3.9.39 \(\int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} (290+58 x-58 \log (5)+(290+116 x-58 \log (5)+e^x (-290 x-58 x^2+58 x \log (5))) \log (x))}{(25 x^2+10 x^3+x^4+(-10 x^2-2 x^3) \log (5)+x^2 \log ^2(5)) \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {58 \left (e^{e^x}+x\right )}{x (-5-x+\log (5)) \log (x)} \]

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Rubi [F]  time = 7.59, 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 {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(290*x + 58*x^2 - 58*x*Log[5] + 58*x^2*Log[x] + E^E^x*(290 + 58*x - 58*Log[5] + (290 + 116*x - 58*Log[5] +
E^x*(-290*x - 58*x^2 + 58*x*Log[5]))*Log[x]))/((25*x^2 + 10*x^3 + x^4 + (-10*x^2 - 2*x^3)*Log[5] + x^2*Log[5]
^2)*Log[x]^2),x]

[Out]

(58*Defer[Int][E^E^x/(x^2*Log[x]^2), x])/(5 - Log[5]) - (58*Defer[Int][E^E^x/(x*Log[x]^2), x])/(5 - Log[5])^2
+ (58*Defer[Int][E^E^x/((5 + x - Log[5])*Log[x]^2), x])/(5 - Log[5])^2 + 58*Defer[Int][1/(x*(5 + x - Log[5])*L
og[x]^2), x] + (58*Defer[Int][E^E^x/(x^2*Log[x]), x])/(5 - Log[5]) - (58*Defer[Int][E^(E^x + x)/(x*Log[x]), x]
)/(5 - Log[5]) + 58*Defer[Int][1/((5 + x - Log[5])^2*Log[x]), x] - (58*Defer[Int][E^E^x/((5 + x - Log[5])^2*Lo
g[x]), x])/(5 - Log[5]) + (58*Defer[Int][E^(E^x + x)/((5 + x - Log[5])*Log[x]), x])/(5 - Log[5])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \left (25+\log ^2(5)\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {58 x^2+x (290-58 \log (5))+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \left (25+\log ^2(5)\right )\right ) \log ^2(x)} \, dx\\ &=\int \frac {58 \left (e^{e^x}+x\right ) (5+x-\log (5))-58 \left (-x^2+e^{e^x+x} x (5+x-\log (5))+e^{e^x} (-5-2 x+\log (5))\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {58 e^{e^x+x}}{x (5+x-\log (5)) \log (x)}+\frac {58 \left (e^{e^x} x+x^2+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )+5 x \left (1-\frac {\log (5)}{5}\right )+2 e^{e^x} x \log (x)+x^2 \log (x)+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)}\right ) \, dx\\ &=-\left (58 \int \frac {e^{e^x+x}}{x (5+x-\log (5)) \log (x)} \, dx\right )+58 \int \frac {e^{e^x} x+x^2+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )+5 x \left (1-\frac {\log (5)}{5}\right )+2 e^{e^x} x \log (x)+x^2 \log (x)+5 e^{e^x} \left (1-\frac {\log (5)}{5}\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=-\left (58 \int \left (-\frac {e^{e^x+x}}{x (-5+\log (5)) \log (x)}+\frac {e^{e^x+x}}{(5+x-\log (5)) (-5+\log (5)) \log (x)}\right ) \, dx\right )+58 \int \frac {\left (e^{e^x}+x\right ) (5+x-\log (5))+\left (x^2+e^{e^x} (5+2 