\(\int \frac {e^{-16 x+4 x \log (x)} (4-96 x-4 x^3-12 x^4+(-4+32 x+4 x^4) \log (x))}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx\) [440]

Optimal result

Integrand size = 65, antiderivative size = 21 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {e^{x (-16+4 \log (x))}}{-1+8 x+x^4} \]

[Out]

exp(x*(4*ln(x)-16))/(x^4+8*x-1)

Rubi [F]

\[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx \]

[In]

Int[(E^(-16*x + 4*x*Log[x])*(4 - 96*x - 4*x^3 - 12*x^4 + (-4 + 32*x + 4*x^4)*Log[x]))/(1 - 16*x + 64*x^2 - 2*x
^4 + 16*x^5 + x^8),x]

[Out]

-8*Defer[Int][E^(4*x*(-4 + Log[x]))/(-1 + 8*x + x^4)^2, x] - 4*Defer[Int][(E^(4*x*(-4 + Log[x]))*x^3)/(-1 + 8*
x + x^4)^2, x] - 12*Defer[Int][E^(4*x*(-4 + Log[x]))/(-1 + 8*x + x^4), x] + 4*Defer[Int][(E^(4*x*(-4 + Log[x])
)*Log[x])/(-1 + 8*x + x^4), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4 x (-4+\log (x))} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{\left (1-8 x-x^4\right )^2} \, dx \\ & = \int \left (\frac {4 e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2}-\frac {96 e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2}-\frac {4 e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2}-\frac {12 e^{4 x (-4+\log (x))} x^4}{\left (-1+8 x+x^4\right )^2}+\frac {4 e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4}\right ) \, dx \\ & = 4 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-4 \int \frac {e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2} \, dx+4 \int \frac {e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4} \, dx-12 \int \frac {e^{4 x (-4+\log (x))} x^4}{\left (-1+8 x+x^4\right )^2} \, dx-96 \int \frac {e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2} \, dx \\ & = 4 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-4 \int \frac {e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2} \, dx+4 \int \frac {e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4} \, dx-12 \int \left (\frac {e^{4 x (-4+\log (x))} (1-8 x)}{\left (-1+8 x+x^4\right )^2}+\frac {e^{4 x (-4+\log (x))}}{-1+8 x+x^4}\right ) \, dx-96 \int \frac {e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2} \, dx \\ & = 4 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-4 \int \frac {e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2} \, dx+4 \int \frac {e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4} \, dx-12 \int \frac {e^{4 x (-4+\log (x))} (1-8 x)}{\left (-1+8 x+x^4\right )^2} \, dx-12 \int \frac {e^{4 x (-4+\log (x))}}{-1+8 x+x^4} \, dx-96 \int \frac {e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2} \, dx \\ & = 4 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-4 \int \frac {e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2} \, dx+4 \int \frac {e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4} \, dx-12 \int \frac {e^{4 x (-4+\log (x))}}{-1+8 x+x^4} \, dx-12 \int \left (\frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2}-\frac {8 e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2}\right ) \, dx-96 \int \frac {e^{4 x (-4+\log (x))} x}{\left (-1+8 x+x^4\right )^2} \, dx \\ & = 4 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-4 \int \frac {e^{4 x (-4+\log (x))} x^3}{\left (-1+8 x+x^4\right )^2} \, dx+4 \int \frac {e^{4 x (-4+\log (x))} \log (x)}{-1+8 x+x^4} \, dx-12 \int \frac {e^{4 x (-4+\log (x))}}{\left (-1+8 x+x^4\right )^2} \, dx-12 \int \frac {e^{4 x (-4+\log (x))}}{-1+8 x+x^4} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {e^{-16 x} x^{4 x}}{-1+8 x+x^4} \]

[In]

Integrate[(E^(-16*x + 4*x*Log[x])*(4 - 96*x - 4*x^3 - 12*x^4 + (-4 + 32*x + 4*x^4)*Log[x]))/(1 - 16*x + 64*x^2
- 2*x^4 + 16*x^5 + x^8),x]

