3.24.98 \(\int \frac {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+(2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)) \log (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2)}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {2 \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x \log (x)} \]

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Rubi [F]  time = 6.68, 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 {-4 x \log (x)+4 e^{2 x+4 e^{2 x} x^2} x^2 (8+8 x) \log (x)+\left (2 x+e^{4 e^{2 x} x^2} (-2-2 \log (x))+2 x \log (x)\right ) \log \left (e^{8 e^{2 x} x^2}-2 e^{4 e^{2 x} x^2} x+x^2\right )}{e^{4 e^{2 x} x^2} x^2 \log ^2(x)-x^3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-2*Defer[Int][Log[(E^(4*E^(2*x)*x^2) - x)^2]/(x^2*Log[x]^2), x] + 32*Defer[Int][E^(2*x*(1 + 2*E^(2*x)*x))/((E^
(4*E^(2*x)*x^2) - x)*Log[x]), x] - 4*Defer[Int][1/((E^(4*E^(2*x)*x^2) - x)*x*Log[x]), x] + 32*Defer[Int][(E^(2
*x*(1 + 2*E^(2*x)*x))*x)/((E^(4*E^(2*x)*x^2) - x)*Log[x]), x] - 2*Defer[Int][Log[(E^(4*E^(2*x)*x^2) - x)^2]/(x
^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (\frac {2 x \left (-1+8 e^{2 x \left (1+2 e^{2 x} x\right )} x (1+x)\right ) \log (x)}{e^{4 e^{2 x} x^2}-x}-\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))\right )}{x^2 \log ^2(x)} \, dx\\ &=2 \int \frac {\frac {2 x \left (-1+8 e^{2 x \left (1+2 e^{2 x} x\right )} x (1+x)\right ) \log (x)}{e^{4 e^{2 x} x^2}-x}-\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))}{x^2 \log ^2(x)} \, dx\\ &=2 \int \left (\frac {16 e^{2 x \left (1+2 e^{2 x} x\right )} (1+x)}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)}-\frac {-e^{4 e^{2 x} x^2} \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )+x \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )-2 x \log (x)-e^{4 e^{2 x} x^2} \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) \log (x)+x \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) \log (x)}{x^2 \left (-e^{4 e^{2 x} x^2}+x\right ) \log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {-e^{4 e^{2 x} x^2} \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )+x \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )-2 x \log (x)-e^{4 e^{2 x} x^2} \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) \log (x)+x \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) \log (x)}{x^2 \left (-e^{4 e^{2 x} x^2}+x\right ) \log ^2(x)} \, dx\right )+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} (1+x)}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {2 x \log (x)+\left (e^{4 e^{2 x} x^2}-x\right ) \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))}{\left (e^{4 e^{2 x} x^2}-x\right ) x^2 \log ^2(x)} \, dx\right )+32 \int \left (\frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)}+\frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)}\right ) \, dx\\ &=-\left (2 \int \frac {\frac {2 x \log (x)}{e^{4 e^{2 x} x^2}-x}+\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))}{x^2 \log ^2(x)} \, dx\right )+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ &=-\left (2 \int \left (\frac {2}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)}+\frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))}{x^2 \log ^2(x)}\right ) \, dx\right )+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right ) (1+\log (x))}{x^2 \log ^2(x)} \, dx\right )-4 \int \frac {1}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ &=-\left (2 \int \left (\frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log ^2(x)}+\frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log (x)}\right ) \, dx\right )-4 \int \frac {1}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ &=-\left (2 \int \frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log ^2(x)} \, dx\right )-2 \int \frac {\log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x^2 \log (x)} \, dx-4 \int \frac {1}{\left (e^{4 e^{2 x} x^2}-x\right ) x \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )}}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx+32 \int \frac {e^{2 x \left (1+2 e^{2 x} x\right )} x}{\left (e^{4 e^{2 x} x^2}-x\right ) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 28, normalized size = 1.00 \begin {gather*} \frac {2 \log \left (\left (e^{4 e^{2 x} x^2}-x\right )^2\right )}{x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.18, size = 99, normalized size = 3.54 \begin {gather*} \frac {2 \, \log \left ({\left (x^{2} e^{\left (4 \, x + 4 \, \log \relax (2) + 4 \, \log \relax (x)\right )} - 2 \, x e^{\left (4 \, x + e^{\left (2 \, x + 2 \, \log \relax (2) + 2 \, \log \relax (x)\right )} + 4 \, \log \relax (2) + 4 \, \log \relax (x)\right )} + e^{\left (4 \, x + 2 \, e^{\left (2 \, x + 2 \, \log \relax (2) + 2 \, \log \relax (x)\right )} + 4 \, \log \relax (2) + 4 \, \log \relax (x)\right )}\right )} e^{\left (-4 \, x - 4 \, \log \relax (2) - 4 \, \log \relax (x)\right )}\right )}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(exp(log(2*x)+x)^2)^2-2*x*exp(exp(log(
2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log(2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(
2*x)+x)^2)-x^3*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.31, size = 37, normalized size = 1.32 \begin {gather*} \frac {2 \, \log \left (x^{2} - 2 \, x e^{\left (4 \, x^{2} e^{\left (2 \, x\right )}\right )} + e^{\left (8 \, x^{2} e^{\left (2 \, x\right )}\right )}\right )}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(exp(log(2*x)+x)^2)^2-2*x*exp(exp(log(
2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log(2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(
2*x)+x)^2)-x^3*log(x)^2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.34, size = 140, normalized size = 5.00

