3.31.67 \(\int \frac {-40-5 x^2+(80+20 x^2) \log (x)+e^{e^x} (-10 x+(30 x+10 e^x x^2) \log (x))}{2 e^{e^x} x^2 \log (x)+(8 x+x^3) \log (x)} \, dx\)

Optimal. Leaf size=34 \[ 4+5 \log \left (\frac {x^2 \left (4-\left (-e^{e^x}-\frac {x}{2}\right ) x\right )}{4 \log (x)}\right ) \]

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Rubi [F]  time = 2.35, 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 {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-40 - 5*x^2 + (80 + 20*x^2)*Log[x] + E^E^x*(-10*x + (30*x + 10*E^x*x^2)*Log[x]))/(2*E^E^x*x^2*Log[x] + (8
*x + x^3)*Log[x]),x]

[Out]

15*Log[x] - 5*Log[Log[x]] - 40*Defer[Int][1/(x*(8 + 2*E^E^x*x + x^2)), x] + 5*Defer[Int][x/(8 + 2*E^E^x*x + x^
2), x] + 10*Defer[Int][(E^(E^x + x)*x)/(8 + 2*E^E^x*x + x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)} \, dx\\ &=\int \left (\frac {10 e^{e^x+x} x}{8+2 e^{e^x} x+x^2}+\frac {5 \left (-8-2 e^{e^x} x-x^2+16 \log (x)+6 e^{e^x} x \log (x)+4 x^2 \log (x)\right )}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)}\right ) \, dx\\ &=5 \int \frac {-8-2 e^{e^x} x-x^2+16 \log (x)+6 e^{e^x} x \log (x)+4 x^2 \log (x)}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \left (\frac {-8+x^2}{x \left (8+2 e^{e^x} x+x^2\right )}+\frac {-1+3 \log (x)}{x \log (x)}\right ) \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \frac {-8+x^2}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx+5 \int \frac {-1+3 \log (x)}{x \log (x)} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \left (-\frac {8}{x \left (8+2 e^{e^x} x+x^2\right )}+\frac {x}{8+2 e^{e^x} x+x^2}\right ) \, dx+5 \operatorname {Subst}\left (\int \frac {-1+3 x}{x} \, dx,x,\log (x)\right )+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \frac {x}{8+2 e^{e^x} x+x^2} \, dx+5 \operatorname {Subst}\left (\int \left (3-\frac {1}{x}\right ) \, dx,x,\log (x)\right )+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx-40 \int \frac {1}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx\\ &=15 \log (x)-5 \log (\log (x))+5 \int \frac {x}{8+2 e^{e^x} x+x^2} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx-40 \int \frac {1}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 26, normalized size = 0.76 \begin {gather*} 5 \left (2 \log (x)+\log \left (8+2 e^{e^x} x+x^2\right )-\log (\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 - 5*x^2 + (80 + 20*x^2)*Log[x] + E^E^x*(-10*x + (30*x + 10*E^x*x^2)*Log[x]))/(2*E^E^x*x^2*Log[x
] + (8*x + x^3)*Log[x]),x]

[Out]

5*(2*Log[x] + Log[8 + 2*E^E^x*x + x^2] - Log[Log[x]])

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fricas [A]  time = 0.68, size = 28, normalized size = 0.82 \begin {gather*} 15 \, \log \relax (x) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{x}\right ) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log(x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x
))+(x^3+8*x)*log(x)),x, algorithm="fricas")

[Out]

15*log(x) + 5*log((x^2 + 2*x*e^(e^x) + 8)/x) - 5*log(log(x))

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giac [A]  time = 0.23, size = 35, normalized size = 1.03 \begin {gather*} -5 \, x + 5 \, \log \left (x^{2} e^{x} + 2 \, x e^{\left (x + e^{x}\right )} + 8 \, e^{x}\right ) + 10 \, \log \relax (x) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log(x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x
))+(x^3+8*x)*log(x)),x, algorithm="giac")

[Out]

-5*x + 5*log(x^2*e^x + 2*x*e^(x + e^x) + 8*e^x) + 10*log(x) - 5*log(log(x))

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maple [A]  time = 0.03, size = 28, normalized size = 0.82

method	result	size
risch	\(15 \ln \relax (x )-5 \ln \left (\ln \relax (x )\right )+5 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {x^{2}+8}{2 x}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*exp(x)*x^2+30*x)*ln(x)-10*x)*exp(exp(x))+(20*x^2+80)*ln(x)-5*x^2-40)/(2*x^2*ln(x)*exp(exp(x))+(x^3+8
*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

15*ln(x)-5*ln(ln(x))+5*ln(exp(exp(x))+1/2*(x^2+8)/x)

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maxima [A]  time = 0.68, size = 29, normalized size = 0.85 \begin {gather*} 15 \, \log \relax (x) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{2 \, x}\right ) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*exp(x)*x^2+30*x)*log(x)-10*x)*exp(exp(x))+(20*x^2+80)*log(x)-5*x^2-40)/(2*x^2*log(x)*exp(exp(x
))+(x^3+8*x)*log(x)),x, algorithm="maxima")

[Out]

15*log(x) + 5*log(1/2*(x^2 + 2*x*e^(e^x) + 8)/x) - 5*log(log(x))

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mupad [B]  time = 2.09, size = 28, normalized size = 0.82 \begin {gather*} 15\,\ln \relax (x)-5\,\ln \left (\ln \relax (x)\right )+5\,\ln \left (\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+x^2+8}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*(10*x - log(x)*(30*x + 10*x^2*exp(x))) + 5*x^2 - log(x)*(20*x^2 + 80) + 40)/(log(x)*(8*x + x
^3) + 2*x^2*exp(exp(x))*log(x)),x)

[Out]

15*log(x) - 5*log(log(x)) + 5*log((2*x*exp(exp(x)) + x^2 + 8)/x)

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sympy [A]  time = 0.42, size = 27, normalized size = 0.79 \begin {gather*} 15 \log {\relax (x )} + 5 \log {\left (e^{e^{x}} + \frac {x^{2} + 8}{2 x} \right )} - 5 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*exp(x)*x**2+30*x)*ln(x)-10*x)*exp(exp(x))+(20*x**2+80)*ln(x)-5*x**2-40)/(2*x**2*ln(x)*exp(exp(
x))+(x**3+8*x)*ln(x)),x)

[Out]

15*log(x) + 5*log(exp(exp(x)) + (x**2 + 8)/(2*x)) - 5*log(log(x))

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