[Out]

x^(4*x)/(E^(16*x)*(-1 + 8*x + x^4))

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {x^{4 x} {\mathrm e}^{-16 x}}{x^{4}+8 x -1}\)	\(21\)
parallelrisch	\(\frac {{\mathrm e}^{4 x \ln \left (x \right )-16 x}}{x^{4}+8 x -1}\)	\(22\)

[In]

int(((4*x^4+32*x-4)*ln(x)-12*x^4-4*x^3-96*x+4)*exp(4*x*ln(x)-16*x)/(x^8+16*x^5-2*x^4+64*x^2-16*x+1),x,method=_
RETURNVERBOSE)

[Out]

1/(x^4+8*x-1)*x^(4*x)*exp(-16*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {e^{\left (4 \, x \log \left (x\right ) - 16 \, x\right )}}{x^{4} + 8 \, x - 1} \]

[In]

integrate(((4*x^4+32*x-4)*log(x)-12*x^4-4*x^3-96*x+4)*exp(4*x*log(x)-16*x)/(x^8+16*x^5-2*x^4+64*x^2-16*x+1),x,
algorithm="fricas")

[Out]

e^(4*x*log(x) - 16*x)/(x^4 + 8*x - 1)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {e^{4 x \log {\left (x \right )} - 16 x}}{x^{4} + 8 x - 1} \]

[In]

integrate(((4*x**4+32*x-4)*ln(x)-12*x**4-4*x**3-96*x+4)*exp(4*x*ln(x)-16*x)/(x**8+16*x**5-2*x**4+64*x**2-16*x+
1),x)

[Out]

exp(4*x*log(x) - 16*x)/(x**4 + 8*x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {e^{\left (4 \, x \log \left (x\right ) - 16 \, x\right )}}{x^{4} + 8 \, x - 1} \]

[In]

integrate(((4*x^4+32*x-4)*log(x)-12*x^4-4*x^3-96*x+4)*exp(4*x*log(x)-16*x)/(x^8+16*x^5-2*x^4+64*x^2-16*x+1),x,
algorithm="maxima")

[Out]

e^(4*x*log(x) - 16*x)/(x^4 + 8*x - 1)

Giac [F]

\[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\int { -\frac {4 \, {\left (3 \, x^{4} + x^{3} - {\left (x^{4} + 8 \, x - 1\right )} \log \left (x\right ) + 24 \, x - 1\right )} e^{\left (4 \, x \log \left (x\right ) - 16 \, x\right )}}{x^{8} + 16 \, x^{5} - 2 \, x^{4} + 64 \, x^{2} - 16 \, x + 1} \,d x } \]

[In]

integrate(((4*x^4+32*x-4)*log(x)-12*x^4-4*x^3-96*x+4)*exp(4*x*log(x)-16*x)/(x^8+16*x^5-2*x^4+64*x^2-16*x+1),x,
algorithm="giac")

[Out]

integrate(-4*(3*x^4 + x^3 - (x^4 + 8*x - 1)*log(x) + 24*x - 1)*e^(4*x*log(x) - 16*x)/(x^8 + 16*x^5 - 2*x^4 + 6
4*x^2 - 16*x + 1), x)

Mupad [B] (verification not implemented)

Time = 8.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-16 x+4 x \log (x)} \left (4-96 x-4 x^3-12 x^4+\left (-4+32 x+4 x^4\right ) \log (x)\right )}{1-16 x+64 x^2-2 x^4+16 x^5+x^8} \, dx=\frac {x^{4\,x}\,{\mathrm {e}}^{-16\,x}}{x^4+8\,x-1} \]

[In]

int(-(exp(4*x*log(x) - 16*x)*(96*x - log(x)*(32*x + 4*x^4 - 4) + 4*x^3 + 12*x^4 - 4))/(64*x^2 - 16*x - 2*x^4 +
16*x^5 + x^8 + 1),x)

[Out]

(x^(4*x)*exp(-16*x))/(8*x + x^4 - 1)