3.4.98 \(\int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+(-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+(12 e^{e^{2 x}} x^2-6 x \log (4)) \log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x})) \log (-5 x+\log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x}))}{(-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+(2 e^{e^{2 x}} x^2-x \log (4)) \log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x})) \log (-5 x+\log (\frac {e^{-e^{2 x}} (2 e^{e^{2 x}} x-\log (4))}{x}))} \, dx\)

Optimal. Leaf size=31 \[ x+5 \left (x+\log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right )\right ) \]

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Rubi [F]  time = 18.93, 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 {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + (-60*E^E^(2*x)*x^3 + 30*x^2*Log[4] + (12*E^
E^(2*x)*x^2 - 6*x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])
/(E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^2 - x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4
])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]

[Out]

6*x + 25*Defer[Int][1/((5*x - Log[2 - Log[4]/(E^E^(2*x)*x)])*Log[-5*x + Log[2 - Log[4]/(E^E^(2*x)*x)]]), x] -
10*Log[4]*Defer[Int][E^(2*x)/((2*E^E^(2*x)*x - Log[4])*(5*x - Log[2 - Log[4]/(E^E^(2*x)*x)])*Log[-5*x + Log[2
- Log[4]/(E^E^(2*x)*x)]]), x] - 5*Log[4]*Defer[Int][1/(x*(2*E^E^(2*x)*x - Log[4])*(5*x - Log[2 - Log[4]/(E^E^(
2*x)*x)])*Log[-5*x + Log[2 - Log[4]/(E^E^(2*x)*x)]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 e^{e^{2 x}} x^2-10 e^{2 x} x \log (4)-5 (1+5 x) \log (4)+6 x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\\ &=\int \left (-\frac {10 e^{2 x} \log (4)}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}+\frac {50 e^{e^{2 x}} x^2-5 \log (4)-25 x \log (4)+60 e^{e^{2 x}} x^3 \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-30 x^2 \log (4) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-12 e^{e^{2 x}} x^2 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )+6 x \log (4) \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx\\ &=-\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \frac {50 e^{e^{2 x}} x^2-5 \log (4)-25 x \log (4)+60 e^{e^{2 x}} x^3 \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-30 x^2 \log (4) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-12 e^{e^{2 x}} x^2 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )+6 x \log (4) \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\\ &=-\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \frac {50 e^{e^{2 x}} x^2-5 (1+5 x) \log (4)+6 x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\\ &=-\left ((10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )+\int \left (-\frac {5 \log (4)}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}+\frac {25+30 x \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-6 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx\\ &=-\left ((5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx+\int \frac {25+30 x \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )-6 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\\ &=-\left ((5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\right )-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx+\int \left (6+\frac {25}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )}\right ) \, dx\\ &=6 x+25 \int \frac {1}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx-(5 \log (4)) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx-(10 \log (4)) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 31, normalized size = 1.00 \begin {gather*} 6 x+5 \log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + (-60*E^E^(2*x)*x^3 + 30*x^2*Log[4] +
(12*E^E^(2*x)*x^2 - 6*x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - L
og[4])/(E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^2 - x*Log[4])*Log[(2*E^E^(2*x)*x -
Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]

[Out]

6*x + 5*Log[Log[-5*x + Log[2 - Log[4]/(E^E^(2*x)*x)]]]

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fricas [A]  time = 0.80, size = 37, normalized size = 1.19 \begin {gather*} 6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \left (\frac {2 \, {\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \relax (2)\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}}{x}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 0.71, size = 39, normalized size = 1.26 \begin {gather*} 6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \relax (2) + \log \left ({\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \relax (2)\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}\right ) - \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="giac")

[Out]

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

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maple [C]  time = 0.62, size = 317, normalized size = 10.23

method	result	size
risch	\(6 x +5 \ln \left (\ln \left (\ln \relax (2)+i \pi -\ln \relax (x )-\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}}\right )+\ln \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )\right )+\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )\right )+\mathrm {csgn}\left (i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right )\right )\right )}{2}+i \pi \mathrm {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )^{2} \left (\mathrm {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )-1\right )-5 x \right )\right )\)	\(317\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x^2*exp(exp(x)^2)-12*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-60*x^3*exp(exp(x)^2)+6
0*x^2*ln(2))*ln(ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*ln(2)*exp(x)^2+
2*(25*x+5)*ln(2))/((2*x^2*exp(exp(x)^2)-2*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-10*x^3*exp(
exp(x)^2)+10*x^2*ln(2))/ln(ln((2*x*exp(exp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-5*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.74, size = 37, normalized size = 1.19 \begin {gather*} 6 \, x + 5 \, \log \left (\log \left (-5 \, x - e^{\left (2 \, x\right )} + \log \relax (2) + \log \left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \relax (2)\right ) - \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-60*x^3*exp(ex
p(x)^2)+60*x^2*log(2))*log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*lo
g(2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(
x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algor
ithm="maxima")

[Out]

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

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mupad [B]  time = 1.68, size = 32, normalized size = 1.03 \begin {gather*} 6\,x+5\,\ln \left (\ln \left (\ln \left (\frac {2\,x-2\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,\ln \relax (2)}{x}\right )-5\,x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(25*x + 5) - log(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x) - 5*x)*(log(-(exp(-exp
(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x)*(12*x*log(2) - 12*x^2*exp(exp(2*x))) - 60*x^2*log(2) + 60*x^3*exp(ex
p(2*x))) - 50*x^2*exp(exp(2*x)) + 20*x*exp(2*x)*log(2))/(log(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x)
)))/x) - 5*x)*(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x)*(2*x*log(2) - 2*x^2*exp(exp(2*x))) - 10
*x^2*log(2) + 10*x^3*exp(exp(2*x)))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x**2*exp(exp(x)**2)-12*x*ln(2))*ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-60*x**3*exp(
exp(x)**2)+60*x**2*ln(2))*ln(ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x)-50*x**2*exp(exp(x)**2)+20*
x*ln(2)*exp(x)**2+2*(25*x+5)*ln(2))/((2*x**2*exp(exp(x)**2)-2*x*ln(2))*ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(e
xp(x)**2))-10*x**3*exp(exp(x)**2)+10*x**2*ln(2))/ln(ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x),x)

[Out]

Timed out

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