3.5.14 \(\int \frac {e^{2 e^x} (6-3 e^4)+12 x^2-3 e^4 x^2-3 x^2 \log (2)+(-16 x+6 e^4 x+2 x \log (2)) \log (3)+(6-3 e^4) \log ^2(3)+e^{e^x} (-16 x+6 e^4 x+2 x \log (2)+e^x (-2 x^2+x^2 \log (2))+(12-6 e^4) \log (3))}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} (-2 x^5+2 x^4 \log (3))} \, dx\)

Optimal. Leaf size=30 \[ \frac {-2+e^4+\frac {x (2-\log (2))}{e^{e^x}-x+\log (3)}}{x^3} \]

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Rubi [F]  time = 8.31, 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^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-16*x + 6*E^4*x + 2*x*Log[2])*Log[3] + (6 -
3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4*x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3]))/(E^(
2*E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4*Log[3])),x]

[Out]

6*Log[3]^2*Defer[Int][1/(x^4*(-E^E^x + x - Log[3])^2), x] - 3*E^4*Log[3]^2*Defer[Int][1/(x^4*(-E^E^x + x - Log
[3])^2), x] + 12*Log[3]*Defer[Int][E^E^x/(x^4*(-E^E^x + x - Log[3])^2), x] - 6*Log[3]*Defer[Int][E^(4 + E^x)/(
x^4*(-E^E^x + x - Log[3])^2), x] - Log[3]*(16 - Log[4])*Defer[Int][1/(x^3*(-E^E^x + x - Log[3])^2), x] + 3*E^4
*Log[9]*Defer[Int][1/(x^3*(-E^E^x + x - Log[3])^2), x] - (16 - Log[4])*Defer[Int][E^E^x/(x^3*(-E^E^x + x - Log
[3])^2), x] + 6*Defer[Int][E^(4 + E^x)/(x^3*(-E^E^x + x - Log[3])^2), x] - 3*E^4*Defer[Int][1/(x^2*(-E^E^x + x
- Log[3])^2), x] + (12 - Log[8])*Defer[Int][1/(x^2*(-E^E^x + x - Log[3])^2), x] + 3*(2 - E^4)*Defer[Int][E^(2
*E^x)/(x^4*(E^E^x - x + Log[3])^2), x] - (2 - Log[2])*Defer[Int][E^(E^x + x)/(x^2*(E^E^x - x + Log[3])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 e^x} \left (6-3 e^4\right )+\left (12-3 e^4\right ) x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx\\ &=\int \frac {e^{2 e^x} \left (6-3 e^4\right )+x^2 \left (12-3 e^4-3 \log (2)\right )+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx\\ &=\int \frac {6 e^{2 e^x} \left (1-\frac {e^4}{2}\right )+e^{e^x+x} x^2 (-2+\log (2))+6 e^{4+e^x} (x-\log (3))-3 e^4 (x-\log (3))^2+6 \log ^2(3)+e^{e^x} (12 \log (3)+x (-16+\log (4)))+x \log (3) (-16+\log (4))-x^2 (-12+\log (8))}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx\\ &=\int \left (-\frac {3 e^4 (x-\log (3))^2}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2}+\frac {6 \log ^2(3)}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2}-\frac {3 e^{2 e^x} \left (-2+e^4\right )}{x^4 \left (e^{e^x}-x+\log (3)\right )^2}+\frac {e^{e^x+x} (-2+\log (2))}{x^2 \left (e^{e^x}-x+\log (3)\right )^2}+\frac {6 e^{4+e^x} (x-\log (3))}{x^4 \left (e^{e^x}-x+\log (3)\right )^2}+\frac {e^{e^x} (12 \log (3)-x (16-\log (4)))}{x^4 \left (e^{e^x}-x+\log (3)\right )^2}+\frac {\log (3) (-16+\log (4))}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2}-\frac {-12+\log (8)}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2}\right ) \, dx\\ &=6 \int \frac {e^{4+e^x} (x-\log (3))}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx-\left (3 e^4\right ) \int \frac {(x-\log (3))^2}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\left (3 \left (2-e^4\right )\right ) \int \frac {e^{2 e^x}}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx+(-2+\log (2)) \int \frac {e^{e^x+x}}{x^2 \left (e^{e^x}-x+\log (3)\right )^2} \, dx+\left (6 \log ^2(3)\right ) \int \frac {1}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx-(\log (3) (16-\log (4))) \int \frac {1}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+(12-\log (8)) \int \frac {1}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\int \frac {e^{e^x} (12 \log (3)-x (16-\log (4)))}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx\\ &=6 \int \left (\frac {e^{4+e^x}}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2}-\frac {e^{4+e^x} \log (3)}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2}\right ) \, dx-\left (3 e^4\right ) \int \left (\frac {1}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2}+\frac {\log ^2(3)}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2}-\frac {\log (9)}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2}\right ) \, dx+\left (3 \left (2-e^4\right )\right ) \int \frac {e^{2 e^x}}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx+(-2+\log (2)) \int \frac {e^{e^x+x}}{x^2 \left (e^{e^x}-x+\log (3)\right )^2} \, dx+\left (6 \log ^2(3)\right ) \int \frac {1}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx-(\log (3) (16-\log (4))) \int \frac {1}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+(12-\log (8)) \int \frac {1}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\int \left (\frac {12 e^{e^x} \log (3)}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2}+\frac {e^{e^x} (-16+\log (4))}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2}\right ) \, dx\\ &=6 \int \frac {e^{4+e^x}}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx-\left (3 e^4\right ) \int \frac {1}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\left (3 \left (2-e^4\right )\right ) \int \frac {e^{2 e^x}}{x^4 \left (e^{e^x}-x+\log (3)\right )^2} \, dx+(-2+\log (2)) \int \frac {e^{e^x+x}}{x^2 \left (e^{e^x}-x+\log (3)\right )^2} \, dx-(6 \log (3)) \int \frac {e^{4+e^x}}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+(12 \log (3)) \int \frac {e^{e^x}}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\left (6 \log ^2(3)\right ) \int \frac {1}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx-\left (3 e^4 \log ^2(3)\right ) \int \frac {1}{x^4 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx-(\log (3) (16-\log (4))) \int \frac {1}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+(-16+\log (4)) \int \frac {e^{e^x}}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+(12-\log (8)) \int \frac {1}{x^2 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx+\left (3 e^4 \log (9)\right ) \int \frac {1}{x^3 \left (-e^{e^x}+x-\log (3)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 64, normalized size = 2.13 \begin {gather*} \frac {-2+e^4-\frac {x \left (2-\log (2)+e^x (x (-2+\log (2))-\log (2) \log (3)+\log (9))\right )}{\left (-1+e^x (x-\log (3))\right ) \left (e^{e^x}-x+\log (3)\right )}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-16*x + 6*E^4*x + 2*x*Log[2])*Log[3] +
(6 - 3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4*x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3])
)/(E^(2*E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4*Log[3])),x]

