Integrand size = 21, antiderivative size = 76 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\frac {4 \sqrt {-x+x^2}}{x}+4 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )+4 \log (x)-4 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{x} \]
4*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))+4*ln(x)-4*ln(1+8*x)-ln(-1+4*x+4*(x^2 -x)^(1/2))/x+4*(x^2-x)^(1/2)/x
Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.87 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=2 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )+\frac {4 \sqrt {(-1+x) x}+\frac {4 \sqrt {-(-1+x)^2 x^2} \arcsin \left (\sqrt {1-x}\right )}{-1+x}+4 x \log (x)-2 x \log (1+8 x)-4 x \log \left (1-4 x-4 \sqrt {(-1+x) x}\right )+4 x \log \left (1-2 x-2 \sqrt {(-1+x) x}\right )-\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x} \]
2*ArcTanh[(1 - 10*x)/(6*Sqrt[(-1 + x)*x])] + (4*Sqrt[(-1 + x)*x] + (4*Sqrt [-((-1 + x)^2*x^2)]*ArcSin[Sqrt[1 - x]])/(-1 + x) + 4*x*Log[x] - 2*x*Log[1 + 8*x] - 4*x*Log[1 - 4*x - 4*Sqrt[(-1 + x)*x]] + 4*x*Log[1 - 2*x - 2*Sqrt [(-1 + x)*x]] - Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])/x
Time = 0.48 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, 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.
\(\displaystyle \int \frac {\log \left (4 x+4 \sqrt {(x-1) x}-1\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 3017 |
\(\displaystyle \int \frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x^2}dx\) |
\(\Big \downarrow \) 3015 |
\(\displaystyle -8 \int -\frac {1}{4 x \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {1}{x \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {128 \sqrt {x^2-x}}{3 (-8 x-1)}-\frac {16}{8 x+1}+\frac {5 \sqrt {x^2-x}}{x}+\frac {2}{x}-\frac {\sqrt {x^2-x}}{x^2}\right )dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (2 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )+\frac {2 \sqrt {x^2-x}}{x}+2 \log (x)-2 \log (8 x+1)\right )-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x}\) |
2*((2*Sqrt[-x + x^2])/x + 2*ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])] + 2*Log [x] - 2*Log[1 + 8*x]) - Log[-1 + 4*x + 4*Sqrt[-x + x^2]]/x
3.2.6.3.1 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 = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.74
method | result | size |
parts | \(-\frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{x}+\frac {4 \sqrt {x^{2}-x}}{x}+4 \,\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )+10 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )-4 \ln \left (1+8 x \right )+4 \ln \left (x \right )-16 \sqrt {x^{2}-x}+2 \sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}-10 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )\) | \(132\) |
-ln(-1+4*x+4*((-1+x)*x)^(1/2))/x+4*(x^2-x)^(1/2)/x+4*arctanh(32/3*(1/8-5/4 *x)/(64*(x+1/8)^2-80*x-1)^(1/2))+10*ln(-1/2+x+(x^2-x)^(1/2))-4*ln(1+8*x)+4 *ln(x)-16*(x^2-x)^(1/2)+2*(64*(x+1/8)^2-80*x-1)^(1/2)-10*ln(-1/2+x+((x+1/8 )^2-5/4*x-1/64)^(1/2))
Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.51 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=-\frac {7 \, x \log \left (8 \, x + 1\right ) + 2 \, {\left (x + 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 8 \, x \log \left (x\right ) + x \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + 7 \, x \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - 7 \, x \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) - 8 \, x - 8 \, \sqrt {x^{2} - x}}{2 \, x} \]
-1/2*(7*x*log(8*x + 1) + 2*(x + 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 8*x*lo g(x) + x*log(-2*x + 2*sqrt(x^2 - x) + 1) + 7*x*log(-2*x + 2*sqrt(x^2 - x) - 1) - 7*x*log(-4*x + 4*sqrt(x^2 - x) + 1) - 8*x - 8*sqrt(x^2 - x))/x
\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int \frac {\log {\left (4 x + 4 \sqrt {x^{2} - x} - 1 \right )}}{x^{2}}\, dx \]
\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int { \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{x^{2}} \,d x } \]
Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=-\frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{x} + \frac {4}{x - \sqrt {x^{2} - x}} - 4 \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + 4 \, \log \left ({\left | x \right |}\right ) - 4 \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) + 4 \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
-log(4*x + 4*sqrt((x - 1)*x) - 1)/x + 4/(x - sqrt(x^2 - x)) - 4*log(abs(8* x + 1)) + 4*log(abs(x)) - 4*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) + 4*log(a bs(-4*x + 4*sqrt(x^2 - x) + 1))
Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^2} \, dx=\int \frac {\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right )}{x^2} \,d x \]