3.55.79 \(\int \frac {e^{e^x} (e^x (-1-x)+e^{2 x} x)+e^x (-x^2+x^3)+16 e^x \log ^3(x)+e^x (-4-4 x) \log ^4(x)}{8 e^{2 e^x}+e^x x^3+8 x^4+(-4 e^x x-64 x^2) \log ^4(x)+128 \log ^8(x)+e^{e^x} (-e^x x-16 x^2+64 \log ^4(x))} \, dx\)

Optimal. Leaf size=28 \[ \log \left (8-\frac {e^x x}{e^{e^x}-x^2+4 \log ^4(x)}\right ) \]

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Rubi [F]  time = 100.47, 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^{e^x} \left (e^x (-1-x)+e^{2 x} x\right )+e^x \left (-x^2+x^3\right )+16 e^x \log ^3(x)+e^x (-4-4 x) \log ^4(x)}{8 e^{2 e^x}+e^x x^3+8 x^4+\left (-4 e^x x-64 x^2\right ) \log ^4(x)+128 \log ^8(x)+e^{e^x} \left (-e^x x-16 x^2+64 \log ^4(x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^x*(E^x*(-1 - x) + E^(2*x)*x) + E^x*(-x^2 + x^3) + 16*E^x*Log[x]^3 + E^x*(-4 - 4*x)*Log[x]^4)/(8*E^(2*
E^x) + E^x*x^3 + 8*x^4 + (-4*E^x*x - 64*x^2)*Log[x]^4 + 128*Log[x]^8 + E^E^x*(-(E^x*x) - 16*x^2 + 64*Log[x]^4)
),x]

[Out]

-Defer[Int][(E^x*x^2)/((-8*E^E^x + E^x*x + 8*x^2 - 32*Log[x]^4)*(-E^E^x + x^2 - 4*Log[x]^4)), x] + Defer[Int][
(E^x*x^3)/((-8*E^E^x + E^x*x + 8*x^2 - 32*Log[x]^4)*(-E^E^x + x^2 - 4*Log[x]^4)), x] - 4*Defer[Int][(E^x*x*Log
[x]^4)/((-8*E^E^x + E^x*x + 8*x^2 - 32*Log[x]^4)*(-E^E^x + x^2 - 4*Log[x]^4)), x] - Defer[Int][E^(E^x + x)/(E^
E^x - x^2 + 4*Log[x]^4), x] - Defer[Int][E^(E^x + x)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32
*Log[x]^4)), x] + 8*Defer[Int][E^(2*E^x + x)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^
4)), x] - Defer[Int][(E^(E^x + x)*x)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4)), x]
- 8*Defer[Int][(E^(E^x + x)*x^2)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4)), x] + 16
*Defer[Int][(E^x*Log[x]^3)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4)), x] - 4*Defer[
Int][(E^x*Log[x]^4)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4)), x] + 32*Defer[Int][(
E^(E^x + x)*Log[x]^4)/((E^E^x - x^2 + 4*Log[x]^4)*(8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (e^{e^x+x} x+(-1+x) x^2-e^{e^x} (1+x)+16 \log ^3(x)-4 (1+x) \log ^4(x)\right )}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx\\ &=\int \left (-\frac {e^{e^x+x}}{e^{e^x}-x^2+4 \log ^4(x)}+\frac {e^x \left (-e^{e^x}+8 e^{2 e^x}-e^{e^x} x-x^2-8 e^{e^x} x^2+x^3+16 \log ^3(x)-4 \log ^4(x)+32 e^{e^x} \log ^4(x)-4 x \log ^4(x)\right )}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}\right ) \, dx\\ &=-\int \frac {e^{e^x+x}}{e^{e^x}-x^2+4 \log ^4(x)} \, dx+\int \frac {e^x \left (-e^{e^x}+8 e^{2 e^x}-e^{e^x} x-x^2-8 e^{e^x} x^2+x^3+16 \log ^3(x)-4 \log ^4(x)+32 e^{e^x} \log ^4(x)-4 x \log ^4(x)\right )}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx\\ &=-\int \frac {e^{e^x+x}}{e^{e^x}-x^2+4 \log ^4(x)} \, dx+\int \left (-\frac {e^x x^2}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )}+\frac {e^x x^3}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )}-\frac {4 e^x x \log ^4(x)}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )}-\frac {e^{e^x+x}}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}+\frac {8 e^{2 e^x+x}}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}-\frac {e^{e^x+x} x}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}-\frac {8 e^{e^x+x} x^2}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}+\frac {16 e^x \log ^3(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}-\frac {4 e^x \log ^4(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}+\frac {32 e^{e^x+x} \log ^4(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )}\right ) \, dx\\ &=-\left (4 \int \frac {e^x x \log ^4(x)}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )} \, dx\right )-4 \int \frac {e^x \log ^4(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx+8 \int \frac {e^{2 e^x+x}}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx-8 \int \frac {e^{e^x+x} x^2}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx+16 \int \frac {e^x \log ^3(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx+32 \int \frac {e^{e^x+x} \log ^4(x)}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx-\int \frac {e^x x^2}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )} \, dx+\int \frac {e^x x^3}{\left (-8 e^{e^x}+e^x x+8 x^2-32 \log ^4(x)\right ) \left (-e^{e^x}+x^2-4 \log ^4(x)\right )} \, dx-\int \frac {e^{e^x+x}}{e^{e^x}-x^2+4 \log ^4(x)} \, dx-\int \frac {e^{e^x+x}}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx-\int \frac {e^{e^x+x} x}{\left (e^{e^x}-x^2+4 \log ^4(x)\right ) \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 47, normalized size = 1.68 \begin {gather*} -\log \left (e^{e^x}-x^2+4 \log ^4(x)\right )+\log \left (8 e^{e^x}-e^x x-8 x^2+32 \log ^4(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^x*(E^x*(-1 - x) + E^(2*x)*x) + E^x*(-x^2 + x^3) + 16*E^x*Log[x]^3 + E^x*(-4 - 4*x)*Log[x]^4)/(8
*E^(2*E^x) + E^x*x^3 + 8*x^4 + (-4*E^x*x - 64*x^2)*Log[x]^4 + 128*Log[x]^8 + E^E^x*(-(E^x*x) - 16*x^2 + 64*Log
[x]^4)),x]