[Out]

(-2 + E^4 - (x*(2 - Log[2] + E^x*(x*(-2 + Log[2]) - Log[2]*Log[3] + Log[9])))/((-1 + E^x*(x - Log[3]))*(E^E^x
- x + Log[3])))/x^3

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fricas [A]  time = 0.83, size = 51, normalized size = 1.70 \begin {gather*} \frac {x e^{4} - {\left (e^{4} - 2\right )} e^{\left (e^{x}\right )} - {\left (e^{4} - 2\right )} \log \relax (3) + x \log \relax (2) - 4 \, x}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="fricas")

[Out]

(x*e^4 - (e^4 - 2)*e^(e^x) - (e^4 - 2)*log(3) + x*log(2) - 4*x)/(x^4 - x^3*e^(e^x) - x^3*log(3))

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giac [B]  time = 0.48, size = 56, normalized size = 1.87 \begin {gather*} \frac {x e^{4} - e^{4} \log \relax (3) + x \log \relax (2) - 4 \, x - e^{\left (e^{x} + 4\right )} + 2 \, e^{\left (e^{x}\right )} + 2 \, \log \relax (3)}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="giac")

[Out]

(x*e^4 - e^4*log(3) + x*log(2) - 4*x - e^(e^x + 4) + 2*e^(e^x) + 2*log(3))/(x^4 - x^3*e^(e^x) - x^3*log(3))

