Integrand size = 103, antiderivative size = 37 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=\frac {4 \left (3 e^{-x}+2 x+\frac {1}{2} \left (5-\log (x)-\log \left (\log \left ((4-x)^2\right )\right )\right )\right )}{x} \]
Time = 5.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=-\frac {2 \left (-5-6 e^{-x}+\log (x)+\log \left (\log \left ((-4+x)^2\right )\right )\right )}{x} \]
Integrate[(-4*E^x*x + (48 + E^x*(48 - 12*x) + 36*x - 12*x^2 + E^x*(-8 + 2* x)*Log[x])*Log[16 - 8*x + x^2] + E^x*(-8 + 2*x)*Log[16 - 8*x + x^2]*Log[Lo g[16 - 8*x + x^2]])/(E^x*(-4*x^2 + x^3)*Log[16 - 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.
\(\displaystyle \int \frac {e^{-x} \left (\left (-12 x^2+36 x+e^x (48-12 x)+e^x (2 x-8) \log (x)+48\right ) \log \left (x^2-8 x+16\right )+e^x (2 x-8) \log \left (x^2-8 x+16\right ) \log \left (\log \left (x^2-8 x+16\right )\right )-4 e^x x\right )}{\left (x^3-4 x^2\right ) \log \left (x^2-8 x+16\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (\left (-12 x^2+36 x+e^x (48-12 x)+e^x (2 x-8) \log (x)+48\right ) \log \left (x^2-8 x+16\right )+e^x (2 x-8) \log \left (x^2-8 x+16\right ) \log \left (\log \left (x^2-8 x+16\right )\right )-4 e^x x\right )}{(x-4) x^2 \log \left (x^2-8 x+16\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (-6 e^{-x} x-6 e^{-x}-\frac {2 x}{(x-4) \log \left ((x-4)^2\right )}+\log (x)+\log \left (\log \left ((x-4)^2\right )\right )-6\right )}{x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {6 e^{-x} x-\frac {2 x}{(4-x) \log \left ((x-4)^2\right )}+6 e^{-x}-\log (x)-\log \left (\log \left ((x-4)^2\right )\right )+6}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {6 e^{-x} x-\frac {2 x}{(4-x) \log \left ((x-4)^2\right )}+6 e^{-x}-\log (x)-\log \left (\log \left ((x-4)^2\right )\right )+6}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -2 \int \left (\frac {6 e^{-x} (x+1)}{x^2}+\frac {6 \log \left ((x-4)^2\right ) x-\log \left ((x-4)^2\right ) \log (x) x-\log \left ((x-4)^2\right ) \log \left (\log \left ((x-4)^2\right )\right ) x+2 x-24 \log \left ((x-4)^2\right )+4 \log \left ((x-4)^2\right ) \log (x)+4 \log \left ((x-4)^2\right ) \log \left (\log \left ((x-4)^2\right )\right )}{(x-4) x^2 \log \left ((x-4)^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (-\int \frac {\log \left (\log \left ((x-4)^2\right )\right )}{x^2}dx-\frac {1}{2} \int \frac {1}{x \log \left ((x-4)^2\right )}dx-\frac {6 e^{-x}}{x}-\frac {5}{x}+\frac {\log (x)}{x}+\frac {1}{4} \log \left (\log \left ((x-4)^2\right )\right )\right )\) |
Int[(-4*E^x*x + (48 + E^x*(48 - 12*x) + 36*x - 12*x^2 + E^x*(-8 + 2*x)*Log [x])*Log[16 - 8*x + x^2] + E^x*(-8 + 2*x)*Log[16 - 8*x + x^2]*Log[Log[16 - 8*x + x^2]])/(E^x*(-4*x^2 + x^3)*Log[16 - 8*x + x^2]),x]
3.10.90.3.1 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_)*((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 = 4.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(-\frac {\left (-192-20 \,{\mathrm e}^{x} x +32 \,{\mathrm e}^{x} \ln \left (x \right )+32 \,{\mathrm e}^{x} \ln \left (\ln \left (x^{2}-8 x +16\right )\right )-160 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{16 x}\) | \(41\) |
risch | \(-\frac {2 \left (\ln \left (x \right )-5+\ln \left (2 \ln \left (x -4\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (x -4\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left (x -4\right )^{2}\right )+\operatorname {csgn}\left (i \left (x -4\right )\right )\right )}^{2}}{2}\right )-6 \,{\mathrm e}^{-x}\right )}{x}\) | \(58\) |
int(((2*x-8)*exp(x)*ln(x^2-8*x+16)*ln(ln(x^2-8*x+16))+((2*x-8)*exp(x)*ln(x )+(-12*x+48)*exp(x)-12*x^2+36*x+48)*ln(x^2-8*x+16)-4*exp(x)*x)/(x^3-4*x^2) /exp(x)/ln(x^2-8*x+16),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=-\frac {2 \, {\left (e^{x} \log \left (x\right ) + e^{x} \log \left (\log \left (x^{2} - 8 \, x + 16\right )\right ) - 5 \, e^{x} - 6\right )} e^{\left (-x\right )}}{x} \]
integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*ex p(x)*log(x)+(-12*x+48)*exp(x)-12*x^2+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/ (x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm=\
Time = 0.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=- \frac {2 \log {\left (x \right )}}{x} - \frac {2 \log {\left (\log {\left (x^{2} - 8 x + 16 \right )} \right )}}{x} + \frac {10}{x} + \frac {12 e^{- x}}{x} \]
integrate(((2*x-8)*exp(x)*ln(x**2-8*x+16)*ln(ln(x**2-8*x+16))+((2*x-8)*exp (x)*ln(x)+(-12*x+48)*exp(x)-12*x**2+36*x+48)*ln(x**2-8*x+16)-4*exp(x)*x)/( x**3-4*x**2)/exp(x)/ln(x**2-8*x+16),x)
Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=\frac {2 \, {\left (6 \, e^{\left (-x\right )} - \log \left (2\right ) - \log \left (x\right ) - \log \left (\log \left (x - 4\right )\right ) + 5\right )}}{x} \]
integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*ex p(x)*log(x)+(-12*x+48)*exp(x)-12*x^2+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/ (x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm=\
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=\frac {2 \, {\left (6 \, e^{\left (-x\right )} - \log \left (x\right ) - \log \left (\log \left (x^{2} - 8 \, x + 16\right )\right ) + 5\right )}}{x} \]
integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*ex p(x)*log(x)+(-12*x+48)*exp(x)-12*x^2+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/ (x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm=\
Time = 9.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx=\frac {12\,{\mathrm {e}}^{-x}}{x}-\frac {2\,\ln \left (x\right )}{x}+\frac {10}{x}+\frac {\ln \left (\ln \left (x^2-8\,x+16\right )\right )\,\left (8\,x-2\,x^2\right )}{x^2\,\left (x-4\right )} \]
int(-(exp(-x)*(log(x^2 - 8*x + 16)*(36*x - exp(x)*(12*x - 48) - 12*x^2 + e xp(x)*log(x)*(2*x - 8) + 48) - 4*x*exp(x) + exp(x)*log(x^2 - 8*x + 16)*log (log(x^2 - 8*x + 16))*(2*x - 8)))/(log(x^2 - 8*x + 16)*(4*x^2 - x^3)),x)