Integrand size = 21, antiderivative size = 101 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=-\frac {2}{x}-\frac {10 \sqrt {-x+x^2}}{x}-\frac {2 \left (-x+x^2\right )^{3/2}}{3 x^3}-16 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )-16 \log (x)+16 \log (1+8 x)-\frac {\log \left (-1+4 x+4 \sqrt {-x+x^2}\right )}{2 x^2} \]
-2/x-2/3*(x^2-x)^(3/2)/x^3-16*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))-16*ln(x) +16*ln(1+8*x)-1/2*ln(-1+4*x+4*(x^2-x)^(1/2))/x^2-10*(x^2-x)^(1/2)/x
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=-\frac {12 x-4 \sqrt {(-1+x) x}+64 x \sqrt {(-1+x) x}+96 x^2 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {(-1+x) x}}\right )+96 x^2 \log (x)-96 x^2 \log (1+8 x)+3 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{6 x^2} \]
-1/6*(12*x - 4*Sqrt[(-1 + x)*x] + 64*x*Sqrt[(-1 + x)*x] + 96*x^2*ArcTanh[( 1 - 10*x)/(6*Sqrt[(-1 + x)*x])] + 96*x^2*Log[x] - 96*x^2*Log[1 + 8*x] + 3* Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]])/x^2
Time = 0.49 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, 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.
\(\displaystyle \int \frac {\log \left (4 x+4 \sqrt {(x-1) x}-1\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 3017 |
\(\displaystyle \int \frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{x^3}dx\) |
\(\Big \downarrow \) 3015 |
\(\displaystyle -4 \int -\frac {1}{4 x^2 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^2 \left (\sqrt {x^2-x} (2 x+1)+2 \left (x-x^2\right )\right )}dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{2 x^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x}{3 \sqrt {x^2-x}}+\frac {1024 \sqrt {x^2-x}}{3 (8 x+1)}+\frac {128}{8 x+1}-\frac {43 \sqrt {x^2-x}}{x}-\frac {16}{x}+\frac {5 \sqrt {x^2-x}}{x^2}+\frac {2}{x^2}-\frac {\sqrt {x^2-x}}{x^3}\right )dx-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -16 \text {arctanh}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )-\frac {10 \sqrt {x^2-x}}{x}-\frac {\log \left (4 \sqrt {x^2-x}+4 x-1\right )}{2 x^2}-\frac {2 \left (x^2-x\right )^{3/2}}{3 x^3}-\frac {2}{x}-16 \log (x)+16 \log (8 x+1)\) |
-2/x - (10*Sqrt[-x + x^2])/x - (2*(-x + x^2)^(3/2))/(3*x^3) - 16*ArcTanh[( 1 - 10*x)/(6*Sqrt[-x + x^2])] - 16*Log[x] + 16*Log[1 + 8*x] - Log[-1 + 4*x + 4*Sqrt[-x + x^2]]/(2*x^2)
3.2.7.3.1 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]])
Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(87)=174\).
Time = 0.08 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.84
method | result | size |
parts | \(-\frac {\ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )}{2 x^{2}}+\frac {2 \sqrt {x^{2}-x}}{3 x^{2}}-\frac {80 \sqrt {x^{2}-x}}{3 x}-16 \,\operatorname {arctanh}\left (\frac {\frac {4}{3}-\frac {40 x}{3}}{\sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}}\right )+\frac {32 \ln \left (1+8 x \right ) x -32 \ln \left (x \right ) x -4}{x}-16 \ln \left (1+8 x \right )+16 \ln \left (x \right )+\frac {2}{x}-\frac {16 \left (x^{2}-x \right )^{\frac {3}{2}}}{x^{2}}+80 \sqrt {x^{2}-x}-40 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x}\right )-8 \sqrt {64 \left (x +\frac {1}{8}\right )^{2}-80 x -1}+40 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +\frac {1}{8}\right )^{2}-\frac {5 x}{4}-\frac {1}{64}}\right )\) | \(186\) |
-1/2*ln(-1+4*x+4*((-1+x)*x)^(1/2))/x^2+2/3*(x^2-x)^(1/2)/x^2-80/3*(x^2-x)^ (1/2)/x-16*arctanh(32/3*(1/8-5/4*x)/(64*(x+1/8)^2-80*x-1)^(1/2))+4/x*(8*ln (1+8*x)*x-8*ln(x)*x-1)-16*ln(1+8*x)+16*ln(x)+2/x-16/x^2*(x^2-x)^(3/2)+80*( x^2-x)^(1/2)-40*ln(-1/2+x+(x^2-x)^(1/2))-8*(64*(x+1/8)^2-80*x-1)^(1/2)+40* ln(-1/2+x+((x+1/8)^2-5/4*x-1/64)^(1/2))
Time = 0.38 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.37 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=\frac {189 \, x^{2} \log \left (8 \, x + 1\right ) - 192 \, x^{2} \log \left (x\right ) + 3 \, x^{2} \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) + 189 \, x^{2} \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) - 189 \, x^{2} \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) - 128 \, x^{2} + 6 \, {\left (x^{2} - 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 8 \, \sqrt {x^{2} - x} {\left (16 \, x - 1\right )} - 24 \, x}{12 \, x^{2}} \]
1/12*(189*x^2*log(8*x + 1) - 192*x^2*log(x) + 3*x^2*log(-2*x + 2*sqrt(x^2 - x) + 1) + 189*x^2*log(-2*x + 2*sqrt(x^2 - x) - 1) - 189*x^2*log(-4*x + 4 *sqrt(x^2 - x) + 1) - 128*x^2 + 6*(x^2 - 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 8*sqrt(x^2 - x)*(16*x - 1) - 24*x)/x^2
Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=\int { \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{x^{3}} \,d x } \]
Time = 0.44 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.29 \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=-\frac {2}{x} - \frac {\log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )}{2 \, x^{2}} - \frac {2 \, {\left (18 \, {\left (x - \sqrt {x^{2} - x}\right )}^{2} - 3 \, x + 3 \, \sqrt {x^{2} - x} + 1\right )}}{3 \, {\left (x - \sqrt {x^{2} - x}\right )}^{3}} + 16 \, \log \left ({\left | 8 \, x + 1 \right |}\right ) - 16 \, \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) - 16 \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]
-2/x - 1/2*log(4*x + 4*sqrt((x - 1)*x) - 1)/x^2 - 2/3*(18*(x - sqrt(x^2 - x))^2 - 3*x + 3*sqrt(x^2 - x) + 1)/(x - sqrt(x^2 - x))^3 + 16*log(abs(8*x + 1)) - 16*log(abs(x)) + 16*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) - 16*log( abs(-4*x + 4*sqrt(x^2 - x) + 1))
Timed out. \[ \int \frac {\log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )}{x^3} \, dx=\int \frac {\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right )}{x^3} \,d x \]