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maple [A]  time = 0.07, size = 33, normalized size = 1.10

method	result	size
risch	\(\frac {{\mathrm e}^{4}}{x^{3}}-\frac {2}{x^{3}}-\frac {\ln \relax (2)-2}{x^{2} \left (\ln \relax (3)+{\mathrm e}^{{\mathrm e}^{x}}-x \right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*ln(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*ln(3)+2*x*ln(2)+6*x*exp(4)-16*x)*exp
(exp(x))+(-3*exp(4)+6)*ln(3)^2+(2*x*ln(2)+6*x*exp(4)-16*x)*ln(3)-3*x^2*ln(2)-3*x^2*exp(4)+12*x^2)/(x^4*exp(exp
(x))^2+(2*x^4*ln(3)-2*x^5)*exp(exp(x))+x^4*ln(3)^2-2*x^5*ln(3)+x^6),x,method=_RETURNVERBOSE)

[Out]

exp(4)/x^3-2/x^3-(ln(2)-2)/x^2/(ln(3)+exp(exp(x))-x)

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maxima [A]  time = 0.91, size = 50, normalized size = 1.67 \begin {gather*} \frac {x {\left (e^{4} + \log \relax (2) - 4\right )} - {\left (e^{4} - 2\right )} e^{\left (e^{x}\right )} - e^{4} \log \relax (3) + 2 \, \log \relax (3)}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp(4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3)^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)
/(x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log(3)+x^6),x, algorithm="maxima")

[Out]

(x*(e^4 + log(2) - 4) - (e^4 - 2)*e^(e^x) - e^4*log(3) + 2*log(3))/(x^4 - x^3*e^(e^x) - x^3*log(3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (3\,{\mathrm {e}}^4-6\right )+{\ln \relax (3)}^2\,\left (3\,{\mathrm {e}}^4-6\right )-\ln \relax (3)\,\left (6\,x\,{\mathrm {e}}^4-16\,x+2\,x\,\ln \relax (2)\right )+3\,x^2\,{\mathrm {e}}^4+3\,x^2\,\ln \relax (2)-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (6\,x\,{\mathrm {e}}^4-16\,x+2\,x\,\ln \relax (2)+{\mathrm {e}}^x\,\left (x^2\,\ln \relax (2)-2\,x^2\right )-\ln \relax (3)\,\left (6\,{\mathrm {e}}^4-12\right )\right )-12\,x^2}{x^4\,{\ln \relax (3)}^2-2\,x^5\,\ln \relax (3)+x^6+x^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (2\,x^4\,\ln \relax (3)-2\,x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x*exp(4) - 16*x + 2*x*log(2)) + 3*x^2
*exp(4) + 3*x^2*log(2) - exp(exp(x))*(6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6
*exp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp(x)) + exp(exp(x))*(2*x^4*log(3) -
2*x^5)),x)

[Out]

-int((exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x*exp(4) - 16*x + 2*x*log(2)) + 3*x^2
*exp(4) + 3*x^2*log(2) - exp(exp(x))*(6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6
*exp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp(x)) + exp(exp(x))*(2*x^4*log(3) -
2*x^5)), x)

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sympy [A]  time = 0.21, size = 34, normalized size = 1.13 \begin {gather*} \frac {2 - \log {\relax (2 )}}{- x^{3} + x^{2} e^{e^{x}} + x^{2} \log {\relax (3 )}} - \frac {6 - 3 e^{4}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(4)+6)*exp(exp(x))**2+((x**2*ln(2)-2*x**2)*exp(x)+(-6*exp(4)+12)*ln(3)+2*x*ln(2)+6*x*exp(4)-
16*x)*exp(exp(x))+(-3*exp(4)+6)*ln(3)**2+(2*x*ln(2)+6*x*exp(4)-16*x)*ln(3)-3*x**2*ln(2)-3*x**2*exp(4)+12*x**2)
/(x**4*exp(exp(x))**2+(2*x**4*ln(3)-2*x**5)*exp(exp(x))+x**4*ln(3)**2-2*x**5*ln(3)+x**6),x)

[Out]

(2 - log(2))/(-x**3 + x**2*exp(exp(x)) + x**2*log(3)) - (6 - 3*exp(4))/(3*x**3)

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