3.87.70 \(\int \frac {2 x+4 x^2+2 x^3+(2 x+6 x^2+4 x^3) \log (x) \log (\log (x)) \log (\log ^2(\log (x)))}{-e^{25} \log (x) \log (\log (x))+(x^2+2 x^3+x^4) \log (x) \log (\log (x)) \log (\log ^2(\log (x)))} \, dx\)

Optimal. Leaf size=20 \[ \log \left (e^{25}-\left (x+x^2\right )^2 \log \left (\log ^2(\log (x))\right )\right ) \]

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Rubi [F]  time = 24.60, 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 {2 x+4 x^2+2 x^3+\left (2 x+6 x^2+4 x^3\right ) \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )}{-e^{25} \log (x) \log (\log (x))+\left (x^2+2 x^3+x^4\right ) \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x + 4*x^2 + 2*x^3 + (2*x + 6*x^2 + 4*x^3)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2])/(-(E^25*Log[x]*Log[Log
[x]]) + (x^2 + 2*x^3 + x^4)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2]),x]

[Out]

2*Log[x] + 2*Log[1 + x] + 4*E^25*Defer[Int][1/((-1 - x)*(E^25 - x^2*(1 + x)^2*Log[Log[Log[x]]^2])), x] + 2*E^2
5*Defer[Int][1/((1 + x)*(E^25 - x^2*(1 + x)^2*Log[Log[Log[x]]^2])), x] + 2*Defer[Int][x/(Log[x]*Log[Log[x]]*(-
E^25 + x^2*(1 + x)^2*Log[Log[Log[x]]^2])), x] + 4*Defer[Int][x^2/(Log[x]*Log[Log[x]]*(-E^25 + x^2*(1 + x)^2*Lo
g[Log[Log[x]]^2])), x] + 2*Defer[Int][x^3/(Log[x]*Log[Log[x]]*(-E^25 + x^2*(1 + x)^2*Log[Log[Log[x]]^2])), x]
- 2*E^25*Defer[Int][(E^25*x - x^3*(1 + x)^2*Log[Log[Log[x]]^2])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x (1+x) \left (-1-x-\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )-2 x \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{e^{25} \log (x) \log (\log (x))-\left (x^2+2 x^3+x^4\right ) \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )} \, dx\\ &=2 \int \frac {x (1+x) \left (-1-x-\log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )-2 x \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )\right )}{e^{25} \log (x) \log (\log (x))-\left (x^2+2 x^3+x^4\right ) \log (x) \log (\log (x)) \log \left (\log ^2(\log (x))\right )} \, dx\\ &=2 \int \left (\frac {1+2 x}{x (1+x)}+\frac {x^2+3 x^3+3 x^4+x^5+e^{25} \log (x) \log (\log (x))+2 e^{25} x \log (x) \log (\log (x))}{x (1+x) \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )}\right ) \, dx\\ &=2 \int \frac {1+2 x}{x (1+x)} \, dx+2 \int \frac {x^2+3 x^3+3 x^4+x^5+e^{25} \log (x) \log (\log (x))+2 e^{25} x \log (x) \log (\log (x))}{x (1+x) \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )} \, dx\\ &=2 \int \left (\frac {1}{x}+\frac {1}{1+x}\right ) \, dx+2 \int \frac {-x^2 (1+x)^3-e^{25} (1+2 x) \log (x) \log (\log (x))}{x (1+x) \log (x) \log (\log (x)) \left (e^{25}-x^2 (1+x)^2 \log \left (\log ^2(\log (x))\right )\right )} \, dx\\ &=2 \log (x)+2 \log (1+x)+2 \int \left (\frac {-x^2-3 x^3-3 x^4-x^5-e^{25} \log (x) \log (\log (x))-2 e^{25} x \log (x) \log (\log (x))}{(1+x) \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )}+\frac {x^2+3 x^3+3 x^4+x^5+e^{25} \log (x) \log (\log (x))+2 e^{25} x \log (x) \log (\log (x))}{x \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )}\right ) \, dx\\ &=2 \log (x)+2 \log (1+x)+2 \int \frac {-x^2-3 x^3-3 x^4-x^5-e^{25} \log (x) \log (\log (x))-2 e^{25} x \log (x) \log (\log (x))}{(1+x) \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )} \, dx+2 \int \frac {x^2+3 x^3+3 x^4+x^5+e^{25} \log (x) \log (\log (x))+2 e^{25} x \log (x) \log (\log (x))}{x \log (x) \log (\log (x)) \left (-e^{25}+x^2 \log \left (\log ^2(\log (x))\right )+2 x^3 \log \left (\log ^2(\log (x))\right )+x^4 \log \left (\log ^2(\log (x))\right )\right )} \, dx\\ &=2 \log (x)+2 \log (1+x)+2 \int \frac {-x^2 (1+x)^3-e^{25} (1+2 x) \log (x) \log (\log (x))}{x \log (x) \log (\log (x)) \left (e^{25}-x^2 (1+x)^2 \log \left (\log ^2(\log (x))\right )\right )} \, dx+2 \int \frac {x^2 (1+x)^3+e^{25} (1+2 x) \log (x) \log (\log (x))}{(1+x) \log (x) \log (\log (x)) \left (e^{25}-x^2 (1+x)^2 \log \left (\log ^2(\log (x))\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 38, normalized size = 1.90 \begin {gather*} \log \left (e^{25}-x^2 \log \left (\log ^2(\log (x))\right )-2 x^3 \log \left (\log ^2(\log (x))\right )-x^4 \log \left (\log ^2(\log (x))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + 4*x^2 + 2*x^3 + (2*x + 6*x^2 + 4*x^3)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2])/(-(E^25*Log[x]*L
og[Log[x]]) + (x^2 + 2*x^3 + x^4)*Log[x]*Log[Log[x]]*Log[Log[Log[x]]^2]),x]

