3.41.81 \(\int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} (10 e^5 x-5 x^2-50 x^4+(e^5-20 x^3) \log (\log (4))-2 x^2 \log ^2(\log (4)))}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx\)

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

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Rubi [F]  time = 12.65, 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 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-50*x^4 - 20*x^3*Log[Log[4]] - 2*x^2*Log[Log[4]]^2 + E^((-E^5 + x - 10*x^3 - 2*x^2*Log[Log[4]])/(10*x^2 +
2*x*Log[Log[4]]))*(10*E^5*x - 5*x^2 - 50*x^4 + (E^5 - 20*x^3)*Log[Log[4]] - 2*x^2*Log[Log[4]]^2))/(50*x^4 + 2
0*x^3*Log[Log[4]] + 2*x^2*Log[Log[4]]^2),x]

[Out]

-x - Defer[Int][E^((-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/Log[4]^(x/(5*x + Log[Log[4]])), x] + Defer[I
nt][E^(5 + (-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/(x^2*Log[4]^(x/(5*x + Log[Log[4]]))), x]/(2*Log[Log[
4]]) - (5*(1 + (5*E^5)/Log[Log[4]])*Defer[Int][E^((-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/(Log[4]^(x/(5
*x + Log[Log[4]]))*(5*x + Log[Log[4]])^2), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{x^2 \left (50 x^2+20 x \log (\log (4))+2 \log ^2(\log (4))\right )} \, dx\\ &=\int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{2 x^2 (5 x+\log (\log (4)))^2} \, dx\\ &=\frac {1}{2} \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{x^2 (5 x+\log (\log (4)))^2} \, dx\\ &=\frac {1}{2} \int \left (-2+\frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (10 e^5 x-50 x^4+e^5 \log (\log (4))-20 x^3 \log (\log (4))-x^2 \left (5+2 \log ^2(\log (4))\right )\right )}{x^2 (5 x+\log (\log (4)))^2}\right ) \, dx\\ &=-x+\frac {1}{2} \int \frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (10 e^5 x-50 x^4+e^5 \log (\log (4))-20 x^3 \log (\log (4))-x^2 \left (5+2 \log ^2(\log (4))\right )\right )}{x^2 (5 x+\log (\log (4)))^2} \, dx\\ &=-x+\frac {1}{2} \int \left (-2 e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)+\frac {\exp \left (5+\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}\right ) \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{x^2 \log (\log (4))}-\frac {5 e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (5 e^5+\log (\log (4))\right )}{\log (\log (4)) (5 x+\log (\log (4)))^2}\right ) \, dx\\ &=-x-\frac {1}{2} \left (5 \left (1+\frac {5 e^5}{\log (\log (4))}\right )\right ) \int \frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{(5 x+\log (\log (4)))^2} \, dx+\frac {\int \frac {\exp \left (5+\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}\right ) \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{x^2} \, dx}{2 \log (\log (4))}-\int e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 47, normalized size = 1.38 \begin {gather*} \frac {1}{2} \left (2 e^{-\frac {e^5+x \left (-1+10 x^2+2 x \log (\log (4))\right )}{2 x (5 x+\log (\log (4)))}}-2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x^4 - 20*x^3*Log[Log[4]] - 2*x^2*Log[Log[4]]^2 + E^((-E^5 + x - 10*x^3 - 2*x^2*Log[Log[4]])/(10
*x^2 + 2*x*Log[Log[4]]))*(10*E^5*x - 5*x^2 - 50*x^4 + (E^5 - 20*x^3)*Log[Log[4]] - 2*x^2*Log[Log[4]]^2))/(50*x
^4 + 20*x^3*Log[Log[4]] + 2*x^2*Log[Log[4]]^2),x]

[Out]

(2/E^((E^5 + x*(-1 + 10*x^2 + 2*x*Log[Log[4]]))/(2*x*(5*x + Log[Log[4]]))) - 2*x)/2

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fricas [A]  time = 0.66, size = 43, normalized size = 1.26 \begin {gather*} -x + e^{\left (-\frac {10 \, x^{3} + 2 \, x^{2} \log \left (2 \, \log \relax (2)\right ) - x + e^{5}}{2 \, {\left (5 \, x^{2} + x \log \left (2 \, \log \relax (2)\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="fricas")

[Out]

