Integrand size = 184, antiderivative size = 27 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=(4-x) \left (-e^x+\log \left (\frac {\log (\log (x (3+x) (5+x)))}{x}\right )\right ) \]
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=e^x (-4+x)-4 \log (x)+4 \log \left (\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )\right )-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right ) \]
Integrate[(60 + 49*x - 4*x^2 - 3*x^3 + (-60 - 17*x + 4*x^2 + x^3 + E^x*(-4 5*x - 9*x^2 + 5*x^3 + x^4))*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]] + (-15*x - 8*x^2 - x^3)*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^ 2 + x^3]]*Log[Log[Log[15*x + 8*x^2 + x^3]]/x])/((15*x + 8*x^2 + x^3)*Log[1 5*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]]),x]
E^x*(-4 + x) - 4*Log[x] + 4*Log[Log[Log[x*(15 + 8*x + x^2)]]] - x*Log[Log[ Log[x*(15 + 8*x + 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.
\(\displaystyle \int \frac {-3 x^3-4 x^2+\left (-x^3-8 x^2-15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right ) \log \left (\frac {\log \left (\log \left (x^3+8 x^2+15 x\right )\right )}{x}\right )+\left (x^3+4 x^2+e^x \left (x^4+5 x^3-9 x^2-45 x\right )-17 x-60\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )+49 x+60}{\left (x^3+8 x^2+15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-3 x^3-4 x^2+\left (-x^3-8 x^2-15 x\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right ) \log \left (\frac {\log \left (\log \left (x^3+8 x^2+15 x\right )\right )}{x}\right )+\left (x^3+4 x^2+e^x \left (x^4+5 x^3-9 x^2-45 x\right )-17 x-60\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )+49 x+60}{x \left (x^2+8 x+15\right ) \log \left (x^3+8 x^2+15 x\right ) \log \left (\log \left (x^3+8 x^2+15 x\right )\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {x^2}{x^2+8 x+15}+\frac {4 x}{x^2+8 x+15}-\frac {17}{x^2+8 x+15}-\frac {60}{\left (x^2+8 x+15\right ) x}-\frac {3 x^2}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}-\frac {4 x}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}-\log \left (\frac {\log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}{x}\right )+\frac {49}{(x+3) (x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}+\frac {60}{(x+3) (x+5) x \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}+e^x (x-3)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {1}{x \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx+4 \int \frac {1}{(x+3) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx+4 \int \frac {1}{(x+5) \log \left (x \left (x^2+8 x+15\right )\right ) \log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}dx-x \log \left (\frac {\log \left (\log \left (x \left (x^2+8 x+15\right )\right )\right )}{x}\right )-e^x (3-x)-e^x-4 \log (x)\) |
Int[(60 + 49*x - 4*x^2 - 3*x^3 + (-60 - 17*x + 4*x^2 + x^3 + E^x*(-45*x - 9*x^2 + 5*x^3 + x^4))*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]] + (-15*x - 8*x^2 - x^3)*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^ 3]]*Log[Log[Log[15*x + 8*x^2 + x^3]]/x])/((15*x + 8*x^2 + x^3)*Log[15*x + 8*x^2 + x^3]*Log[Log[15*x + 8*x^2 + x^3]]),x]
3.16.7.3.1 Defintions of rubi rules used
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_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.35 (sec) , antiderivative size = 733, normalized size of antiderivative = 27.15
\[\text {Expression too large to display}\]
int(((-x^3-8*x^2-15*x)*ln(x^3+8*x^2+15*x)*ln(ln(x^3+8*x^2+15*x))*ln(ln(ln( x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-17*x-60)*ln(x ^3+8*x^2+15*x)*ln(ln(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/(x^3+8*x^2+15*x )/ln(x^3+8*x^2+15*x)/ln(ln(x^3+8*x^2+15*x)),x)