[Out]

Log[E^25 - x^2*Log[Log[Log[x]]^2] - 2*x^3*Log[Log[Log[x]]^2] - x^4*Log[Log[Log[x]]^2]]

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fricas [B]  time = 0.55, size = 49, normalized size = 2.45 \begin {gather*} 2 \, \log \left (x^{2} + x\right ) + \log \left (\frac {{\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \left (\log \left (\log \relax (x)\right )^{2}\right ) - e^{25}}{x^{4} + 2 \, x^{3} + x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+6*x^2+2*x)*log(x)*log(log(x))*log(log(log(x))^2)+2*x^3+4*x^2+2*x)/((x^4+2*x^3+x^2)*log(x)*lo
g(log(x))*log(log(log(x))^2)-exp(25)*log(x)*log(log(x))),x, algorithm="fricas")

[Out]

2*log(x^2 + x) + log(((x^4 + 2*x^3 + x^2)*log(log(log(x))^2) - e^25)/(x^4 + 2*x^3 + x^2))

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giac [A]  time = 0.34, size = 37, normalized size = 1.85 \begin {gather*} \log \left (-x^{4} \log \left (\log \left (\log \relax (x)\right )^{2}\right ) - 2 \, x^{3} \log \left (\log \left (\log \relax (x)\right )^{2}\right ) - x^{2} \log \left (\log \left (\log \relax (x)\right )^{2}\right ) + e^{25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+6*x^2+2*x)*log(x)*log(log(x))*log(log(log(x))^2)+2*x^3+4*x^2+2*x)/((x^4+2*x^3+x^2)*log(x)*lo
g(log(x))*log(log(log(x))^2)-exp(25)*log(x)*log(log(x))),x, algorithm="giac")

[Out]

log(-x^4*log(log(log(x))^2) - 2*x^3*log(log(log(x))^2) - x^2*log(log(log(x))^2) + e^25)