x-\log (5))\right ) \log (x)}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx\\ &=58 \int \left (\frac {x+5 \left (1-\frac {\log (5)}{5}\right )+x \log (x)}{x (5+x-\log (5))^2 \log ^2(x)}+\frac {e^{e^x} \left (x+5 \left (1-\frac {\log (5)}{5}\right )+2 x \log (x)+5 \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)}\right ) \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {x+5 \left (1-\frac {\log (5)}{5}\right )+x \log (x)}{x (5+x-\log (5))^2 \log ^2(x)} \, dx+58 \int \frac {e^{e^x} \left (x+5 \left (1-\frac {\log (5)}{5}\right )+2 x \log (x)+5 \left (1-\frac {\log (5)}{5}\right ) \log (x)\right )}{x^2 (5+x-\log (5))^2 \log ^2(x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \left (\frac {1}{x (5+x-\log (5)) \log ^2(x)}+\frac {1}{(5+x-\log (5))^2 \log (x)}\right ) \, dx+58 \int \left (\frac {e^{e^x}}{x^2 (5+x-\log (5)) \log ^2(x)}+\frac {e^{e^x} (5+2 x-\log (5))}{x^2 (5+x-\log (5))^2 \log (x)}\right ) \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {e^{e^x}}{x^2 (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx+58 \int \frac {e^{e^x} (5+2 x-\log (5))}{x^2 (5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \left (-\frac {e^{e^x}}{x (-5+\log (5))^2 \log ^2(x)}+\frac {e^{e^x}}{(5+x-\log (5)) (-5+\log (5))^2 \log ^2(x)}-\frac {e^{e^x}}{x^2 (-5+\log (5)) \log ^2(x)}\right ) \, dx+58 \int \left (-\frac {e^{e^x}}{x^2 (-5+\log (5)) \log (x)}+\frac {e^{e^x}}{(5+x-\log (5))^2 (-5+\log (5)) \log (x)}\right ) \, dx+58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ &=58 \int \frac {1}{x (5+x-\log (5)) \log ^2(x)} \, dx+58 \int \frac {1}{(5+x-\log (5))^2 \log (x)} \, dx-\frac {58 \int \frac {e^{e^x}}{x \log ^2(x)} \, dx}{(5-\log (5))^2}+\frac {58 \int \frac {e^{e^x}}{(5+x-\log (5)) \log ^2(x)} \, dx}{(5-\log (5))^2}+\frac {58 \int \frac {e^{e^x}}{x^2 \log ^2(x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x}}{x^2 \log (x)} \, dx}{5-\log (5)}-\frac {58 \int \frac {e^{e^x+x}}{x \log (x)} \, dx}{5-\log (5)}-\frac {58 \int \frac {e^{e^x}}{(5+x-\log (5))^2 \log (x)} \, dx}{5-\log (5)}+\frac {58 \int \frac {e^{e^x+x}}{(5+x-\log (5)) \log (x)} \, dx}{5-\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.92, size = 25, normalized size = 1.00 \begin {gather*} -\frac {58 \left (e^{e^x}+x\right )}{x (5+x-\log (5)) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(290*x + 58*x^2 - 58*x*Log[5] + 58*x^2*Log[x] + E^E^x*(290 + 58*x - 58*Log[5] + (290 + 116*x - 58*Lo
g[5] + E^x*(-290*x - 58*x^2 + 58*x*Log[5]))*Log[x]))/((25*x^2 + 10*x^3 + x^4 + (-10*x^2 - 2*x^3)*Log[5] + x^2*
Log[5]^2)*Log[x]^2),x]

