Integrand size = 19, antiderivative size = 127 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {x}{32}-\frac {x^2}{8}-\frac {11}{32} \sqrt {-x+x^2}+\frac {1}{16} (1-2 x) \sqrt {-x+x^2}+\frac {1}{256} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {33}{128} \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )-\frac {1}{256} \log (1+8 x)+\frac {1}{2} x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]
1/32*x-1/8*x^2+1/256*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))-33/128*arctanh(x/ (x^2-x)^(1/2))-1/256*ln(1+8*x)+1/2*x^2*ln(-1+4*x+4*(x^2-x)^(1/2))-11/32*(x ^2-x)^(1/2)+1/16*(1-2*x)*(x^2-x)^(1/2)
Time = 0.41 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.50 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {8 \sqrt {1-x} x^{3/2}-32 \sqrt {1-x} x^{5/2}-32 \sqrt {1-x} x^{3/2} \sqrt {(-1+x) x}-72 \sqrt {-(-1+x)^2 x^2}-66 \sqrt {(-1+x) x} \arcsin \left (\sqrt {1-x}\right )+\sqrt {-((-1+x) x)} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )-\sqrt {-((-1+x) x)} \log (1+8 x)+128 \sqrt {1-x} x^{5/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{256 \sqrt {-((-1+x) x)}} \]
(8*Sqrt[1 - x]*x^(3/2) - 32*Sqrt[1 - x]*x^(5/2) - 32*Sqrt[1 - x]*x^(3/2)*S qrt[(-1 + x)*x] - 72*Sqrt[-((-1 + x)^2*x^2)] - 66*Sqrt[(-1 + x)*x]*ArcSin[ Sqrt[1 - x]] + Sqrt[-((-1 + x)*x)]*ArcTanh[(1 - 10*x)/(6*Sqrt[(-1 + x)*x]) ] - Sqrt[-((-1 + x)*x)]*Log[1 + 8*x] + 128*Sqrt[1 - x]*x^(5/2)*Log[-1 + 4* x + 4*Sqrt[(-1 + x)*x]])/(256*Sqrt[-((-1 + x)*x)])
Time = 0.45 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3017, 3015, 27, 7293, 2009}
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.
\(\displaystyle \int x \log \left (4 x+4 \sqrt {(x-1) x}-1\right ) \, dx\) |
\(\Big \downarrow \) 3017 |
\(\displaystyle \int x \log \left (4 \sqrt {x^2-x}+4 x-1\right )dx\) |
\(\Big \downarrow \) 3015 |
\(\displaystyle 4 \int -\frac {x^2}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+\frac {1}{2} x^2 \log \left (4 \sqrt {x^2-x}+4 x-1\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} x^2 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\int \frac {x^2}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} x^2 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {x}{4}+\frac {\sqrt {x^2-x}}{12 (8 x+1)}+\frac {\sqrt {x^2-x}}{4}+\frac {1}{32 (8 x+1)}-\frac {1}{32}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{256} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )-\frac {33}{128} \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )-\frac {x^2}{8}+\frac {1}{16} (1-2 x) \sqrt {x^2-x}-\frac {11 \sqrt {x^2-x}}{32}+\frac {1}{2} x^2 \log \left (4 \sqrt {x^2-x}+4 x-1\right )+\frac {x}{32}-\frac {1}{256} \log (8 x+1)\) |
x/32 - x^2/8 - (11*Sqrt[-x + x^2])/32 + ((1 - 2*x)*Sqrt[-x + x^2])/16 + Ar cTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]/256 - (33*ArcTanh[x/Sqrt[-x + x^2]])/ 128 - Log[1 + 8*x]/256 + (x^2*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/2
3.2.3.3.1 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[Log[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]] *((g_.)*(x_))^(m_.), x_Symbol] :> Simp[(g*x)^(m + 1)*(Log[d + e*x + f*Sqrt[ a + b*x + c*x^2]]/(g*(m + 1))), x] + Simp[f^2*((b^2 - 4*a*c)/(2*g*(m + 1))) Int[(g*x)^(m + 1)/((2*d*e - b*f^2)*(a + b*x + c*x^2) - f*(b*d - 2*a*e + (2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[e^2 - c*f^2, 0] && NeQ[m, -1] && IntegerQ[2*m]