method	result	size
risch	\(\frac {4 \ln \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )}{x \ln \relax (x )}-\frac {i \pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right ) \left (\mathrm {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )\right )^{2}-2 \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right ) \mathrm {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )\right )+\mathrm {csgn}\left (i \left (-{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}}+x \right )^{2}\right )^{2}\right )}{x \ln \relax (x )}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*ln(x)-2)*exp(exp(ln(2*x)+x)^2)+2*x*ln(x)+2*x)*ln(exp(exp(ln(2*x)+x)^2)^2-2*x*exp(exp(ln(2*x)+x)^2)+x
^2)+(8*x+8)*ln(x)*exp(ln(2*x)+x)^2*exp(exp(ln(2*x)+x)^2)-4*x*ln(x))/(x^2*ln(x)^2*exp(exp(ln(2*x)+x)^2)-x^3*ln(
x)^2),x,method=_RETURNVERBOSE)

[Out]

4/x/ln(x)*ln(-exp(4*exp(2*x)*x^2)+x)-I/x*Pi*csgn(I*(-exp(4*exp(2*x)*x^2)+x)^2)*(csgn(I*(-exp(4*exp(2*x)*x^2)+x
))^2-2*csgn(I*(-exp(4*exp(2*x)*x^2)+x)^2)*csgn(I*(-exp(4*exp(2*x)*x^2)+x))+csgn(I*(-exp(4*exp(2*x)*x^2)+x)^2)^
2)/ln(x)

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maxima [A]  time = 0.72, size = 24, normalized size = 0.86 \begin {gather*} \frac {4 \, \log \left (x - e^{\left (4 \, x^{2} e^{\left (2 \, x\right )}\right )}\right )}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*log(x)-2)*exp(exp(log(2*x)+x)^2)+2*x*log(x)+2*x)*log(exp(exp(log(2*x)+x)^2)^2-2*x*exp(exp(log(
2*x)+x)^2)+x^2)+(8*x+8)*log(x)*exp(log(2*x)+x)^2*exp(exp(log(2*x)+x)^2)-4*x*log(x))/(x^2*log(x)^2*exp(exp(log(
2*x)+x)^2)-x^3*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.67, size = 37, normalized size = 1.32 \begin {gather*} \frac {2\,\ln \left ({\mathrm {e}}^{8\,x^2\,{\mathrm {e}}^{2\,x}}-2\,x\,{\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x}}+x^2\right )}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(2*exp(2*x + 2*log(2*x))) - 2*x*exp(exp(2*x + 2*log(2*x))) + x^2)*(2*x - exp(exp(2*x + 2*log(2*x)
))*(2*log(x) + 2) + 2*x*log(x)) - 4*x*log(x) + exp(2*x + 2*log(2*x))*exp(exp(2*x + 2*log(2*x)))*log(x)*(8*x +
8))/(x^3*log(x)^2 - x^2*exp(exp(2*x + 2*log(2*x)))*log(x)^2),x)

[Out]

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

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sympy [A]  time = 2.79, size = 37, normalized size = 1.32 \begin {gather*} \frac {2 \log {\left (x^{2} - 2 x e^{4 x^{2} e^{2 x}} + e^{8 x^{2} e^{2 x}} \right )}}{x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*ln(x)-2)*exp(exp(ln(2*x)+x)**2)+2*x*ln(x)+2*x)*ln(exp(exp(ln(2*x)+x)**2)**2-2*x*exp(exp(ln(2*x
)+x)**2)+x**2)+(8*x+8)*ln(x)*exp(ln(2*x)+x)**2*exp(exp(ln(2*x)+x)**2)-4*x*ln(x))/(x**2*ln(x)**2*exp(exp(ln(2*x
)+x)**2)-x**3*ln(x)**2),x)

[Out]

2*log(x**2 - 2*x*exp(4*x**2*exp(2*x)) + exp(8*x**2*exp(2*x)))/(x*log(x))

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