[Out]

(-58*(E^E^x + x))/(x*(5 + x - Log[5])*Log[x])

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fricas [A]  time = 0.91, size = 25, normalized size = 1.00 \begin {gather*} -\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x \log \relax (5) + 5 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x)-58*log(5)+58*x+290)*exp(exp(x))+58*
x^2*log(x)-58*x*log(5)+58*x^2+290*x)/(x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algor
ithm="fricas")

[Out]

-58*(x + e^(e^x))/((x^2 - x*log(5) + 5*x)*log(x))

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giac [A]  time = 0.52, size = 39, normalized size = 1.56 \begin {gather*} -\frac {58 \, {\left (x e^{x} + e^{\left (x + e^{x}\right )}\right )}}{x^{2} e^{x} \log \relax (x) - x e^{x} \log \relax (5) \log \relax (x) + 5 \, x e^{x} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x)-58*log(5)+58*x+290)*exp(exp(x))+58*
x^2*log(x)-58*x*log(5)+58*x^2+290*x)/(x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algor
ithm="giac")

[Out]

-58*(x*e^x + e^(x + e^x))/(x^2*e^x*log(x) - x*e^x*log(5)*log(x) + 5*x*e^x*log(x))

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maple [A]  time = 0.06, size = 38, normalized size = 1.52

method	result	size
risch	\(\frac {58}{\left (\ln \relax (5)-x -5\right ) \ln \relax (x )}+\frac {58 \,{\mathrm e}^{{\mathrm e}^{x}}}{x \left (\ln \relax (5)-x -5\right ) \ln \relax (x )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((58*x*ln(5)-58*x^2-290*x)*exp(x)-58*ln(5)+116*x+290)*ln(x)-58*ln(5)+58*x+290)*exp(exp(x))+58*x^2*ln(x)-
58*x*ln(5)+58*x^2+290*x)/(x^2*ln(5)^2+(-2*x^3-10*x^2)*ln(5)+x^4+10*x^3+25*x^2)/ln(x)^2,x,method=_RETURNVERBOSE
)

[Out]

58/(ln(5)-x-5)/ln(x)+58/x/(ln(5)-x-5)/ln(x)*exp(exp(x))

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maxima [A]  time = 0.57, size = 24, normalized size = 0.96 \begin {gather*} -\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x {\left (\log \relax (5) - 5\right )}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x)-58*log(5)+58*x+290)*exp(exp(x))+58*
x^2*log(x)-58*x*log(5)+58*x^2+290*x)/(x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algor
ithm="maxima")

[Out]

-58*(x + e^(e^x))/((x^2 - x*(log(5) - 5))*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {290\,x+58\,x^2\,\ln \relax (x)-58\,x\,\ln \relax (5)+58\,x^2+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (58\,x-58\,\ln \relax (5)+\ln \relax (x)\,\left (116\,x-58\,\ln \relax (5)-{\mathrm {e}}^x\,\left (290\,x-58\,x\,\ln \relax (5)+58\,x^2\right )+290\right )+290\right )}{{\ln \relax (x)}^2\,\left (x^2\,{\ln \relax (5)}^2-\ln \relax (5)\,\left (2\,x^3+10\,x^2\right )+25\,x^2+10\,x^3+x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((290*x + 58*x^2*log(x) - 58*x*log(5) + 58*x^2 + exp(exp(x))*(58*x - 58*log(5) + log(x)*(116*x - 58*log(5)
- exp(x)*(290*x - 58*x*log(5) + 58*x^2) + 290) + 290))/(log(x)^2*(x^2*log(5)^2 - log(5)*(10*x^2 + 2*x^3) + 25*
x^2 + 10*x^3 + x^4)),x)

[Out]

int((290*x + 58*x^2*log(x) - 58*x*log(5) + 58*x^2 + exp(exp(x))*(58*x - 58*log(5) + log(x)*(116*x - 58*log(5)
- exp(x)*(290*x - 58*x*log(5) + 58*x^2) + 290) + 290))/(log(x)^2*(x^2*log(5)^2 - log(5)*(10*x^2 + 2*x^3) + 25*
x^2 + 10*x^3 + x^4)), x)

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sympy [B]  time = 0.47, size = 41, normalized size = 1.64 \begin {gather*} - \frac {58 e^{e^{x}}}{x^{2} \log {\relax (x )} - x \log {\relax (5 )} \log {\relax (x )} + 5 x \log {\relax (x )}} - \frac {58}{\left (x - \log {\relax (5 )} + 5\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((58*x*ln(5)-58*x**2-290*x)*exp(x)-58*ln(5)+116*x+290)*ln(x)-58*ln(5)+58*x+290)*exp(exp(x))+58*x**
2*ln(x)-58*x*ln(5)+58*x**2+290*x)/(x**2*ln(5)**2+(-2*x**3-10*x**2)*ln(5)+x**4+10*x**3+25*x**2)/ln(x)**2,x)

[Out]

-58*exp(exp(x))/(x**2*log(x) - x*log(5)*log(x) + 5*x*log(x)) - 58/((x - log(5) + 5)*log(x))

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