3.64.70 \(\int \frac {144+e^{e^{2 x}} (-4+(4-8 e^{2 x} x) \log (x))}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {x}{9-x+\frac {1}{4} e^{e^{2 x}} \log (x)} \]

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Rubi [F]  time = 2.14, 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 {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(144 + E^E^(2*x)*(-4 + (4 - 8*E^(2*x)*x)*Log[x]))/(1296 - 288*x + 16*x^2 + E^E^(2*x)*(72 - 8*x)*Log[x] + E
^(2*E^(2*x))*Log[x]^2),x]

[Out]

16*Defer[Int][x/(-36 + 4*x - E^E^(2*x)*Log[x])^2, x] - 16*Defer[Int][x/(Log[x]*(-36 + 4*x - E^E^(2*x)*Log[x])^
2), x] - 4*Defer[Int][(-36 + 4*x - E^E^(2*x)*Log[x])^(-1), x] + 144*Defer[Int][1/(Log[x]*(36 - 4*x + E^E^(2*x)
*Log[x])^2), x] - 8*Defer[Int][(E^(E^(2*x) + 2*x)*x*Log[x])/(36 - 4*x + E^E^(2*x)*Log[x])^2, x] - 4*Defer[Int]
[1/(Log[x]*(36 - 4*x + E^E^(2*x)*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx\\ &=\int \left (\frac {4 \left (36-e^{e^{2 x}}+e^{e^{2 x}} \log (x)\right )}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}-\frac {8 e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx\\ &=4 \int \frac {36-e^{e^{2 x}}+e^{e^{2 x}} \log (x)}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx\\ &=4 \int \left (\frac {-1+\log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )}+\frac {4 (9-x+x \log (x))}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx\\ &=4 \int \frac {-1+\log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \frac {9-x+x \log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx\\ &=4 \int \left (-\frac {1}{-36+4 x-e^{e^{2 x}} \log (x)}-\frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )}\right ) \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \left (\frac {x}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}-\frac {x}{\log (x) \left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}+\frac {9}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {1}{-36+4 x-e^{e^{2 x}} \log (x)} \, dx\right )-4 \int \frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \frac {x}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx-16 \int \frac {x}{\log (x) \left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx+144 \int \frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.52, size = 20, normalized size = 0.91 \begin {gather*} \frac {4 x}{36-4 x+e^{e^{2 x}} \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(144 + E^E^(2*x)*(-4 + (4 - 8*E^(2*x)*x)*Log[x]))/(1296 - 288*x + 16*x^2 + E^E^(2*x)*(72 - 8*x)*Log[
x] + E^(2*E^(2*x))*Log[x]^2),x]

[Out]

(4*x)/(36 - 4*x + E^E^(2*x)*Log[x])

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fricas [A]  time = 0.68, size = 18, normalized size = 0.82 \begin {gather*} \frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (x) - 4 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="fricas")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

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giac [A]  time = 0.20, size = 18, normalized size = 0.82 \begin {gather*} \frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (x) - 4 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="giac")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

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maple [A]  time = 0.06, size = 20, normalized size = 0.91

method	result	size
risch	\(-\frac {4 x}{-\ln \relax (x ) {\mathrm e}^{{\mathrm e}^{2 x}}+4 x -36}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x*exp(x)^2+4)*ln(x)-4)*exp(exp(x)^2)+144)/(ln(x)^2*exp(exp(x)^2)^2+(-8*x+72)*ln(x)*exp(exp(x)^2)+16*
x^2-288*x+1296),x,method=_RETURNVERBOSE)

[Out]

-4*x/(-ln(x)*exp(exp(2*x))+4*x-36)

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maxima [A]  time = 0.52, size = 18, normalized size = 0.82 \begin {gather*} \frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (x) - 4 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="maxima")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\left (\ln \relax (x)\,\left (8\,x\,{\mathrm {e}}^{2\,x}-4\right )+4\right )-144}{{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\ln \relax (x)}^2-288\,x+16\,x^2-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\ln \relax (x)\,\left (8\,x-72\right )+1296} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(2*x))*(log(x)*(8*x*exp(2*x) - 4) + 4) - 144)/(exp(2*exp(2*x))*log(x)^2 - 288*x + 16*x^2 - exp(ex
p(2*x))*log(x)*(8*x - 72) + 1296),x)

[Out]

int(-(exp(exp(2*x))*(log(x)*(8*x*exp(2*x) - 4) + 4) - 144)/(exp(2*exp(2*x))*log(x)^2 - 288*x + 16*x^2 - exp(ex
p(2*x))*log(x)*(8*x - 72) + 1296), x)

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sympy [A]  time = 0.34, size = 17, normalized size = 0.77 \begin {gather*} \frac {4 x}{- 4 x + e^{e^{2 x}} \log {\relax (x )} + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x*exp(x)**2+4)*ln(x)-4)*exp(exp(x)**2)+144)/(ln(x)**2*exp(exp(x)**2)**2+(-8*x+72)*ln(x)*exp(ex
p(x)**2)+16*x**2-288*x+1296),x)

[Out]

4*x/(-4*x + exp(exp(2*x))*log(x) + 36)

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