\(\int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} (4 x+16 x^3-64 \log (x)+32 \log ^2(x)) \, dx\) [5502]

Optimal result

Integrand size = 41, antiderivative size = 27 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 e^{3+4 x^2-\frac {4 \left (x+4 \log ^2(x)\right )}{x}} x^2 \]

[Out]

2*exp(3-4/x*(4*ln(x)^2+x)+4*x^2)*x^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(27)=54\).

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.78, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2326} \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=-\frac {16 e^{-\frac {-4 x^3+x+16 \log ^2(x)}{x}} \left (x^3+2 \log ^2(x)-4 \log (x)\right )}{\frac {-12 x^2+\frac {32 \log (x)}{x}+1}{x}-\frac {-4 x^3+x+16 \log ^2(x)}{x^2}} \]

[In]

Int[E^((-x + 4*x^3 - 16*Log[x]^2)/x)*(4*x + 16*x^3 - 64*Log[x] + 32*Log[x]^2),x]

[Out]

(-16*(x^3 - 4*Log[x] + 2*Log[x]^2))/(E^((x - 4*x^3 + 16*Log[x]^2)/x)*((1 - 12*x^2 + (32*Log[x])/x)/x - (x - 4*
x^3 + 16*Log[x]^2)/x^2))

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}& = -\frac {16 e^{-\frac {x-4 x^3+16 \log ^2(x)}{x}} \left (x^3-4 \log (x)+2 \log ^2(x)\right )}{\frac {1-12 x^2+\frac {32 \log (x)}{x}}{x}-\frac {x-4 x^3+16 \log ^2(x)}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 e^{-1+4 x^2-\frac {16 \log ^2(x)}{x}} x^2 \]

[In]

Integrate[E^((-x + 4*x^3 - 16*Log[x]^2)/x)*(4*x + 16*x^3 - 64*Log[x] + 32*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method	result	size
risch	\(2 x^{2} {\mathrm e}^{-\frac {-4 x^{3}+16 \ln \left (x \right )^{2}+x}{x}}\)	\(25\)
parallelrisch	\(2 x^{2} {\mathrm e}^{-\frac {-4 x^{3}+16 \ln \left (x \right )^{2}+x}{x}}\)	\(25\)

[In]

int((32*ln(x)^2-64*ln(x)+16*x^3+4*x)*exp((-16*ln(x)^2+4*x^3-x)/x),x,method=_RETURNVERBOSE)

[Out]

2*x^2*exp(-(-4*x^3+16*ln(x)^2+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 \, x^{2} e^{\left (\frac {4 \, x^{3} - 16 \, \log \left (x\right )^{2} - x}{x}\right )} \]

[In]

integrate((32*log(x)^2-64*log(x)+16*x^3+4*x)*exp((-16*log(x)^2+4*x^3-x)/x),x, algorithm="fricas")

[Out]

2*x^2*e^((4*x^3 - 16*log(x)^2 - x)/x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 x^{2} e^{\frac {4 x^{3} - x - 16 \log {\left (x \right )}^{2}}{x}} \]

[In]

integrate((32*ln(x)**2-64*ln(x)+16*x**3+4*x)*exp((-16*ln(x)**2+4*x**3-x)/x),x)

[Out]

2*x**2*exp((4*x**3 - x - 16*log(x)**2)/x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 \, x^{2} e^{\left (4 \, x^{2} - \frac {16 \, \log \left (x\right )^{2}}{x} - 1\right )} \]

[In]

integrate((32*log(x)^2-64*log(x)+16*x^3+4*x)*exp((-16*log(x)^2+4*x^3-x)/x),x, algorithm="maxima")

[Out]

2*x^2*e^(4*x^2 - 16*log(x)^2/x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2 \, x^{2} e^{\left (\frac {4 \, x^{3} - 16 \, \log \left (x\right )^{2} - x}{x}\right )} \]

[In]

integrate((32*log(x)^2-64*log(x)+16*x^3+4*x)*exp((-16*log(x)^2+4*x^3-x)/x),x, algorithm="giac")

[Out]

2*x^2*e^((4*x^3 - 16*log(x)^2 - x)/x)

Mupad [B] (verification not implemented)

Time = 10.83 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int e^{\frac {-x+4 x^3-16 \log ^2(x)}{x}} \left (4 x+16 x^3-64 \log (x)+32 \log ^2(x)\right ) \, dx=2\,x^2\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{4\,x^2}\,{\mathrm {e}}^{-\frac {16\,{\ln \left (x\right )}^2}{x}} \]

[In]

int(exp(-(x + 16*log(x)^2 - 4*x^3)/x)*(4*x - 64*log(x) + 32*log(x)^2 + 16*x^3),x)

[Out]

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