Integrand size = 98, antiderivative size = 25 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=\log \left ((-4+\log (x)) \left (\log (5)+\frac {\log ^2(x)}{x}\right ) \log \left (\left (4+e^5\right ) x\right )\right ) \]
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=-\log (x)+\log (4-\log (x))+\log \left (x \log (5)+\log ^2(x)\right )+\log \left (\log \left (\left (4+e^5\right ) x\right )\right ) \]
Integrate[(-4*x*Log[5] + x*Log[5]*Log[x] - 4*Log[x]^2 + Log[x]^3 + (x*Log[ 5] - 8*Log[x] + 7*Log[x]^2 - Log[x]^3)*Log[4*x + E^5*x])/((-4*x^2*Log[5] + x^2*Log[5]*Log[x] - 4*x*Log[x]^2 + x*Log[x]^3)*Log[4*x + E^5*x]),x]
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.
\(\displaystyle \int \frac {\log ^3(x)-4 \log ^2(x)+\left (-\log ^3(x)+7 \log ^2(x)-8 \log (x)+x \log (5)\right ) \log \left (e^5 x+4 x\right )+x \log (5) \log (x)-4 x \log (5)}{\left (x^2 \log (5) \log (x)-4 x^2 \log (5)+x \log ^3(x)-4 x \log ^2(x)\right ) \log \left (e^5 x+4 x\right )} \, dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {\log ^3(x)-4 \log ^2(x)+\left (-\log ^3(x)+7 \log ^2(x)-8 \log (x)+x \log (5)\right ) \log \left (e^5 x+4 x\right )+x \log (5) \log (x)-4 x \log (5)}{\left (x^2 \log (5) \log (x)-4 x^2 \log (5)+x \log ^3(x)-4 x \log ^2(x)\right ) \log \left (\left (4+e^5\right ) x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\log ^3(x)+4 \log ^2(x)-\left (-\log ^3(x)+7 \log ^2(x)-8 \log (x)+x \log (5)\right ) \log \left (e^5 x+4 x\right )-x \log (5) \log (x)+4 x \log (5)}{x (4-\log (x)) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 \log ^2(x)}{x (\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}+\frac {\log (5) \log (x)}{(\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}-\frac {4 \log (5)}{(\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}+\frac {\log ^3(x)}{x (\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}+\frac {-\log ^3(x)+7 \log ^2(x)-8 \log (x)+x \log (5)}{x (\log (x)-4) \left (\log ^2(x)+x \log (5)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \log (5) \int \frac {1}{(\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}dx+\log (5) \int \frac {\log (x)}{(\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}dx-4 \int \frac {\log ^2(x)}{x (\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}dx+\int \frac {\log ^3(x)}{x (\log (x)-4) \left (\log ^2(x)+x \log (5)\right ) \log \left (\left (4+e^5\right ) x\right )}dx+\log \left (\log ^2(x)+x \log (5)\right )-\log (x)+\log (4-\log (x))\) |
Int[(-4*x*Log[5] + x*Log[5]*Log[x] - 4*Log[x]^2 + Log[x]^3 + (x*Log[5] - 8 *Log[x] + 7*Log[x]^2 - Log[x]^3)*Log[4*x + E^5*x])/((-4*x^2*Log[5] + x^2*L og[5]*Log[x] - 4*x*Log[x]^2 + x*Log[x]^3)*Log[4*x + E^5*x]),x]
3.11.35.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*( a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p}, x] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Time = 0.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24
method | result | size |
default | \(\ln \left (\ln \left (x \,{\mathrm e}^{5}+4 x \right )\right )-\ln \left (x \right )+\ln \left (\ln \left (x \right )-4\right )+\ln \left (x \ln \left (5\right )+\ln \left (x \right )^{2}\right )\) | \(31\) |
risch | \(-\ln \left (x \right )+\ln \left (\ln \left (x \right )^{2} \ln \left (5\right ) x +\ln \left (x \right )^{4}-4 x \ln \left (5\right ) \ln \left (x \right )-4 \ln \left (x \right )^{3}\right )\) | \(33\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (\ln \left (x \right )-4\right )+\ln \left (\ln \left (\left (4+{\mathrm e}^{5}\right ) x \right )\right )+\ln \left (\frac {x \ln \left (5\right )+\ln \left (x \right )^{2}}{\ln \left (5\right )}\right )\) | \(34\) |
int(((-ln(x)^3+7*ln(x)^2-8*ln(x)+x*ln(5))*ln(x*exp(5)+4*x)+ln(x)^3-4*ln(x) ^2+x*ln(5)*ln(x)-4*x*ln(5))/(x*ln(x)^3-4*x*ln(x)^2+x^2*ln(5)*ln(x)-4*x^2*l n(5))/ln(x*exp(5)+4*x),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=\log \left (x \log \left (5\right ) + \log \left (x\right )^{2}\right ) - \log \left (x\right ) + \log \left (\log \left (x\right ) + \log \left (e^{5} + 4\right )\right ) + \log \left (\log \left (x\right ) - 4\right ) \]
integrate(((-log(x)^3+7*log(x)^2-8*log(x)+x*log(5))*log(x*exp(5)+4*x)+log( x)^3-4*log(x)^2+x*log(5)*log(x)-4*x*log(5))/(x*log(x)^3-4*x*log(x)^2+x^2*l og(5)*log(x)-4*x^2*log(5))/log(x*exp(5)+4*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (24) = 48\).
Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.04 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=- \log {\left (x \right )} + \log {\left (- 4 x \log {\left (5 \right )} \log {\left (4 + e^{5} \right )} + \left (- 4 x \log {\left (5 \right )} + x \log {\left (5 \right )} \log {\left (4 + e^{5} \right )}\right ) \log {\left (x \right )} + \left (x \log {\left (5 \right )} - 4 \log {\left (4 + e^{5} \right )}\right ) \log {\left (x \right )}^{2} + \log {\left (x \right )}^{4} + \left (-4 + \log {\left (4 + e^{5} \right )}\right ) \log {\left (x \right )}^{3} \right )} \]
integrate(((-ln(x)**3+7*ln(x)**2-8*ln(x)+x*ln(5))*ln(x*exp(5)+4*x)+ln(x)** 3-4*ln(x)**2+x*ln(5)*ln(x)-4*x*ln(5))/(x*ln(x)**3-4*x*ln(x)**2+x**2*ln(5)* ln(x)-4*x**2*ln(5))/ln(x*exp(5)+4*x),x)
-log(x) + log(-4*x*log(5)*log(4 + exp(5)) + (-4*x*log(5) + x*log(5)*log(4 + exp(5)))*log(x) + (x*log(5) - 4*log(4 + exp(5)))*log(x)**2 + log(x)**4 + (-4 + log(4 + exp(5)))*log(x)**3)
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=\log \left (x \log \left (5\right ) + \log \left (x\right )^{2}\right ) - \log \left (x\right ) + \log \left (\log \left (x\right ) + \log \left (e^{5} + 4\right )\right ) + \log \left (\log \left (x\right ) - 4\right ) \]
integrate(((-log(x)^3+7*log(x)^2-8*log(x)+x*log(5))*log(x*exp(5)+4*x)+log( x)^3-4*log(x)^2+x*log(5)*log(x)-4*x*log(5))/(x*log(x)^3-4*x*log(x)^2+x^2*l og(5)*log(x)-4*x^2*log(5))/log(x*exp(5)+4*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.96 \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=\log \left (x \log \left (5\right ) \log \left (x\right )^{2} + \log \left (x\right )^{4} + x \log \left (5\right ) \log \left (x\right ) \log \left (e^{5} + 4\right ) + \log \left (x\right )^{3} \log \left (e^{5} + 4\right ) - 4 \, x \log \left (5\right ) \log \left (x\right ) - 4 \, \log \left (x\right )^{3} - 4 \, x \log \left (5\right ) \log \left (e^{5} + 4\right ) - 4 \, \log \left (x\right )^{2} \log \left (e^{5} + 4\right )\right ) - \log \left (x\right ) \]
integrate(((-log(x)^3+7*log(x)^2-8*log(x)+x*log(5))*log(x*exp(5)+4*x)+log( x)^3-4*log(x)^2+x*log(5)*log(x)-4*x*log(5))/(x*log(x)^3-4*x*log(x)^2+x^2*l og(5)*log(x)-4*x^2*log(5))/log(x*exp(5)+4*x),x, algorithm=\
log(x*log(5)*log(x)^2 + log(x)^4 + x*log(5)*log(x)*log(e^5 + 4) + log(x)^3 *log(e^5 + 4) - 4*x*log(5)*log(x) - 4*log(x)^3 - 4*x*log(5)*log(e^5 + 4) - 4*log(x)^2*log(e^5 + 4)) - log(x)
Timed out. \[ \int \frac {-4 x \log (5)+x \log (5) \log (x)-4 \log ^2(x)+\log ^3(x)+\left (x \log (5)-8 \log (x)+7 \log ^2(x)-\log ^3(x)\right ) \log \left (4 x+e^5 x\right )}{\left (-4 x^2 \log (5)+x^2 \log (5) \log (x)-4 x \log ^2(x)+x \log ^3(x)\right ) \log \left (4 x+e^5 x\right )} \, dx=\int \frac {\ln \left (4\,x+x\,{\mathrm {e}}^5\right )\,\left ({\ln \left (x\right )}^3-7\,{\ln \left (x\right )}^2+8\,\ln \left (x\right )-x\,\ln \left (5\right )\right )+4\,x\,\ln \left (5\right )+4\,{\ln \left (x\right )}^2-{\ln \left (x\right )}^3-x\,\ln \left (5\right )\,\ln \left (x\right )}{\ln \left (4\,x+x\,{\mathrm {e}}^5\right )\,\left (-\ln \left (5\right )\,x^2\,\ln \left (x\right )+4\,\ln \left (5\right )\,x^2-x\,{\ln \left (x\right )}^3+4\,x\,{\ln \left (x\right )}^2\right )} \,d x \]
int((log(4*x + x*exp(5))*(8*log(x) - x*log(5) - 7*log(x)^2 + log(x)^3) + 4 *x*log(5) + 4*log(x)^2 - log(x)^3 - x*log(5)*log(x))/(log(4*x + x*exp(5))* (4*x*log(x)^2 - x*log(x)^3 + 4*x^2*log(5) - x^2*log(5)*log(x))),x)