Integrand size = 17, antiderivative size = 95 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=-\frac {x}{2}-\frac {1}{2} \sqrt {-x+x^2}-\frac {1}{16} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-\frac {7}{8} \text {arctanh}\left (\frac {x}{\sqrt {-x+x^2}}\right )+\frac {1}{16} \log (1+8 x)+x \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]
-1/2*x-1/16*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))-7/8*arctanh(x/(x^2-x)^(1/2 ))+1/16*ln(1+8*x)+x*ln(-1+4*x+4*(x^2-x)^(1/2))-1/2*(x^2-x)^(1/2)
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\frac {1}{16} \left (-8 x-8 \sqrt {(-1+x) x}+2 \log (1+8 x)-7 \log \left (1-2 x-2 \sqrt {(-1+x) x}\right )+16 x \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )-\log \left (1-10 x+6 \sqrt {(-1+x) x}\right )\right ) \]
(-8*x - 8*Sqrt[(-1 + x)*x] + 2*Log[1 + 8*x] - 7*Log[1 - 2*x - 2*Sqrt[(-1 + x)*x]] + 16*x*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]] - Log[1 - 10*x + 6*Sqrt[ (-1 + x)*x]])/16
Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3017, 3013, 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.
\(\displaystyle \int \log \left (4 x+4 \sqrt {(x-1) x}-1\right ) \, dx\) |
\(\Big \downarrow \) 3017 |
\(\displaystyle \int \log \left (4 \sqrt {x^2-x}+4 x-1\right )dx\) |
\(\Big \downarrow \) 3013 |
\(\displaystyle 8 \int -\frac {x}{4 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx+x \log \left (4 \sqrt {x^2-x}+4 x-1\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle x \log \left (4 \sqrt {x^2-x}+4 x-1\right )-2 \int \frac {x}{\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle x \log \left (4 \sqrt {x^2-x}+4 x-1\right )-2 \int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {2 \sqrt {x^2-x}}{3 (-8 x-1)}-\frac {1}{4 (8 x+1)}+\frac {1}{4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \log \left (4 \sqrt {x^2-x}+4 x-1\right )-2 \left (\frac {1}{32} \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )+\frac {7}{16} \text {arctanh}\left (\frac {x}{\sqrt {x^2-x}}\right )+\frac {\sqrt {x^2-x}}{4}+\frac {x}{4}-\frac {1}{32} \log (8 x+1)\right )\) |
-2*(x/4 + Sqrt[-x + x^2]/4 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]/32 + ( 7*ArcTanh[x/Sqrt[-x + x^2]])/16 - Log[1 + 8*x]/32) + x*Log[-1 + 4*x + 4*Sq rt[-x + x^2]]
3.2.4.3.1 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]] , x_Symbol] :> Simp[x*Log[d + e*x + f*Sqrt[a + b*x + c*x^2]], x] + Simp[f^2 *((b^2 - 4*a*c)/2) Int[x/((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}, x] && EqQ[e^2 - c*f^2, 0]
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 = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84
method | result | size |
default | \(x \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )-\frac {7 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{16}-\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{16}-\frac {\sqrt {x^{2}-x}}{2}-\frac {x}{2}+\frac {\ln \left (1+8 x \right )}{16}\) | \(80\) |
parts | \(x \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )-\frac {19 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )}{32}-\frac {\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )}{16}+\frac {\sqrt {x^{2}-x}}{4}-x \sqrt {x^{2}-x}-\frac {x}{2}+\frac {\ln \left (1+8 x \right )}{16}+\frac {\left (2 x -1\right ) \sqrt {x^{2}-x}}{2}-\frac {\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}{32}+\frac {5 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )}{32}\) | \(142\) |
x*ln(-1+4*x+4*((-1+x)*x)^(1/2))-7/16*ln(-1/2+x+(x^2-x)^(1/2))-1/16*arctanh (32/3*(1/8-5/4*x)/(64*(x+1/8)^2-80*x-1)^(1/2))-1/2*(x^2-x)^(1/2)-1/2*x+1/1 6*ln(1+8*x)
Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx={\left (x + 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - x} - \frac {7}{16} \, \log \left (8 \, x + 1\right ) + \frac {15}{16} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) - \frac {7}{16} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) + \frac {7}{16} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]
(x + 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/2*x - 1/2*sqrt(x^2 - x) - 7/16* log(8*x + 1) + 15/16*log(-2*x + 2*sqrt(x^2 - x) + 1) - 7/16*log(-2*x + 2*s qrt(x^2 - x) - 1) + 7/16*log(-4*x + 4*sqrt(x^2 - x) + 1)
\[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int \log {\left (4 x + 4 \sqrt {x \left (x - 1\right )} - 1 \right )}\, dx \]
\[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int { \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) \,d x } \]
x*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 1/2*x + integrate(1/2*(2*x^2 + x) /(4*x^3 - 5*x^2 + 4*(x^(5/2) - x^(3/2))*sqrt(x - 1) + x), x) - 1/2*log(sqr t(x) + 1) - 1/2*log(sqrt(x) - 1)
Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=x \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} - x} + \frac {1}{16} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {7}{16} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) - \frac {1}{16} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
x*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/2*x - 1/2*sqrt(x^2 - x) + 1/16*log( abs(8*x + 1)) + 7/16*log(abs(-2*x + 2*sqrt(x^2 - x) + 1)) + 1/16*log(abs(- 2*x + 2*sqrt(x^2 - x) - 1)) - 1/16*log(abs(-4*x + 4*sqrt(x^2 - x) + 1))
Timed out. \[ \int \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx=\int \ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]