3.30.53 \(\int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} (-4 e^{2 x}+2 e^{2 x} \log (x))}{e^{16+2 x} x^2+e^{4 e^{-2 x}} (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x))} \, dx\)

Optimal. Leaf size=21 \[ \log \left (e^{16-4 e^{-2 x}} x+(-2+\log (x))^2\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 3.82, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{16+2 x} x+8 e^{16} x^2+e^{4 e^{-2 x}} \left (-4 e^{2 x}+2 e^{2 x} \log (x)\right )}{e^{16+2 x} x^2+e^{4 e^{-2 x}} \left (4 e^{2 x} x-4 e^{2 x} x \log (x)+e^{2 x} x \log ^2(x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-2 x} \left (e^{2 (8+x)} x+8 e^{16} x^2+2 e^{4 e^{-2 x}+2 x} (-2+\log (x))\right )}{x \left (e^{16} x+e^{4 e^{-2 x}} (-2+\log (x))^2\right )} \, dx\\ &=\int \left (\frac {8 e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)}+\frac {-4 e^{4 e^{-2 x}}+e^{16} x+2 e^{4 e^{-2 x}} \log (x)}{x \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx\\ &=8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+\int \frac {-4 e^{4 e^{-2 x}}+e^{16} x+2 e^{4 e^{-2 x}} \log (x)}{x \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx\\ &=8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+\int \left (\frac {2}{x (-2+\log (x))}+\frac {e^{16} (-4+\log (x))}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx\\ &=2 \int \frac {1}{x (-2+\log (x))} \, dx+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \frac {-4+\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-2+\log (x)\right )+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \left (-\frac {4}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}+\frac {\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )}\right ) \, dx\\ &=2 \log (2-\log (x))+8 \int \frac {e^{16-2 x} x}{4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)} \, dx+e^{16} \int \frac {\log (x)}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx-\left (4 e^{16}\right ) \int \frac {1}{(-2+\log (x)) \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 1.87, size = 53, normalized size = 2.52 \begin {gather*} -4 e^{-2 x}+\log \left (4 e^{4 e^{-2 x}}+e^{16} x-4 e^{4 e^{-2 x}} \log (x)+e^{4 e^{-2 x}} \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.32, size = 69, normalized size = 3.29 \begin {gather*} {\left (e^{\left (2 \, x + 16\right )} \log \left (\frac {x e^{16} + {\left (\log \relax (x)^{2} - 4 \, \log \relax (x) + 4\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{\log \relax (x)^{2} - 4 \, \log \relax (x) + 4}\right ) + 2 \, e^{\left (2 \, x + 16\right )} \log \left (\log \relax (x) - 2\right ) - 4 \, e^{16}\right )} e^{\left (-2 \, x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)^2*log(x)-4*exp(x)^2)*exp(4/exp(x)^2)+x*exp(16)*exp(x)^2+8*x^2*exp(16))/((x*exp(x)^2*log(x
)^2-4*x*exp(x)^2*log(x)+4*x*exp(x)^2)*exp(4/exp(x)^2)+x^2*exp(16)*exp(x)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {8 \, x^{2} e^{16} + x e^{\left (2 \, x + 16\right )} + 2 \, {\left (e^{\left (2 \, x\right )} \log \relax (x) - 2 \, e^{\left (2 \, x\right )}\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{x^{2} e^{\left (2 \, x + 16\right )} + {\left (x e^{\left (2 \, x\right )} \log \relax (x)^{2} - 4 \, x e^{\left (2 \, x\right )} \log \relax (x) + 4 \, x e^{\left (2 \, x\right )}\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)^2*log(x)-4*exp(x)^2)*exp(4/exp(x)^2)+x*exp(16)*exp(x)^2+8*x^2*exp(16))/((x*exp(x)^2*log(x
)^2-4*x*exp(x)^2*log(x)+4*x*exp(x)^2)*exp(4/exp(x)^2)+x^2*exp(16)*exp(x)^2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.08, size = 40, normalized size = 1.90

method	result	size
risch	\(2 \ln \left (\ln \relax (x )-2\right )-4 \,{\mathrm e}^{-2 x}+\ln \left ({\mathrm e}^{4 \,{\mathrm e}^{-2 x}}+\frac {x \,{\mathrm e}^{16}}{\ln \relax (x )^{2}-4 \ln \relax (x )+4}\right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)^2*ln(x)-4*exp(x)^2)*exp(4/exp(x)^2)+x*exp(16)*exp(x)^2+8*x^2*exp(16))/((x*exp(x)^2*ln(x)^2-4*x*
exp(x)^2*ln(x)+4*x*exp(x)^2)*exp(4/exp(x)^2)+x^2*exp(16)*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.62, size = 51, normalized size = 2.43 \begin {gather*} -4 \, e^{\left (-2 \, x\right )} + \log \left (\frac {x e^{16} + {\left (\log \relax (x)^{2} - 4 \, \log \relax (x) + 4\right )} e^{\left (4 \, e^{\left (-2 \, x\right )}\right )}}{\log \relax (x)^{2} - 4 \, \log \relax (x) + 4}\right ) + 2 \, \log \left (\log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)^2*log(x)-4*exp(x)^2)*exp(4/exp(x)^2)+x*exp(16)*exp(x)^2+8*x^2*exp(16))/((x*exp(x)^2*log(x
)^2-4*x*exp(x)^2*log(x)+4*x*exp(x)^2)*exp(4/exp(x)^2)+x^2*exp(16)*exp(x)^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {x\,{\mathrm {e}}^{2\,x+16}+8\,x^2\,{\mathrm {e}}^{16}-{\mathrm {e}}^{4\,{\mathrm {e}}^{-2\,x}}\,\left (4\,{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)\right )}{{\mathrm {e}}^{4\,{\mathrm {e}}^{-2\,x}}\,\left (x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (x)}^2-4\,x\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)+4\,x\,{\mathrm {e}}^{2\,x}\right )+x^2\,{\mathrm {e}}^{2\,x+16}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2*exp(16) - exp(4*exp(-2*x))*(4*exp(2*x) - 2*exp(2*x)*log(x)) + x*exp(2*x)*exp(16))/(exp(4*exp(-2*x))
*(4*x*exp(2*x) + x*exp(2*x)*log(x)^2 - 4*x*exp(2*x)*log(x)) + x^2*exp(2*x)*exp(16)),x)

[Out]

int((x*exp(2*x + 16) + 8*x^2*exp(16) - exp(4*exp(-2*x))*(4*exp(2*x) - 2*exp(2*x)*log(x)))/(exp(4*exp(-2*x))*(4
*x*exp(2*x) + x*exp(2*x)*log(x)^2 - 4*x*exp(2*x)*log(x)) + x^2*exp(2*x + 16)), x)

________________________________________________________________________________________

sympy [B]  time = 0.88, size = 41, normalized size = 1.95 \begin {gather*} \log {\left (\frac {x e^{16}}{\log {\relax (x )}^{2} - 4 \log {\relax (x )} + 4} + e^{4 e^{- 2 x}} \right )} + 2 \log {\left (\log {\relax (x )} - 2 \right )} - 4 e^{- 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)**2*ln(x)-4*exp(x)**2)*exp(4/exp(x)**2)+x*exp(16)*exp(x)**2+8*x**2*exp(16))/((x*exp(x)**2*
ln(x)**2-4*x*exp(x)**2*ln(x)+4*x*exp(x)**2)*exp(4/exp(x)**2)+x**2*exp(16)*exp(x)**2),x)

[Out]

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

________________________________________________________________________________________