Integrand size = 54, antiderivative size = 28 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=\frac {\left (2+\log \left (\frac {\log (4) \left (x+\frac {1}{3} (x-\log (x))\right )}{x}\right )\right )^2}{\log ^8(5)} \]
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=\frac {\left (2+\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )\right )^2}{\log ^8(5)} \]
Integrate[(4 - 4*Log[x] + (2 - 2*Log[x])*Log[(4*x*Log[4] - Log[4]*Log[x])/ (3*x)])/(-4*x^2*Log[5]^8 + x*Log[5]^8*Log[x]),x]
Time = 0.49 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {3041, 7292, 27, 25, 7237}
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.
\(\displaystyle \int \frac {-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )+4}{x \log ^8(5) \log (x)-4 x^2 \log ^8(5)} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )+4}{x \left (\log ^8(5) \log (x)-4 x \log ^8(5)\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 (1-\log (x)) \left (-\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )-2\right )}{x \log ^8(5) (4 x-\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int -\frac {(1-\log (x)) \left (\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )+2\right )}{x (4 x-\log (x))}dx}{\log ^8(5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 \int \frac {(1-\log (x)) \left (\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )+2\right )}{x (4 x-\log (x))}dx}{\log ^8(5)}\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {\left (\log \left (\frac {\log (4) (4 x-\log (x))}{3 x}\right )+2\right )^2}{\log ^8(5)}\) |
Int[(4 - 4*Log[x] + (2 - 2*Log[x])*Log[(4*x*Log[4] - Log[4]*Log[x])/(3*x)] )/(-4*x^2*Log[5]^8 + x*Log[5]^8*Log[x]),x]
3.27.68.3.1 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[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.)) ^(p_.), x_Symbol] :> Int[u*x^(p*r)*(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 0.80 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00
method | result | size |
norman | \(\frac {\frac {4 \ln \left (\frac {-2 \ln \left (2\right ) \ln \left (x \right )+8 x \ln \left (2\right )}{3 x}\right )}{\ln \left (5\right )}+\frac {\ln \left (\frac {-2 \ln \left (2\right ) \ln \left (x \right )+8 x \ln \left (2\right )}{3 x}\right )^{2}}{\ln \left (5\right )}}{\ln \left (5\right )^{7}}\) | \(56\) |
default | \(\frac {-4 \ln \left (x \right )+4 \ln \left (-4 x +\ln \left (x \right )\right )+\ln \left (\frac {4 x -\ln \left (x \right )}{x}\right )^{2}-2 \ln \left (\ln \left (2\right )\right ) \left (\ln \left (x \right )-\ln \left (-4 x +\ln \left (x \right )\right )\right )}{\ln \left (5\right )^{8}}-\frac {2 \ln \left (2\right ) \left (\ln \left (x \right )-\ln \left (-4 x +\ln \left (x \right )\right )\right )}{\ln \left (5\right )^{8}}+\frac {2 \ln \left (3\right ) \left (\ln \left (x \right )-\ln \left (-4 x +\ln \left (x \right )\right )\right )}{\ln \left (5\right )^{8}}\) | \(96\) |
risch | \(\frac {\ln \left (x -\frac {\ln \left (x \right )}{4}\right )^{2}}{\ln \left (5\right )^{8}}-\frac {2 \ln \left (x \right ) \ln \left (x -\frac {\ln \left (x \right )}{4}\right )}{\ln \left (5\right )^{8}}+\frac {i \pi \ln \left (-4 x +\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{2}}{\ln \left (5\right )^{8}}+\frac {i \pi \ln \left (-4 x +\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{3}}{\ln \left (5\right )^{8}}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{2}}{\ln \left (5\right )^{8}}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-x +\frac {\ln \left (x \right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )}{\ln \left (5\right )^{8}}-\frac {i \pi \ln \left (-4 x +\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-x +\frac {\ln \left (x \right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )}{\ln \left (5\right )^{8}}+\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (-x +\frac {\ln \left (x \right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{2}}{\ln \left (5\right )^{8}}-\frac {i \pi \ln \left (x \right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{3}}{\ln \left (5\right )^{8}}-\frac {i \pi \ln \left (-4 x +\ln \left (x \right )\right ) \operatorname {csgn}\left (i \left (-x +\frac {\ln \left (x \right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x +\frac {\ln \left (x \right )}{4}\right )}{x}\right )^{2}}{\ln \left (5\right )^{8}}-\frac {2 \ln \left (x \right ) \ln \left (\ln \left (2\right )\right )}{\ln \left (5\right )^{8}}+\frac {2 \ln \left (\ln \left (2\right )\right ) \ln \left (-4 x +\ln \left (x \right )\right )}{\ln \left (5\right )^{8}}+\frac {\ln \left (x \right )^{2}}{\ln \left (5\right )^{8}}-\frac {6 \ln \left (2\right ) \ln \left (x \right )}{\ln \left (5\right )^{8}}+\frac {2 \ln \left (3\right ) \ln \left (x \right )}{\ln \left (5\right )^{8}}+\frac {6 \ln \left (-4 x +\ln \left (x \right )\right ) \ln \left (2\right )}{\ln \left (5\right )^{8}}-\frac {2 \ln \left (-4 x +\ln \left (x \right )\right ) \ln \left (3\right )}{\ln \left (5\right )^{8}}-\frac {4 \ln \left (x \right )}{\ln \left (5\right )^{8}}+\frac {4 \ln \left (-4 x +\ln \left (x \right )\right )}{\ln \left (5\right )^{8}}\) | \(446\) |