-x*ln(ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*( x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))+x *ln(x)+1/2*I*Pi*x*csgn(I/x)*csgn(I*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I *x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+1 5))+csgn(I*(x^2+8*x+15)))))*csgn(I/x*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn (I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x +15))+csgn(I*(x^2+8*x+15)))))-1/2*I*Pi*x*csgn(I/x)*csgn(I/x*ln(ln(x)+ln(x^ 2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I* x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))^2-1/2*I*Pi*x*csgn(I*l n(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x +15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))*csgn(I/x *ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8 *x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x+15)))))^2+1/2* I*Pi*x*csgn(I/x*ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi*csgn(I*x*(x^2+8*x+15))*(- csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*(x^2+8*x +15)))))^3-4*ln(x)+exp(x)*x-4*exp(x)+4*ln(ln(ln(x)+ln(x^2+8*x+15)-1/2*I*Pi *csgn(I*x*(x^2+8*x+15))*(-csgn(I*x*(x^2+8*x+15))+csgn(I*x))*(-csgn(I*x*(x^ 2+8*x+15))+csgn(I*(x^2+8*x+15)))))
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} - {\left (x - 4\right )} \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \]
integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ (x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm =\
Timed out. \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\text {Timed out} \]
integrate(((-x**3-8*x**2-15*x)*ln(x**3+8*x**2+15*x)*ln(ln(x**3+8*x**2+15*x ))*ln(ln(ln(x**3+8*x**2+15*x))/x)+((x**4+5*x**3-9*x**2-45*x)*exp(x)+x**3+4 *x**2-17*x-60)*ln(x**3+8*x**2+15*x)*ln(ln(x**3+8*x**2+15*x))-3*x**3-4*x**2 +49*x+60)/(x**3+8*x**2+15*x)/ln(x**3+8*x**2+15*x)/ln(ln(x**3+8*x**2+15*x)) ,x)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} + {\left (x - 4\right )} \log \left (x\right ) - {\left (x - 4\right )} \log \left (\log \left (\log \left (x + 5\right ) + \log \left (x + 3\right ) + \log \left (x\right )\right )\right ) \]
integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ (x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm =\
\[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\int { -\frac {{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 3 \, x^{3} - {\left (x^{3} + 4 \, x^{2} + {\left (x^{4} + 5 \, x^{3} - 9 \, x^{2} - 45 \, x\right )} e^{x} - 17 \, x - 60\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 4 \, x^{2} - 49 \, x - 60}{{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )} \,d x } \]
integrate(((-x^3-8*x^2-15*x)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))* log(log(log(x^3+8*x^2+15*x))/x)+((x^4+5*x^3-9*x^2-45*x)*exp(x)+x^3+4*x^2-1 7*x-60)*log(x^3+8*x^2+15*x)*log(log(x^3+8*x^2+15*x))-3*x^3-4*x^2+49*x+60)/ (x^3+8*x^2+15*x)/log(x^3+8*x^2+15*x)/log(log(x^3+8*x^2+15*x)),x, algorithm =\
integrate(-((x^3 + 8*x^2 + 15*x)*log(x^3 + 8*x^2 + 15*x)*log(log(log(x^3 + 8*x^2 + 15*x))/x)*log(log(x^3 + 8*x^2 + 15*x)) + 3*x^3 - (x^3 + 4*x^2 + ( x^4 + 5*x^3 - 9*x^2 - 45*x)*e^x - 17*x - 60)*log(x^3 + 8*x^2 + 15*x)*log(l og(x^3 + 8*x^2 + 15*x)) + 4*x^2 - 49*x - 60)/((x^3 + 8*x^2 + 15*x)*log(x^3 + 8*x^2 + 15*x)*log(log(x^3 + 8*x^2 + 15*x))), x)
Time = 9.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.81 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=4\,\ln \left (\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )\right )-4\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (x-4\right )-\frac {\ln \left (\frac {\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )}{x}\right )\,\left (x^4+8\,x^3+15\,x^2\right )}{x\,\left (x^2+8\,x+15\right )} \]
int(-(4*x^2 - 49*x + 3*x^3 + log(log(15*x + 8*x^2 + x^3))*log(15*x + 8*x^2 + x^3)*(17*x + exp(x)*(45*x + 9*x^2 - 5*x^3 - x^4) - 4*x^2 - x^3 + 60) + log(log(15*x + 8*x^2 + x^3))*log(15*x + 8*x^2 + x^3)*log(log(log(15*x + 8* x^2 + x^3))/x)*(15*x + 8*x^2 + x^3) - 60)/(log(log(15*x + 8*x^2 + x^3))*lo g(15*x + 8*x^2 + x^3)*(15*x + 8*x^2 + x^3)),x)