Int[Log[(d_.) + (f_.)*Sqrt[u_] + (e_.)*(x_)]*(v_.), x_Symbol] :> Int[v*Log[ d + e*x + f*Sqrt[ExpandToSum[u, x]]], x] /; FreeQ[{d, e, f}, x] && Quadrati cQ[u, x] && !QuadraticMatchQ[u, x] && (EqQ[v, 1] || MatchQ[v, ((g_.)*x)^(m _.) /; FreeQ[{g, m}, x]])
Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.38
method | result | size |
parts | \(\frac {x^{2} \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{2}-\frac {61 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{512}-\frac {13 \sqrt {x^{2}-x}}{64}+\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{256}+\frac {x \sqrt {x^{2}-x}}{48}-\frac {x^{2} \sqrt {x^{2}-x}}{3}-\frac {x^{2}}{8}+\frac {x}{32}-\frac {\ln \left (1+8 x \right )}{256}+\frac {3 \left (2 x -1\right ) \sqrt {x^{2}-x}}{32}+\frac {\left (x^{2}-x \right )^{\frac {3}{2}}}{3}+\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{512}-\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{512}\) | \(175\) |
1/2*x^2*ln(-1+4*x+4*((-1+x)*x)^(1/2))-61/512*ln(-1/2+x+(x^2-x)^(1/2))-13/6 4*(x^2-x)^(1/2)+1/256*arctanh(32/3*(1/8-5/4*x)/(64*(x+1/8)^2-80*x-1)^(1/2) )+1/48*x*(x^2-x)^(1/2)-1/3*x^2*(x^2-x)^(1/2)-1/8*x^2+1/32*x-1/256*ln(1+8*x )+3/32*(2*x-1)*(x^2-x)^(1/2)+1/3*(x^2-x)^(3/2)+1/512*(64*(x+1/8)^2-80*x-1) ^(1/2)-5/512*ln(-1/2+x+((x+1/8)^2-5/4*x-1/64)^(1/2))
Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {1}{8} \, x^{2} + \frac {1}{2} \, {\left (x^{2} - 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{32} \, \sqrt {x^{2} - x} {\left (4 \, x + 9\right )} + \frac {1}{32} \, x + \frac {63}{256} \, \log \left (8 \, x + 1\right ) - \frac {31}{256} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + \frac {63}{256} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - \frac {63}{256} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]
-1/8*x^2 + 1/2*(x^2 - 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/32*sqrt(x^2 - x)*(4*x + 9) + 1/32*x + 63/256*log(8*x + 1) - 31/256*log(-2*x + 2*sqrt(x^2 - x) + 1) + 63/256*log(-2*x + 2*sqrt(x^2 - x) - 1) - 63/256*log(-4*x + 4* sqrt(x^2 - x) + 1)
\[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int x \log {\left (4 x + 4 \sqrt {x^{2} - x} - 1 \right )}\, dx \]
\[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { x \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \]
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{8} \, x^{2} - \frac {1}{32} \, \sqrt {x^{2} - x} {\left (4 \, x + 9\right )} + \frac {1}{32} \, x - \frac {1}{256} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {33}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) - \frac {1}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + \frac {1}{256} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
1/2*x^2*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/8*x^2 - 1/32*sqrt(x^2 - x)*(4 *x + 9) + 1/32*x - 1/256*log(abs(8*x + 1)) + 33/256*log(abs(-2*x + 2*sqrt( x^2 - x) + 1)) - 1/256*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) + 1/256*log(ab s(-4*x + 4*sqrt(x^2 - x) + 1))
Timed out. \[ \int x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int x\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]