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maple [C]  time = 0.75, size = 229, normalized size = 11.45

method	result	size
risch	\(2 \ln \left (x^{2}+x \right )+\ln \left (\ln \left (\ln \left (\ln \relax (x )\right )\right )-\frac {i \left (\pi \,x^{4} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )-2 \pi \,x^{4} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{2}+\pi \,x^{4} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{3}+2 \pi \,x^{3} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )-4 \pi \,x^{3} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{2}+2 \pi \,x^{3} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{3}+\pi \,x^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )-2 \pi \,x^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{2}+\pi \,x^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{3}-2 i {\mathrm e}^{25}\right )}{4 x^{2} \left (x^{2}+2 x +1\right )}\right )\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+6*x^2+2*x)*ln(x)*ln(ln(x))*ln(ln(ln(x))^2)+2*x^3+4*x^2+2*x)/((x^4+2*x^3+x^2)*ln(x)*ln(ln(x))*ln(ln
(ln(x))^2)-exp(25)*ln(x)*ln(ln(x))),x,method=_RETURNVERBOSE)

[Out]

2*ln(x^2+x)+ln(ln(ln(ln(x)))-1/4*I*(Pi*x^4*csgn(I*ln(ln(x)))^2*csgn(I*ln(ln(x))^2)-2*Pi*x^4*csgn(I*ln(ln(x)))*
csgn(I*ln(ln(x))^2)^2+Pi*x^4*csgn(I*ln(ln(x))^2)^3+2*Pi*x^3*csgn(I*ln(ln(x)))^2*csgn(I*ln(ln(x))^2)-4*Pi*x^3*c
sgn(I*ln(ln(x)))*csgn(I*ln(ln(x))^2)^2+2*Pi*x^3*csgn(I*ln(ln(x))^2)^3+Pi*x^2*csgn(I*ln(ln(x)))^2*csgn(I*ln(ln(
x))^2)-2*Pi*x^2*csgn(I*ln(ln(x)))*csgn(I*ln(ln(x))^2)^2+Pi*x^2*csgn(I*ln(ln(x))^2)^3-2*I*exp(25))/x^2/(x^2+2*x
+1))

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maxima [B]  time = 0.40, size = 51, normalized size = 2.55 \begin {gather*} 2 \, \log \left (x + 1\right ) + 2 \, \log \relax (x) + \log \left (\frac {2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \left (\log \left (\log \relax (x)\right )\right ) - e^{25}}{2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+6*x^2+2*x)*log(x)*log(log(x))*log(log(log(x))^2)+2*x^3+4*x^2+2*x)/((x^4+2*x^3+x^2)*log(x)*lo
g(log(x))*log(log(log(x))^2)-exp(25)*log(x)*log(log(x))),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {2\,x+4\,x^2+2\,x^3+\ln \left ({\ln \left (\ln \relax (x)\right )}^2\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (4\,x^3+6\,x^2+2\,x\right )}{\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{25}\,\ln \relax (x)-\ln \left ({\ln \left (\ln \relax (x)\right )}^2\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (x^4+2\,x^3+x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 4*x^2 + 2*x^3 + log(log(log(x))^2)*log(log(x))*log(x)*(2*x + 6*x^2 + 4*x^3))/(log(log(x))*exp(25)*
log(x) - log(log(log(x))^2)*log(log(x))*log(x)*(x^2 + 2*x^3 + x^4)),x)

[Out]

-int((2*x + 4*x^2 + 2*x^3 + log(log(log(x))^2)*log(log(x))*log(x)*(2*x + 6*x^2 + 4*x^3))/(log(log(x))*exp(25)*
log(x) - log(log(log(x))^2)*log(log(x))*log(x)*(x^2 + 2*x^3 + x^4)), x)

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sympy [A]  time = 1.07, size = 32, normalized size = 1.60 \begin {gather*} 2 \log {\left (x^{2} + x \right )} + \log {\left (\log {\left (\log {\left (\log {\relax (x )} \right )}^{2} \right )} - \frac {e^{25}}{x^{4} + 2 x^{3} + x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+6*x**2+2*x)*ln(x)*ln(ln(x))*ln(ln(ln(x))**2)+2*x**3+4*x**2+2*x)/((x**4+2*x**3+x**2)*ln(x)*l
n(ln(x))*ln(ln(ln(x))**2)-exp(25)*ln(x)*ln(ln(x))),x)

[Out]

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

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