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

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

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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 = {} $$\begin{array}{c}\int \frac{-50x^4-20x^3\log (\log (4))-2x^2{\log }^2(\log (4))+\exp \left(\frac{-e^5+x-10x^3-2x^2\log (\log (4))}{10x^2+2x\log (\log (4))}\right)(10e^{5}x-5x^2-50x^4+(e^5-20x^3)\log (\log (4))-2x^2{\log }^2(\log (4)))}{50x^4+20x^3\log (\log (4))+2x^2{\log }^2(\log (4))}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-50x^4-20x^3\log (\log (4))-2x^2{\log }^2(\log (4))+\exp \left(\frac{-e^5+x-10x^3-2x^2\log (\log (4))}{10x^2+2x\log (\log (4))}\right)(10e^{5}x-5x^2-50x^4+(e^5-20x^3)\log (\log (4))-2x^2{\log }^2(\log (4)))}{x^2(50x^2+20x\log (\log (4))+2{\log }^2(\log (4)))}dx\\ & =\int \frac{-50x^4-20x^3\log (\log (4))-2x^2{\log }^2(\log (4))+\exp \left(\frac{-e^5+x-10x^3-2x^2\log (\log (4))}{10x^2+2x\log (\log (4))}\right)(10e^{5}x-5x^2-50x^4+(e^5-20x^3)\log (\log (4))-2x^2{\log }^2(\log (4)))}{2x^2(5x+\log (\log (4)){)}^2}dx\\ & =\frac{1}{2}\int \frac{-50x^4-20x^3\log (\log (4))-2x^2{\log }^2(\log (4))+\exp \left(\frac{-e^5+x-10x^3-2x^2\log (\log (4))}{10x^2+2x\log (\log (4))}\right)(10e^{5}x-5x^2-50x^4+(e^5-20x^3)\log (\log (4))-2x^2{\log }^2(\log (4)))}{x^2(5x+\log (\log (4)){)}^2}dx\\ & =\frac{1}{2}\int (-2+\frac{e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)(10e^{5}x-50x^4+e^5\log (\log (4))-20x^3\log (\log (4))-x^2(5+2{\log }^2(\log (4))))}{x^2(5x+\log (\log (4)){)}^2})dx\\ & =-x+\frac{1}{2}\int \frac{e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)(10e^{5}x-50x^4+e^5\log (\log (4))-20x^3\log (\log (4))-x^2(5+2{\log }^2(\log (4))))}{x^2(5x+\log (\log (4)){)}^2}dx\\ & =-x+\frac{1}{2}\int (-2e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)+\frac{\exp (5+\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}){\log }^{-\frac{x}{5x+\log (\log (4))}}(4)}{x^2\log (\log (4))}-\frac{5e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)(5e^5+\log (\log (4)))}{\log (\log (4))(5x+\log (\log (4)){)}^2})dx\\ & =-x-\frac{1}{2}\left(5(1+\frac{5e^5}{\log (\log (4))})\right)\int \frac{e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)}{(5x+\log (\log (4)){)}^2}dx+\frac{\int \frac{\exp (5+\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}){\log }^{-\frac{x}{5x+\log (\log (4))}}(4)}{x^2}dx}{2\log (\log (4))}-\int e^{\frac{-e^5+x-10x^3}{2x(5x+\log (\log (4)))}}{\log }^{-\frac{x}{5x+\log (\log (4))}}(4)dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}-x+e^{(-\frac{10x^3+2x^2\log (2\log (2))-x+e^5}{2(5x^2+x\log (2\log (2)))})}\end{array}$$

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 $$\begin{array}{c}\mathit{undef}\end{array}$$

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	$-x+{\mathrm{e}}^{-\frac{2x^2\ln (2)+2x^2\ln (\ln (2))+10x^3+{\mathrm{e}}^5-x}{2x(\ln (2)+\ln (\ln (2))+5x)}}$	$48$
norman	$\frac{(\frac{\ln (2{)}^2}{5}+\frac{2\ln (2)\ln (\ln (2))}{5}+\frac{\ln {(\ln (2))}^2}{5})x+(\ln (2)+\ln (\ln (2)))x{\mathrm{e}}^{\frac{-2x^2\ln (2\ln (2))-{\mathrm{e}}^5-10x^3+x}{2x\ln (2\ln (2))+10x^2}}-5x^3+5x^2{\mathrm{e}}^{\frac{-2x^2\ln (2\ln (2))-{\mathrm{e}}^5-10x^3+x}{2x\ln (2\ln (2))+10x^2}}}{x(\ln (2\ln (2))+5x)}$	$136$

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 $$\begin{array}{c}-\frac{2}{5}(\frac{\log (2\log (2))}{5x+\log (2\log (2))}+\log (5x+\log (2\log (2))))\log (2\log (2))+\frac{2}{5}\log (5x+\log (2\log (2)))\log (2\log (2))-x+\frac{2\log {(2\log (2))}^2}{5(5x+\log (2\log (2)))}+e^{(-x+\frac{5e^5}{2(5x(\log (2)+\log (\log (2)))+\log (2{)}^2+2\log (2)\log (\log (2))+\log {(\log (2))}^2)}+\frac{1}{2(5x+\log (2)+\log (\log (2)))}-\frac{e^5}{2x(\log (2)+\log (\log (2)))})}\end{array}$$

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 $$\begin{array}{c}\int -\frac{20x^3\ln (2\ln (2))+2x^2{\ln (2\ln (2))}^2+50x^4+{\mathrm{e}}^{-\frac{10x^3+2\ln (2\ln (2))x^2-x+{\mathrm{e}}^5}{10x^2+2\ln (2\ln (2))x}}(2x^2{\ln (2\ln (2))}^2-\ln (2\ln (2))({\mathrm{e}}^5-20x^3)-10x{\mathrm{e}}^5+5x^2+50x^4)}{50x^4+20\ln (2\ln (2))x^3+2{\ln (2\ln (2))}^2{x}^2}dx\end{array}$$

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 $$\begin{array}{c}-x+e^{\frac{-10x^3-2x^2\log (2\log (2))+x-e^5}{10x^2+2x\log (2\log (2))}}\end{array}$$

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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