3.55.7 \(\int \frac {1+8 x+e^{e^{x^2}} (-4+(-4-8 e^{x^2} x^2) \log (x))}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} (-200 x^2-800 x^3) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \log (2)+\frac {1}{25 \left (-x+4 x \left (-x+e^{e^{x^2}} \log (x)\right )\right )} \]

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Rubi [F]  time = 2.38, 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 {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + 8*x + E^E^x^2*(-4 + (-4 - 8*E^x^2*x^2)*Log[x]))/(25*x^2 + 200*x^3 + 400*x^4 + E^E^x^2*(-200*x^2 - 800
*x^3)*Log[x] + 400*E^(2*E^x^2)*x^2*Log[x]^2),x]

[Out]

Defer[Int][1/(x^2*(1 + 4*x - 4*E^E^x^2*Log[x])^2), x]/25 + (8*Defer[Int][1/(x*(1 + 4*x - 4*E^E^x^2*Log[x])^2),
x])/25 - (4*Defer[Int][E^E^x^2/(x^2*(-1 - 4*x + 4*E^E^x^2*Log[x])^2), x])/25 - (8*Defer[Int][(E^(E^x^2 + x^2)
*Log[x])/(-1 - 4*x + 4*E^E^x^2*Log[x])^2, x])/25 - (4*Defer[Int][(E^E^x^2*Log[x])/(x^2*(-1 - 4*x + 4*E^E^x^2*L
og[x])^2), x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-4 e^{e^{x^2}}+8 x-4 e^{e^{x^2}} \left (1+2 e^{x^2} x^2\right ) \log (x)}{25 x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ &=\frac {1}{25} \int \frac {1-4 e^{e^{x^2}}+8 x-4 e^{e^{x^2}} \left (1+2 e^{x^2} x^2\right ) \log (x)}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ &=\frac {1}{25} \int \left (\frac {1}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2}+\frac {8}{x \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2}-\frac {4 e^{e^{x^2}}}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}-\frac {8 e^{e^{x^2}+x^2} \log (x)}{\left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}-\frac {4 e^{e^{x^2}} \log (x)}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{25} \int \frac {1}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {4}{25} \int \frac {e^{e^{x^2}}}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {4}{25} \int \frac {e^{e^{x^2}} \log (x)}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx+\frac {8}{25} \int \frac {1}{x \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {8}{25} \int \frac {e^{e^{x^2}+x^2} \log (x)}{\left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.61, size = 25, normalized size = 0.83 \begin {gather*} \frac {1}{25 x \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 8*x + E^E^x^2*(-4 + (-4 - 8*E^x^2*x^2)*Log[x]))/(25*x^2 + 200*x^3 + 400*x^4 + E^E^x^2*(-200*x^2
- 800*x^3)*Log[x] + 400*E^(2*E^x^2)*x^2*Log[x]^2),x]

[Out]

1/(25*x*(-1 - 4*x + 4*E^E^x^2*Log[x]))

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fricas [A]  time = 0.60, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2*exp(x^2)-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*log(x)^2*exp(exp(x^2))^2+(-800*x^3-200
*x^2)*log(x)*exp(exp(x^2))+400*x^4+200*x^3+25*x^2),x, algorithm="fricas")

[Out]

1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)

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giac [A]  time = 0.69, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2*exp(x^2)-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*log(x)^2*exp(exp(x^2))^2+(-800*x^3-200
*x^2)*log(x)*exp(exp(x^2))+400*x^4+200*x^3+25*x^2),x, algorithm="giac")

[Out]

1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.73

method	result	size
risch	\(-\frac {1}{25 x \left (-4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}} \ln \relax (x )+4 x +1\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x^2*exp(x^2)-4)*ln(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*ln(x)^2*exp(exp(x^2))^2+(-800*x^3-200*x^2)*ln
(x)*exp(exp(x^2))+400*x^4+200*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/25/x/(-4*exp(exp(x^2))*ln(x)+4*x+1)

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maxima [A]  time = 0.41, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x^2*exp(x^2)-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*log(x)^2*exp(exp(x^2))^2+(-800*x^3-200
*x^2)*log(x)*exp(exp(x^2))+400*x^4+200*x^3+25*x^2),x, algorithm="maxima")

[Out]

1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)

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mupad [B]  time = 3.76, size = 86, normalized size = 2.87 \begin {gather*} -\frac {\frac {4\,x}{25}+\ln \relax (x)\,\left (\frac {2\,x^2\,{\mathrm {e}}^{x^2}}{25}-\frac {4\,x}{25}+\frac {8\,x^3\,{\mathrm {e}}^{x^2}}{25}\right )+\frac {1}{25}}{\left (4\,x-4\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \relax (x)+1\right )\,\left (x-4\,x^2\,\ln \relax (x)+4\,x^2+2\,x^3\,{\mathrm {e}}^{x^2}\,\ln \relax (x)+8\,x^4\,{\mathrm {e}}^{x^2}\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - exp(exp(x^2))*(log(x)*(8*x^2*exp(x^2) + 4) + 4) + 1)/(25*x^2 + 200*x^3 + 400*x^4 + 400*x^2*exp(2*ex
p(x^2))*log(x)^2 - exp(exp(x^2))*log(x)*(200*x^2 + 800*x^3)),x)

[Out]

-((4*x)/25 + log(x)*((2*x^2*exp(x^2))/25 - (4*x)/25 + (8*x^3*exp(x^2))/25) + 1/25)/((4*x - 4*exp(exp(x^2))*log
(x) + 1)*(x - 4*x^2*log(x) + 4*x^2 + 2*x^3*exp(x^2)*log(x) + 8*x^4*exp(x^2)*log(x)))

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sympy [A]  time = 0.35, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{- 100 x^{2} + 100 x e^{e^{x^{2}}} \log {\relax (x )} - 25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x**2*exp(x**2)-4)*ln(x)-4)*exp(exp(x**2))+8*x+1)/(400*x**2*ln(x)**2*exp(exp(x**2))**2+(-800*x*
*3-200*x**2)*ln(x)*exp(exp(x**2))+400*x**4+200*x**3+25*x**2),x)

[Out]

1/(-100*x**2 + 100*x*exp(exp(x**2))*log(x) - 25*x)

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