-x + e^(-1/2*(10*x^3 + 2*x^2*log(2*log(2)) - x + e^5)/(5*x^2 + x*log(2*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(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.33, size = 48, normalized size = 1.41

method	result	size
risch	\(-x +{\mathrm e}^{-\frac {2 x^{2} \ln \relax (2)+2 x^{2} \ln \left (\ln \relax (2)\right )+10 x^{3}+{\mathrm e}^{5}-x}{2 x \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )+5 x \right )}}\)	\(48\)
norman	\(\frac {\left (\frac {\ln \relax (2)^{2}}{5}+\frac {2 \ln \relax (2) \ln \left (\ln \relax (2)\right )}{5}+\frac {\ln \left (\ln \relax (2)\right )^{2}}{5}\right ) x +\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) x \,{\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \relax (2)\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \relax (2)\right )+10 x^{2}}}-5 x^{3}+5 x^{2} {\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \relax (2)\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \relax (2)\right )+10 x^{2}}}}{x \left (\ln \left (2 \ln \relax (2)\right )+5 x \right )}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2*ln(2*ln(2))^2+(exp(5)-20*x^3)*ln(2*ln(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*ln(2*ln(2))-exp(5
)-10*x^3+x)/(2*x*ln(2*ln(2))+10*x^2))-2*x^2*ln(2*ln(2))^2-20*x^3*ln(2*ln(2))-50*x^4)/(2*x^2*ln(2*ln(2))^2+20*x
^3*ln(2*ln(2))+50*x^4),x,method=_RETURNVERBOSE)

[Out]

-x+exp(-1/2*(2*x^2*ln(2)+2*x^2*ln(ln(2))+10*x^3+exp(5)-x)/x/(ln(2)+ln(ln(2))+5*x))

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maxima [B]  time = 0.71, size = 141, normalized size = 4.15 \begin {gather*} -\frac {2}{5} \, {\left (\frac {\log \left (2 \, \log \relax (2)\right )}{5 \, x + \log \left (2 \, \log \relax (2)\right )} + \log \left (5 \, x + \log \left (2 \, \log \relax (2)\right )\right )\right )} \log \left (2 \, \log \relax (2)\right ) + \frac {2}{5} \, \log \left (5 \, x + \log \left (2 \, \log \relax (2)\right )\right ) \log \left (2 \, \log \relax (2)\right ) - x + \frac {2 \, \log \left (2 \, \log \relax (2)\right )^{2}}{5 \, {\left (5 \, x + \log \left (2 \, \log \relax (2)\right )\right )}} + e^{\left (-x + \frac {5 \, e^{5}}{2 \, {\left (5 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + \log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )}} + \frac {1}{2 \, {\left (5 \, x + \log \relax (2) + \log \left (\log \relax (2)\right )\right )}} - \frac {e^{5}}{2 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {20\,x^3\,\ln \left (2\,\ln \relax (2)\right )+2\,x^2\,{\ln \left (2\,\ln \relax (2)\right )}^2+50\,x^4+{\mathrm {e}}^{-\frac {10\,x^3+2\,\ln \left (2\,\ln \relax (2)\right )\,x^2-x+{\mathrm {e}}^5}{10\,x^2+2\,\ln \left (2\,\ln \relax (2)\right )\,x}}\,\left (2\,x^2\,{\ln \left (2\,\ln \relax (2)\right )}^2-\ln \left (2\,\ln \relax (2)\right )\,\left ({\mathrm {e}}^5-20\,x^3\right )-10\,x\,{\mathrm {e}}^5+5\,x^2+50\,x^4\right )}{50\,x^4+20\,\ln \left (2\,\ln \relax (2)\right )\,x^3+2\,{\ln \left (2\,\ln \relax (2)\right )}^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4 + exp(-(exp(5) - x + 2*x^2*log(2*log(2)) + 10*x^3)
/(2*x*log(2*log(2)) + 10*x^2))*(2*x^2*log(2*log(2))^2 - log(2*log(2))*(exp(5) - 20*x^3) - 10*x*exp(5) + 5*x^2
+ 50*x^4))/(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4),x)

[Out]

int(-(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4 + exp(-(exp(5) - x + 2*x^2*log(2*log(2)) + 10*x^3)
/(2*x*log(2*log(2)) + 10*x^2))*(2*x^2*log(2*log(2))^2 - log(2*log(2))*(exp(5) - 20*x^3) - 10*x*exp(5) + 5*x^2
+ 50*x^4))/(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4), x)

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sympy [A]  time = 1.61, size = 39, normalized size = 1.15 \begin {gather*} - x + e^{\frac {- 10 x^{3} - 2 x^{2} \log {\left (2 \log {\relax (2 )} \right )} + x - e^{5}}{10 x^{2} + 2 x \log {\left (2 \log {\relax (2 )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2*ln(2*ln(2))**2+(exp(5)-20*x**3)*ln(2*ln(2))+10*x*exp(5)-50*x**4-5*x**2)*exp((-2*x**2*ln(2*
ln(2))-exp(5)-10*x**3+x)/(2*x*ln(2*ln(2))+10*x**2))-2*x**2*ln(2*ln(2))**2-20*x**3*ln(2*ln(2))-50*x**4)/(2*x**2
*ln(2*ln(2))**2+20*x**3*ln(2*ln(2))+50*x**4),x)

[Out]

-x + exp((-10*x**3 - 2*x**2*log(2*log(2)) + x - exp(5))/(10*x**2 + 2*x*log(2*log(2))))

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