x - 121976\right )} \sqrt {-2 \, x + 1}}{3781250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/3781250*(12803*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2* x + 1) - 8)/(5*x + 3)) + 55*(445500*x^4 + 1103850*x^3 + 506880*x^2 - 20030 5*x - 121976)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Time = 164.85 (sec) , antiderivative size = 366, normalized size of antiderivative = 3.05 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {81 \left (1 - 2 x\right )^{\frac {5}{2}}}{1250} - \frac {369 \left (1 - 2 x\right )^{\frac {3}{2}}}{1250} + \frac {108 \sqrt {1 - 2 x}}{3125} + \frac {114 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {104 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{3125} \]
81*(1 - 2*x)**(5/2)/1250 - 369*(1 - 2*x)**(3/2)/1250 + 108*sqrt(1 - 2*x)/3 125 + 114*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/34375 - 104*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/ 11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x ) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625 + 88*Piecewise((sqrt (55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2* x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*s qrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16* (sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/3125
Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {81}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {369}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12803}{3781250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {108}{3125} \, \sqrt {-2 \, x + 1} + \frac {263 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 583 \, \sqrt {-2 \, x + 1}}{6875 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
81/1250*(-2*x + 1)^(5/2) - 369/1250*(-2*x + 1)^(3/2) + 12803/3781250*sqrt( 55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 10 8/3125*sqrt(-2*x + 1) + 1/6875*(263*(-2*x + 1)^(3/2) - 583*sqrt(-2*x + 1)) /(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {81}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {369}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12803}{3781250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {108}{3125} \, \sqrt {-2 \, x + 1} + \frac {263 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 583 \, \sqrt {-2 \, x + 1}}{27500 \, {\left (5 \, x + 3\right )}^{2}} \]
81/1250*(2*x - 1)^2*sqrt(-2*x + 1) - 369/1250*(-2*x + 1)^(3/2) + 12803/378 1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*s qrt(-2*x + 1))) + 108/3125*sqrt(-2*x + 1) + 1/27500*(263*(-2*x + 1)^(3/2) - 583*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx=\frac {108\,\sqrt {1-2\,x}}{3125}-\frac {369\,{\left (1-2\,x\right )}^{3/2}}{1250}+\frac {81\,{\left (1-2\,x\right )}^{5/2}}{1250}-\frac {\frac {53\,\sqrt {1-2\,x}}{15625}-\frac {263\,{\left (1-2\,x\right )}^{3/2}}{171875}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,12803{}\mathrm {i}}{1890625} \]