[Out]

-Log[E^E^x - x^2 + 4*Log[x]^4] + Log[8*E^E^x - E^x*x - 8*x^2 + 32*Log[x]^4]

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fricas [A]  time = 0.64, size = 42, normalized size = 1.50 \begin {gather*} \log \left (32 \, \log \relax (x)^{4} - 8 \, x^{2} - x e^{x} + 8 \, e^{\left (e^{x}\right )}\right ) - \log \left (4 \, \log \relax (x)^{4} - x^{2} + e^{\left (e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2+(-x-1)*exp(x))*exp(exp(x))+(-4*x-4)*exp(x)*log(x)^4+16*exp(x)*log(x)^3+(x^3-x^2)*exp(x)
)/(8*exp(exp(x))^2+(64*log(x)^4-exp(x)*x-16*x^2)*exp(exp(x))+128*log(x)^8+(-4*exp(x)*x-64*x^2)*log(x)^4+exp(x)
*x^3+8*x^4),x, algorithm="fricas")

[Out]

log(32*log(x)^4 - 8*x^2 - x*e^x + 8*e^(e^x)) - log(4*log(x)^4 - x^2 + e^(e^x))

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giac [A]  time = 1.05, size = 42, normalized size = 1.50 \begin {gather*} \log \left (32 \, \log \relax (x)^{4} - 8 \, x^{2} - x e^{x} + 8 \, e^{\left (e^{x}\right )}\right ) - \log \left (-4 \, \log \relax (x)^{4} + x^{2} - e^{\left (e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2+(-x-1)*exp(x))*exp(exp(x))+(-4*x-4)*exp(x)*log(x)^4+16*exp(x)*log(x)^3+(x^3-x^2)*exp(x)
)/(8*exp(exp(x))^2+(64*log(x)^4-exp(x)*x-16*x^2)*exp(exp(x))+128*log(x)^8+(-4*exp(x)*x-64*x^2)*log(x)^4+exp(x)
*x^3+8*x^4),x, algorithm="giac")

[Out]

log(32*log(x)^4 - 8*x^2 - x*e^x + 8*e^(e^x)) - log(-4*log(x)^4 + x^2 - e^(e^x))

