Integrand size = 99, antiderivative size = 20 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\log \left (x \sqrt [x \log \left (\frac {16}{e^{25}}+x\right )]{\log (x)}\right ) \]
Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\log (x)+\frac {\log (\log (x))}{x \log \left (\frac {16}{e^{25}}+x\right )} \]
Integrate[((16/E^25 + x)*Log[16/E^25 + x] + ((16*x)/E^25 + x^2)*Log[x]*Log [16/E^25 + x]^2 + (-(x*Log[x]) + (-16/E^25 - x)*Log[x]*Log[16/E^25 + x])*L og[Log[x]])/(((16*x^2)/E^25 + x^3)*Log[x]*Log[16/E^25 + x]^2),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.
\(\displaystyle \int \frac {\left (x^2+\frac {16 x}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )+\left (x+\frac {16}{e^{25}}\right ) \log \left (x+\frac {16}{e^{25}}\right )+\left (\left (-x-\frac {16}{e^{25}}\right ) \log (x) \log \left (x+\frac {16}{e^{25}}\right )-x \log (x)\right ) \log (\log (x))}{\left (x^3+\frac {16 x^2}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2+\frac {16 x}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )+\left (x+\frac {16}{e^{25}}\right ) \log \left (x+\frac {16}{e^{25}}\right )+\left (\left (-x-\frac {16}{e^{25}}\right ) \log (x) \log \left (x+\frac {16}{e^{25}}\right )-x \log (x)\right ) \log (\log (x))}{x^2 \left (x+\frac {16}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x-\frac {e^{25} x \log (\log (x))}{\left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}+\frac {1-\log (x) \log (\log (x))}{\log (x) \log \left (x+\frac {16}{e^{25}}\right )}}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {x \log (x) \log \left (x+\frac {16}{e^{25}}\right )+1}{x^2 \log (x) \log \left (x+\frac {16}{e^{25}}\right )}-\frac {\left (e^{25} x+e^{25} x \log \left (x+\frac {16}{e^{25}}\right )+16 \log \left (x+\frac {16}{e^{25}}\right )\right ) \log (\log (x))}{x^2 \left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x^2 \log (x) \log \left (x+\frac {16}{e^{25}}\right )}dx-\int \frac {\log (\log (x))}{x^2 \log \left (x+\frac {16}{e^{25}}\right )}dx-\frac {1}{16} e^{25} \int \frac {\log (\log (x))}{x \log ^2\left (x+\frac {16}{e^{25}}\right )}dx+\frac {1}{16} e^{50} \int \frac {\log (\log (x))}{\left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}dx+\log (x)\) |
Int[((16/E^25 + x)*Log[16/E^25 + x] + ((16*x)/E^25 + x^2)*Log[x]*Log[16/E^ 25 + x]^2 + (-(x*Log[x]) + (-16/E^25 - x)*Log[x]*Log[16/E^25 + x])*Log[Log [x]])/(((16*x^2)/E^25 + x^3)*Log[x]*Log[16/E^25 + x]^2),x]
3.1.25.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 11.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (\ln \left (x \right )\right )}{x \ln \left (16 \,{\mathrm e}^{-25}+x \right )}+\ln \left (x \right )\) | \(20\) |
parallelrisch | \(\frac {\left (2 \,{\mathrm e}^{4 \ln \left (2\right )-25} \ln \left ({\mathrm e}^{4 \ln \left (2\right )-25}+x \right ) \ln \left (x \right ) x +2 \,{\mathrm e}^{4 \ln \left (2\right )-25} \ln \left (\ln \left (x \right )\right )\right ) {\mathrm e}^{25}}{32 x \ln \left ({\mathrm e}^{4 \ln \left (2\right )-25}+x \right )}\) | \(62\) |
int((((-exp(4*ln(2)-25)-x)*ln(x)*ln(exp(4*ln(2)-25)+x)-x*ln(x))*ln(ln(x))+ (x*exp(4*ln(2)-25)+x^2)*ln(x)*ln(exp(4*ln(2)-25)+x)^2+(exp(4*ln(2)-25)+x)* ln(exp(4*ln(2)-25)+x))/(x^2*exp(4*ln(2)-25)+x^3)/ln(x)/ln(exp(4*ln(2)-25)+ x)^2,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {x \log \left (x + e^{\left (4 \, \log \left (2\right ) - 25\right )}\right ) \log \left (x\right ) + \log \left (\log \left (x\right )\right )}{x \log \left (x + e^{\left (4 \, \log \left (2\right ) - 25\right )}\right )} \]
integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) *log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( x)/log(exp(4*log(2)-25)+x)^2,x, algorithm=\
Exception generated. \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((((-exp(4*ln(2)-25)-x)*ln(x)*ln(exp(4*ln(2)-25)+x)-x*ln(x))*ln(l n(x))+(x*exp(4*ln(2)-25)+x**2)*ln(x)*ln(exp(4*ln(2)-25)+x)**2+(exp(4*ln(2) -25)+x)*ln(exp(4*ln(2)-25)+x))/(x**2*exp(4*ln(2)-25)+x**3)/ln(x)/ln(exp(4* ln(2)-25)+x)**2,x)
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {\log \left (\log \left (x\right )\right )}{x \log \left (x e^{25} + 16\right ) - 25 \, x} + \log \left (x\right ) \]
integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) *log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( x)/log(exp(4*log(2)-25)+x)^2,x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {x \log \left (x e^{25} + 16\right ) \log \left (x\right ) - 25 \, x \log \left (x\right ) + \log \left (\log \left (x\right )\right )}{x \log \left (x e^{25} + 16\right ) - 25 \, x} \]
integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) *log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( x)/log(exp(4*log(2)-25)+x)^2,x, algorithm=\
Time = 13.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\ln \left (x\right )+\frac {\ln \left (\ln \left (x\right )\right )}{x\,\ln \left (x+16\,{\mathrm {e}}^{-25}\right )} \]