int(((-2*ln(x)+2)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+4-4*ln(x))/(x*ln(5) ^8*ln(x)-4*x^2*ln(5)^8),x,method=_RETURNVERBOSE)
(4/ln(5)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+1/ln(5)*ln(1/3*(-2*ln(2)*ln( x)+8*x*ln(2))/x)^2)/ln(5)^7
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=\frac {\log \left (\frac {2 \, {\left (4 \, x \log \left (2\right ) - \log \left (2\right ) \log \left (x\right )\right )}}{3 \, x}\right )^{2} + 4 \, \log \left (\frac {2 \, {\left (4 \, x \log \left (2\right ) - \log \left (2\right ) \log \left (x\right )\right )}}{3 \, x}\right )}{\log \left (5\right )^{8}} \]
integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log( x))/(x*log(5)^8*log(x)-4*x^2*log(5)^8),x, algorithm=\
(log(2/3*(4*x*log(2) - log(2)*log(x))/x)^2 + 4*log(2/3*(4*x*log(2) - log(2 )*log(x))/x))/log(5)^8
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=- \frac {4 \log {\left (x \right )}}{\log {\left (5 \right )}^{8}} + \frac {\log {\left (\frac {\frac {8 x \log {\left (2 \right )}}{3} - \frac {2 \log {\left (2 \right )} \log {\left (x \right )}}{3}}{x} \right )}^{2}}{\log {\left (5 \right )}^{8}} + \frac {4 \log {\left (- 4 x + \log {\left (x \right )} \right )}}{\log {\left (5 \right )}^{8}} \]
integrate(((-2*ln(x)+2)*ln(1/3*(-2*ln(2)*ln(x)+8*x*ln(2))/x)+4-4*ln(x))/(x *ln(5)**8*ln(x)-4*x**2*ln(5)**8),x)
-4*log(x)/log(5)**8 + log((8*x*log(2)/3 - 2*log(2)*log(x)/3)/x)**2/log(5)* *8 + 4*log(-4*x + log(x))/log(5)**8
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.46 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=-\frac {2 \, {\left (i \, \pi - \log \left (3\right ) + \log \left (2\right ) + \log \left (\log \left (2\right )\right ) + 2\right )} \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, {\left (-i \, \pi + \log \left (3\right ) - \log \left (2\right ) + \log \left (x\right ) - \log \left (\log \left (2\right )\right ) - 2\right )} \log \left (-4 \, x + \log \left (x\right )\right ) - \log \left (-4 \, x + \log \left (x\right )\right )^{2}}{\log \left (5\right )^{8}} \]
integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log( x))/(x*log(5)^8*log(x)-4*x^2*log(5)^8),x, algorithm=\
-(2*(I*pi - log(3) + log(2) + log(log(2)) + 2)*log(x) - log(x)^2 + 2*(-I*p i + log(3) - log(2) + log(x) - log(log(2)) - 2)*log(-4*x + log(x)) - log(- 4*x + log(x))^2)/log(5)^8
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.14 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=\frac {\log \left (4 \, x \log \left (2\right ) - \log \left (2\right ) \log \left (x\right )\right )^{2}}{\log \left (5\right )^{8}} + \frac {2 \, {\left (\log \left (3\right ) - \log \left (2\right ) - 2\right )} \log \left (x\right )}{\log \left (5\right )^{8}} - \frac {2 \, \log \left (4 \, x \log \left (2\right ) - \log \left (2\right ) \log \left (x\right )\right ) \log \left (x\right )}{\log \left (5\right )^{8}} + \frac {\log \left (x\right )^{2}}{\log \left (5\right )^{8}} - \frac {2 \, {\left (\log \left (3\right ) - \log \left (2\right ) - 2\right )} \log \left (-4 \, x + \log \left (x\right )\right )}{\log \left (5\right )^{8}} \]
integrate(((-2*log(x)+2)*log(1/3*(-2*log(2)*log(x)+8*x*log(2))/x)+4-4*log( x))/(x*log(5)^8*log(x)-4*x^2*log(5)^8),x, algorithm=\
log(4*x*log(2) - log(2)*log(x))^2/log(5)^8 + 2*(log(3) - log(2) - 2)*log(x )/log(5)^8 - 2*log(4*x*log(2) - log(2)*log(x))*log(x)/log(5)^8 + log(x)^2/ log(5)^8 - 2*(log(3) - log(2) - 2)*log(-4*x + log(x))/log(5)^8
Time = 11.44 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {4-4 \log (x)+(2-2 \log (x)) \log \left (\frac {4 x \log (4)-\log (4) \log (x)}{3 x}\right )}{-4 x^2 \log ^8(5)+x \log ^8(5) \log (x)} \, dx=\frac {{\ln \left (\frac {\frac {8\,x\,\ln \left (2\right )}{3}-\frac {2\,\ln \left (2\right )\,\ln \left (x\right )}{3}}{x}\right )}^2+4\,\ln \left (\ln \left (x\right )-4\,x\right )-4\,\ln \left (x\right )}{{\ln \left (5\right )}^8} \]