3.26.69 \(\int \frac {16 \log ^2(x^2)+e^{2 e^{e^{\frac {x}{\log (x^2)}}}+4 x} ((1+4 x) \log ^2(x^2)+e^{e^{\frac {x}{\log (x^2)}}+\frac {x}{\log (x^2)}} (-4 x+2 x \log (x^2)))+e^{e^{e^{\frac {x}{\log (x^2)}}}+2 x} ((8+16 x) \log ^2(x^2)+e^{e^{\frac {x}{\log (x^2)}}+\frac {x}{\log (x^2)}} (-16 x+8 x \log (x^2)))}{\log ^2(x^2)} \, dx\)

Optimal. Leaf size=26 \[ -4+\left (4+e^{e^{e^{\frac {x}{\log \left (x^2\right )}}}+2 x}\right )^2 x \]

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Rubi [F]  time = 4.99, 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 {16 \log ^2\left (x^2\right )+e^{2 e^{e^{\frac {x}{\log \left (x^2\right )}}}+4 x} \left ((1+4 x) \log ^2\left (x^2\right )+e^{e^{\frac {x}{\log \left (x^2\right )}}+\frac {x}{\log \left (x^2\right )}} \left (-4 x+2 x \log \left (x^2\right )\right )\right )+e^{e^{e^{\frac {x}{\log \left (x^2\right )}}}+2 x} \left ((8+16 x) \log ^2\left (x^2\right )+e^{e^{\frac {x}{\log \left (x^2\right )}}+\frac {x}{\log \left (x^2\right )}} \left (-16 x+8 x \log \left (x^2\right )\right )\right )}{\log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 24, normalized size = 0.92 \begin {gather*} \left (4+e^{e^{e^{\frac {x}{\log \left (x^2\right )}}}+2 x}\right )^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + E^(E^E^(x/Log[x^2]) + 2*x))^2*x

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fricas [B]  time = 0.55, size = 108, normalized size = 4.15 \begin {gather*} x e^{\left (2 \, {\left (2 \, x e^{\left (\frac {x}{\log \left (x^{2}\right )}\right )} + e^{\left (\frac {e^{\left (\frac {x}{\log \left (x^{2}\right )}\right )} \log \left (x^{2}\right ) + x}{\log \left (x^{2}\right )}\right )}\right )} e^{\left (-\frac {x}{\log \left (x^{2}\right )}\right )}\right )} + 8 \, x e^{\left ({\left (2 \, x e^{\left (\frac {x}{\log \left (x^{2}\right )}\right )} + e^{\left (\frac {e^{\left (\frac {x}{\log \left (x^{2}\right )}\right )} \log \left (x^{2}\right ) + x}{\log \left (x^{2}\right )}\right )}\right )} e^{\left (-\frac {x}{\log \left (x^{2}\right )}\right )}\right )} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(x^2)-4*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(4*x+1)*log(x^2)^2)*exp(exp(exp(x/log(x^2)
))+2*x)^2+((8*x*log(x^2)-16*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(16*x+8)*log(x^2)^2)*exp(exp(exp(x/log(x^2
)))+2*x)+16*log(x^2)^2)/log(x^2)^2,x, algorithm="fricas")

[Out]

x*e^(2*(2*x*e^(x/log(x^2)) + e^((e^(x/log(x^2))*log(x^2) + x)/log(x^2)))*e^(-x/log(x^2))) + 8*x*e^((2*x*e^(x/l
og(x^2)) + e^((e^(x/log(x^2))*log(x^2) + x)/log(x^2)))*e^(-x/log(x^2))) + 16*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(x^2)-4*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(4*x+1)*log(x^2)^2)*exp(exp(exp(x/log(x^2)
))+2*x)^2+((8*x*log(x^2)-16*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(16*x+8)*log(x^2)^2)*exp(exp(exp(x/log(x^2
)))+2*x)+16*log(x^2)^2)/log(x^2)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.39, size = 42, normalized size = 1.62

method	result	size
risch	\(x \,{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{\frac {x}{\ln \left (x^{2}\right )}}}+4 x}+8 x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x}{\ln \left (x^{2}\right )}}}+2 x}+16 x\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x*ln(x^2)-4*x)*exp(x/ln(x^2))*exp(exp(x/ln(x^2)))+(4*x+1)*ln(x^2)^2)*exp(exp(exp(x/ln(x^2)))+2*x)^2+(
(8*x*ln(x^2)-16*x)*exp(x/ln(x^2))*exp(exp(x/ln(x^2)))+(16*x+8)*ln(x^2)^2)*exp(exp(exp(x/ln(x^2)))+2*x)+16*ln(x
^2)^2)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(2*exp(exp(x/ln(x^2)))+4*x)+8*x*exp(exp(exp(x/ln(x^2)))+2*x)+16*x

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maxima [A]  time = 0.77, size = 39, normalized size = 1.50 \begin {gather*} x e^{\left (4 \, x + 2 \, e^{\left (e^{\left (\frac {x}{2 \, \log \relax (x)}\right )}\right )}\right )} + 8 \, x e^{\left (2 \, x + e^{\left (e^{\left (\frac {x}{2 \, \log \relax (x)}\right )}\right )}\right )} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*log(x^2)-4*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(4*x+1)*log(x^2)^2)*exp(exp(exp(x/log(x^2)
))+2*x)^2+((8*x*log(x^2)-16*x)*exp(x/log(x^2))*exp(exp(x/log(x^2)))+(16*x+8)*log(x^2)^2)*exp(exp(exp(x/log(x^2
)))+2*x)+16*log(x^2)^2)/log(x^2)^2,x, algorithm="maxima")

[Out]

x*e^(4*x + 2*e^(e^(1/2*x/log(x)))) + 8*x*e^(2*x + e^(e^(1/2*x/log(x)))) + 16*x

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mupad [B]  time = 1.69, size = 41, normalized size = 1.58 \begin {gather*} 16\,x+x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {x}{\ln \left (x^2\right )}}}}\,{\mathrm {e}}^{4\,x}+8\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{\frac {x}{\ln \left (x^2\right )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x + 2*exp(exp(x/log(x^2))))*(log(x^2)^2*(4*x + 1) - exp(exp(x/log(x^2)))*exp(x/log(x^2))*(4*x - 2*x
*log(x^2))) + 16*log(x^2)^2 + exp(2*x + exp(exp(x/log(x^2))))*(log(x^2)^2*(16*x + 8) - exp(exp(x/log(x^2)))*ex
p(x/log(x^2))*(16*x - 8*x*log(x^2))))/log(x^2)^2,x)

[Out]

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

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sympy [A]  time = 18.87, size = 39, normalized size = 1.50 \begin {gather*} 8 x e^{2 x + e^{e^{\frac {x}{\log {\left (x^{2} \right )}}}}} + x e^{4 x + 2 e^{e^{\frac {x}{\log {\left (x^{2} \right )}}}}} + 16 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*ln(x**2)-4*x)*exp(x/ln(x**2))*exp(exp(x/ln(x**2)))+(4*x+1)*ln(x**2)**2)*exp(exp(exp(x/ln(x**2
)))+2*x)**2+((8*x*ln(x**2)-16*x)*exp(x/ln(x**2))*exp(exp(x/ln(x**2)))+(16*x+8)*ln(x**2)**2)*exp(exp(exp(x/ln(x
**2)))+2*x)+16*ln(x**2)**2)/ln(x**2)**2,x)

[Out]

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

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