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maple [A]  time = 0.05, size = 41, normalized size = 1.46

method	result	size
risch	\(\ln \left (4 \ln \relax (x )^{4}-x^{2}-\frac {{\mathrm e}^{x} x}{8}+{\mathrm e}^{{\mathrm e}^{x}}\right )-\ln \left ({\mathrm e}^{{\mathrm e}^{x}}+4 \ln \relax (x )^{4}-x^{2}\right )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(x)^2+(-x-1)*exp(x))*exp(exp(x))+(-4*x-4)*exp(x)*ln(x)^4+16*exp(x)*ln(x)^3+(x^3-x^2)*exp(x))/(8*exp
(exp(x))^2+(64*ln(x)^4-exp(x)*x-16*x^2)*exp(exp(x))+128*ln(x)^8+(-4*exp(x)*x-64*x^2)*ln(x)^4+exp(x)*x^3+8*x^4)
,x,method=_RETURNVERBOSE)

[Out]

ln(4*ln(x)^4-x^2-1/8*exp(x)*x+exp(exp(x)))-ln(exp(exp(x))+4*ln(x)^4-x^2)

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maxima [A]  time = 0.47, size = 40, normalized size = 1.43 \begin {gather*} \log \left (4 \, \log \relax (x)^{4} - x^{2} - \frac {1}{8} \, x e^{x} + e^{\left (e^{x}\right )}\right ) - \log \left (4 \, \log \relax (x)^{4} - x^{2} + e^{\left (e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)^2+(-x-1)*exp(x))*exp(exp(x))+(-4*x-4)*exp(x)*log(x)^4+16*exp(x)*log(x)^3+(x^3-x^2)*exp(x)
)/(8*exp(exp(x))^2+(64*log(x)^4-exp(x)*x-16*x^2)*exp(exp(x))+128*log(x)^8+(-4*exp(x)*x-64*x^2)*log(x)^4+exp(x)
*x^3+8*x^4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {-{\mathrm {e}}^x\,\left (4\,x+4\right )\,{\ln \relax (x)}^4+16\,{\mathrm {e}}^x\,{\ln \relax (x)}^3+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (x+1\right )\right )-{\mathrm {e}}^x\,\left (x^2-x^3\right )}{8\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+x^3\,{\mathrm {e}}^x+128\,{\ln \relax (x)}^8-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x\,{\mathrm {e}}^x-64\,{\ln \relax (x)}^4+16\,x^2\right )+8\,x^4-{\ln \relax (x)}^4\,\left (4\,x\,{\mathrm {e}}^x+64\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x))*(x*exp(2*x) - exp(x)*(x + 1)) + 16*exp(x)*log(x)^3 - exp(x)*(x^2 - x^3) - exp(x)*log(x)^4*(4*
x + 4))/(8*exp(2*exp(x)) + x^3*exp(x) + 128*log(x)^8 - exp(exp(x))*(x*exp(x) - 64*log(x)^4 + 16*x^2) + 8*x^4 -
log(x)^4*(4*x*exp(x) + 64*x^2)),x)

[Out]

int((exp(exp(x))*(x*exp(2*x) - exp(x)*(x + 1)) + 16*exp(x)*log(x)^3 - exp(x)*(x^2 - x^3) - exp(x)*log(x)^4*(4*
x + 4))/(8*exp(2*exp(x)) + x^3*exp(x) + 128*log(x)^8 - exp(exp(x))*(x*exp(x) - 64*log(x)^4 + 16*x^2) + 8*x^4 -
log(x)^4*(4*x*exp(x) + 64*x^2)), x)

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sympy [A]  time = 1.25, size = 39, normalized size = 1.39 \begin {gather*} - \log {\left (- x^{2} + e^{e^{x}} + 4 \log {\relax (x )}^{4} \right )} + \log {\left (- x^{2} - \frac {x e^{x}}{8} + e^{e^{x}} + 4 \log {\relax (x )}^{4} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(x)**2+(-x-1)*exp(x))*exp(exp(x))+(-4*x-4)*exp(x)*ln(x)**4+16*exp(x)*ln(x)**3+(x**3-x**2)*exp
(x))/(8*exp(exp(x))**2+(64*ln(x)**4-exp(x)*x-16*x**2)*exp(exp(x))+128*ln(x)**8+(-4*exp(x)*x-64*x**2)*ln(x)**4+
exp(x)*x**3+8*x**4),x)

[Out]

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

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