3.96.66 \(\int \frac {e^{\frac {-x+\log (x)}{(5+e^x) \log (\log (25 x^2))}} (10 x \log (2)+2 e^x x \log (2)+(-10 \log (2)-2 e^x \log (2)) \log (x)+((5-5 x) \log (2)+e^x (1-x+x^2) \log (2)-e^x x \log (2) \log (x)) \log (25 x^2) \log (\log (25 x^2)))}{(25 x+10 e^x x+e^{2 x} x) \log (25 x^2) \log ^2(\log (25 x^2))} \, dx\)

Optimal. Leaf size=30 \[ \left (-5+e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}}\right ) \log (2) \]

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Rubi [F]  time = 50.04, 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^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2*E^x*x*Log[2] + (-10*Log[2] - 2*E^x*Log[2]
)*Log[x] + ((5 - 5*x)*Log[2] + E^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2]]))/
((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]

[Out]

2*Log[2]*Defer[Int][E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))/((5 + E^x)*Log[25*x^2]*Log[Log[25*x^2]]^2),
x] - 2*Log[2]*Defer[Int][(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*Log[x])/((5 + E^x)*x*Log[25*x^2]*Log
[Log[25*x^2]]^2), x] - Log[2]*Defer[Int][E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))/((5 + E^x)*Log[Log[25*
x^2]]), x] + Log[2]*Defer[Int][E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))/((5 + E^x)*x*Log[Log[25*x^2]]),
x] - 5*Log[2]*Defer[Int][(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*x)/((5 + E^x)^2*Log[Log[25*x^2]]), x]
+ Log[2]*Defer[Int][(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*x)/((5 + E^x)*Log[Log[25*x^2]]), x] + 5*L
og[2]*Defer[Int][(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*Log[x])/((5 + E^x)^2*Log[Log[25*x^2]]), x] -
Log[2]*Defer[Int][(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*Log[x])/((5 + E^x)*Log[Log[25*x^2]]), x]

Rubi steps

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

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Mathematica [F]  time = 1.96, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2*E^x*x*Log[2] + (-10*Log[2] - 2*E^x*
Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^
2]]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]

[Out]

Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2*E^x*x*Log[2] + (-10*Log[2] - 2*E^x*
Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^
2]]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2), x]

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fricas [A]  time = 0.54, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (-\frac {x - \log \relax (x)}{{\left (e^{x} + 5\right )} \log \left (2 \, \log \relax (5) + 2 \, \log \relax (x)\right )}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2))*log(25*x^2)*log(log(25*x^2))+(-2*
exp(x)*log(2)-10*log(2))*log(x)+2*x*log(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*
exp(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="fricas")

[Out]

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

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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(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2))*log(25*x^2)*log(log(25*x^2))+(-2*
exp(x)*log(2)-10*log(2))*log(x)+2*x*log(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*
exp(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="giac")

[Out]

undef

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maple [C]  time = 0.30, size = 58, normalized size = 1.93

method	result	size
risch	\(\ln \relax (2) {\mathrm e}^{\frac {\ln \relax (x )-x}{\left ({\mathrm e}^{x}+5\right ) \ln \left (2 \ln \relax (5)+2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(2)*exp(x)*ln(x)+(x^2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))*ln(25*x^2)*ln(ln(25*x^2))+(-2*exp(x)*ln(2)-
10*ln(2))*ln(x)+2*x*ln(2)*exp(x)+10*x*ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x^2)))/(x*exp(x)^2+10*exp(x)*x+
25*x)/ln(25*x^2)/ln(ln(25*x^2))^2,x,method=_RETURNVERBOSE)

[Out]

ln(2)*exp((ln(x)-x)/(exp(x)+5)/ln(2*ln(5)+2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2))*log(25*x^2)*log(log(25*x^2))+(-2*
exp(x)*log(2)-10*log(2))*log(x)+2*x*log(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*
exp(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 9.03, size = 53, normalized size = 1.77 \begin {gather*} x^{\frac {1}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {x}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x - log(x))/(log(log(25*x^2))*(exp(x) + 5)))*(log(x)*(10*log(2) + 2*exp(x)*log(2)) - 10*x*log(2) -
2*x*exp(x)*log(2) + log(log(25*x^2))*log(25*x^2)*(log(2)*(5*x - 5) - exp(x)*log(2)*(x^2 - x + 1) + x*exp(x)*l
og(2)*log(x))))/(log(log(25*x^2))^2*log(25*x^2)*(25*x + x*exp(2*x) + 10*x*exp(x))),x)

[Out]

x^(1/(5*log(log(25*x^2)) + log(log(25*x^2))*exp(x)))*exp(-x/(5*log(log(25*x^2)) + log(log(25*x^2))*exp(x)))*lo
g(2)

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sympy [A]  time = 56.47, size = 24, normalized size = 0.80 \begin {gather*} e^{\frac {- x + \log {\relax (x )}}{\left (e^{x} + 5\right ) \log {\left (2 \log {\relax (x )} + \log {\left (25 \right )} \right )}}} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(2)*exp(x)*ln(x)+(x**2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))*ln(25*x**2)*ln(ln(25*x**2))+(-2*exp(
x)*ln(2)-10*ln(2))*ln(x)+2*x*ln(2)*exp(x)+10*x*ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x**2)))/(x*exp(x)**2+1
0*exp(x)*x+25*x)/ln(25*x**2)/ln(ln(25*x**2))**2,x)

[Out]

exp((-x + log(x))/((exp(x) + 5)*log(2*log(x) + log(25))))*log(2)

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