3.102.32 \(\int \frac {-20 x+12 e^{x/5} x+(-60 e^{x/5}+20 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log (\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))+(15 e^{x/5}-5 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log ^2(\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))}{(60 e^{x/5}-20 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log ^2(\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))} \, dx\)

Optimal. Leaf size=34 \[ \frac {x}{4}-\frac {x}{\log \left (\log \left (\log \left (\frac {1}{4} e^2 \left (-3 e^{x/5}+x\right ) \log (2)\right )\right )\right )} \]

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Rubi [F]  time = 7.16, 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 {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20*x + 12*E^(x/5)*x + (-60*E^(x/5) + 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(
2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]] + (15*E^(x/5) - 5*
x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Lo
g[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]]^2)/((60*E^(x/5) - 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2
])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]
]]^2),x]

[Out]

x/4 + Defer[Int][x/((Log[-3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2))*Log[Log[-3*E^(x/5) + x] + 2*(1 + Log[Log[2
]/4]/2)]*Log[Log[Log[-3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2)]]^2), x]/5 + Defer[Int][x^2/((3*E^(x/5) - x)*(L
og[-3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2))*Log[Log[-3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2)]*Log[Log[Log[-
3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2)]]^2), x]/5 + Defer[Int][x/((-3*E^(x/5) + x)*(Log[-3*E^(x/5) + x] + 2*
(1 + Log[Log[2]/4]/2))*Log[Log[-3*E^(x/5) + x] + 2*(1 + Log[Log[2]/4]/2)]*Log[Log[Log[-3*E^(x/5) + x] + 2*(1 +
Log[Log[2]/4]/2)]]^2), x] - 5*Defer[Subst][Defer[Int][Log[Log[Log[-3*E^x + 5*x] + 2*(1 + Log[Log[2]/4]/2)]]^(
-1), x], x, x/5]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{4}+\frac {\left (-5+3 e^{x/5}\right ) x}{5 \left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}-\frac {1}{\log \left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {\left (-5+3 e^{x/5}\right ) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-\int \frac {1}{\log \left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx\\ &=\frac {x}{4}+\frac {1}{5} \int \left (\frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}+\frac {(-5+x) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {(-5+x) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \left (\frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}+\frac {5 x}{\left (-3 e^{x/5}+x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )+\int \frac {x}{\left (-3 e^{x/5}+x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 34, normalized size = 1.00 \begin {gather*} \frac {x}{4}-\frac {x}{\log \left (\log \left (2+\log \left (-3 e^{x/5}+x\right )+\log \left (\frac {\log (2)}{4}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x + 12*E^(x/5)*x + (-60*E^(x/5) + 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(
-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]] + (15*E^(x/5
) - 5*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[
Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]]^2)/((60*E^(x/5) - 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x
*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[
2])/4]]]^2),x]

[Out]

x/4 - x/Log[Log[2 + Log[-3*E^(x/5) + x] + Log[Log[2]/4]]]

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fricas [A]  time = 0.55, size = 52, normalized size = 1.53 \begin {gather*} \frac {x \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \relax (2) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \relax (2)\right )\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \relax (2) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \relax (2)\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(
2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*ex
p(1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/
4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60
*exp(1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(log(-3/4*exp(2)*log(2)*exp(1/5*x)
+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm="fricas
")

[Out]

1/4*(x*log(log(log(1/4*x*e^2*log(2) - 3/4*e^(1/5*x + 2)*log(2)))) - 4*x)/log(log(log(1/4*x*e^2*log(2) - 3/4*e^
(1/5*x + 2)*log(2))))

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giac [A]  time = 0.83, size = 50, normalized size = 1.47 \begin {gather*} \frac {x \log \left (\log \left (-2 \, \log \relax (2) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \relax (2)\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \relax (2) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \relax (2)\right ) + 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(
2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*ex
p(1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/
4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60
*exp(1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(log(-3/4*exp(2)*log(2)*exp(1/5*x)
+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm="giac")

[Out]

1/4*(x*log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2)) - 4*x)/log(log(-2*log(2) + log(x - 3*e^(1/
5*x)) + log(log(2)) + 2))

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maple [A]  time = 0.07, size = 31, normalized size = 0.91

method	result	size
risch	\(\frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (-\frac {3 \ln \relax (2) {\mathrm e}^{2+\frac {x}{5}}}{4}+\frac {x \,{\mathrm e}^{2} \ln \relax (2)}{4}\right )\right )\right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*exp(1/5*x)-5*x)*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x
)+1/4*x*exp(2)*ln(2)))*ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))^2+(-60*exp(1/5*x)+20*x)*ln(
-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln
(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))+12*x*exp(1/5*x)-20*x)/(60*exp(1/5*x)-20*x)/ln(-3/4*exp(
2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))/ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))/ln(ln(ln(-3/4*
exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x-x/ln(ln(ln(-3/4*ln(2)*exp(2+1/5*x)+1/4*x*exp(2)*ln(2))))

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maxima [A]  time = 0.64, size = 50, normalized size = 1.47 \begin {gather*} \frac {x \log \left (\log \left (-2 \, \log \relax (2) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \relax (2)\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \relax (2) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \relax (2)\right ) + 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(
2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*ex
p(1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/
4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60
*exp(1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(log(-3/4*exp(2)*log(2)*exp(1/5*x)
+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm="maxima
")

[Out]

1/4*(x*log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2)) - 4*x)/log(log(-2*log(2) + log(x - 3*e^(1/
5*x)) + log(log(2)) + 2))

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mupad [B]  time = 14.09, size = 30, normalized size = 0.88 \begin {gather*} \frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^2\,\ln \relax (2)}{4}-\frac {3\,{\mathrm {e}}^2\,\ln \relax (2)\,{\left ({\mathrm {e}}^x\right )}^{1/5}}{4}\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x - 12*x*exp(x/5) - log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4))*log(log(log((x*exp(2)
*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)))*log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)*(20*x - 6
0*exp(x/5)) + log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4))*log(log(log((x*exp(2)*log(2))/4 - (
3*exp(x/5)*exp(2)*log(2))/4)))^2*log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)*(5*x - 15*exp(x/5)))/
(log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4))*log(log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*ex
p(2)*log(2))/4)))^2*log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)*(20*x - 60*exp(x/5))),x)

[Out]

x/4 - x/log(log(log((x*exp(2)*log(2))/4 - (3*exp(2)*log(2)*exp(x)^(1/5))/4)))

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sympy [A]  time = 15.50, size = 34, normalized size = 1.00 \begin {gather*} \frac {x}{4} - \frac {x}{\log {\left (\log {\left (\log {\left (\frac {x e^{2} \log {\relax (2 )}}{4} - \frac {3 e^{2} e^{\frac {x}{5}} \log {\relax (2 )}}{4} \right )} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*exp(1/5*x)-5*x)*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp
(1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))**2+(-60*exp(1/5*x)+20
*x)*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))
)*ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))+12*x*exp(1/5*x)-20*x)/(60*exp(1/5*x)-20*x)/ln(-3
/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))/ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))/ln(ln(l
n(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))**2,x)

[Out]

x/4 - x/log(log(log(x*exp(2)*log(2)/4 - 3*exp(2)*exp(x/5)*log(2)/4)))

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