\(\int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+(512-256 x+32 x^2) \log (x))}{16 x^3-8 x^4+x^5} \, dx\) [9385]

Optimal result

Integrand size = 73, antiderivative size = 27 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=-5+\frac {e^{4+2 e^{\frac {1}{4-x}}+16 \log ^2(x)}}{x^2} \]

[Out]

exp(4*ln(x)^2)^4/x^2*exp(exp(1/(-x+4))+2)^2-5

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(27)=54\).

Time = 0.67 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {1608, 27, 2326} \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{2 e^{\frac {1}{4-x}}+16 \log ^2(x)+4} \left (16 \left (x^2-8 x+16\right ) \log (x)+e^{\frac {1}{4-x}} x\right )}{(4-x)^2 x^3 \left (\frac {e^{\frac {1}{4-x}}}{(4-x)^2}+\frac {16 \log (x)}{x}\right )} \]

[In]

Int[(E^(4 + 2/E^(-4 + x)^(-1) + 16*Log[x]^2)*(-32 + 16*x + (2*x)/E^(-4 + x)^(-1) - 2*x^2 + (512 - 256*x + 32*x
^2)*Log[x]))/(16*x^3 - 8*x^4 + x^5),x]

[Out]

(E^(4 + 2*E^(4 - x)^(-1) + 16*Log[x]^2)*(E^(4 - x)^(-1)*x + 16*(16 - 8*x + x^2)*Log[x]))/((4 - x)^2*x^3*(E^(4
- x)^(-1)/(4 - x)^2 + (16*Log[x])/x))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{x^3 \left (16-8 x+x^2\right )} \, dx \\ & = \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{(-4+x)^2 x^3} \, dx \\ & = \frac {e^{4+2 e^{\frac {1}{4-x}}+16 \log ^2(x)} \left (e^{\frac {1}{4-x}} x+16 \left (16-8 x+x^2\right ) \log (x)\right )}{(4-x)^2 x^3 \left (\frac {e^{\frac {1}{4-x}}}{(4-x)^2}+\frac {16 \log (x)}{x}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{2 \left (2+e^{\frac {1}{4-x}}+8 \log ^2(x)\right )}}{x^2} \]

[In]

Integrate[(E^(4 + 2/E^(-4 + x)^(-1) + 16*Log[x]^2)*(-32 + 16*x + (2*x)/E^(-4 + x)^(-1) - 2*x^2 + (512 - 256*x
+ 32*x^2)*Log[x]))/(16*x^3 - 8*x^4 + x^5),x]

[Out]

E^(2*(2 + E^(4 - x)^(-1) + 8*Log[x]^2))/x^2

Maple [A] (verified)

Time = 190.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{-\frac {1}{x -4}}+4+16 \ln \left (x \right )^{2}}}{x^{2}}\)	\(24\)
parallelrisch	\(\frac {{\mathrm e}^{16 \ln \left (x \right )^{2}} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {1}{x -4}}+4}}{x^{2}}\)	\(27\)

[In]

int(((32*x^2-256*x+512)*ln(x)+2*x*exp(-1/(x-4))-2*x^2+16*x-32)*exp(exp(-1/(x-4))+2)^2*exp(4*ln(x)^2)^4/(x^5-8*
x^4+16*x^3),x,method=_RETURNVERBOSE)

[Out]

1/x^2*exp(2*exp(-1/(x-4))+4+16*ln(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{2}} \]

[In]

integrate(((32*x^2-256*x+512)*log(x)+2*x*exp(-1/(x-4))-2*x^2+16*x-32)*exp(exp(-1/(x-4))+2)^2*exp(4*log(x)^2)^4
/(x^5-8*x^4+16*x^3),x, algorithm="fricas")

[Out]

e^(16*log(x)^2 + 2*e^(-1/(x - 4)) + 4)/x^2

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{4 + 2 e^{- \frac {1}{x - 4}}} e^{16 \log {\left (x \right )}^{2}}}{x^{2}} \]

[In]

integrate(((32*x**2-256*x+512)*ln(x)+2*x*exp(-1/(x-4))-2*x**2+16*x-32)*exp(exp(-1/(x-4))+2)**2*exp(4*ln(x)**2)
**4/(x**5-8*x**4+16*x**3),x)

[Out]

exp(4 + 2*exp(-1/(x - 4)))*exp(16*log(x)**2)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{2}} \]

[In]

integrate(((32*x^2-256*x+512)*log(x)+2*x*exp(-1/(x-4))-2*x^2+16*x-32)*exp(exp(-1/(x-4))+2)^2*exp(4*log(x)^2)^4
/(x^5-8*x^4+16*x^3),x, algorithm="maxima")

[Out]

e^(16*log(x)^2 + 2*e^(-1/(x - 4)) + 4)/x^2

Giac [F]

\[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\int { -\frac {2 \, {\left (x^{2} - x e^{\left (-\frac {1}{x - 4}\right )} - 16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (x\right ) - 8 \, x + 16\right )} e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{5} - 8 \, x^{4} + 16 \, x^{3}} \,d x } \]

[In]

integrate(((32*x^2-256*x+512)*log(x)+2*x*exp(-1/(x-4))-2*x^2+16*x-32)*exp(exp(-1/(x-4))+2)^2*exp(4*log(x)^2)^4
/(x^5-8*x^4+16*x^3),x, algorithm="giac")

[Out]

integrate(-2*(x^2 - x*e^(-1/(x - 4)) - 16*(x^2 - 8*x + 16)*log(x) - 8*x + 16)*e^(16*log(x)^2 + 2*e^(-1/(x - 4)
) + 4)/(x^5 - 8*x^4 + 16*x^3), x)

Mupad [B] (verification not implemented)

Time = 14.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {{\mathrm {e}}^{16\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-\frac {1}{x-4}}}}{x^2} \]

[In]

int((exp(16*log(x)^2)*exp(2*exp(-1/(x - 4)) + 4)*(16*x + log(x)*(32*x^2 - 256*x + 512) + 2*x*exp(-1/(x - 4)) -
2*x^2 - 32))/(16*x^3 - 8*x^4 + x^5),x)

[Out]

(exp(16*log(x)^2)*exp(4)*exp(2*exp(-1/(x - 4))))/x^2