3.55.53 \(\int \frac {-4 x-4 x^2+8 x^3 \log (x)+(8 x^2-4 x^3) \log ^2(x)-4 x^2 \log ^3(x)+(8 x^2+8 x \log (x)) \log (x+\log (x))}{x^3 \log ^4(x)+x^2 \log ^5(x)+(-2 x^2 \log ^2(x)-2 x \log ^3(x)) \log (x+\log (x))+(x+\log (x)) \log ^2(x+\log (x))} \, dx\)

Optimal. Leaf size=20 \[ \frac {4 x^2}{-x \log ^2(x)+\log (x+\log (x))} \]

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Rubi [F]  time = 2.48, 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 {-4 x-4 x^2+8 x^3 \log (x)+\left (8 x^2-4 x^3\right ) \log ^2(x)-4 x^2 \log ^3(x)+\left (8 x^2+8 x \log (x)\right ) \log (x+\log (x))}{x^3 \log ^4(x)+x^2 \log ^5(x)+\left (-2 x^2 \log ^2(x)-2 x \log ^3(x)\right ) \log (x+\log (x))+(x+\log (x)) \log ^2(x+\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x - 4*x^2 + 8*x^3*Log[x] + (8*x^2 - 4*x^3)*Log[x]^2 - 4*x^2*Log[x]^3 + (8*x^2 + 8*x*Log[x])*Log[x + Lo
g[x]])/(x^3*Log[x]^4 + x^2*Log[x]^5 + (-2*x^2*Log[x]^2 - 2*x*Log[x]^3)*Log[x + Log[x]] + (x + Log[x])*Log[x +
Log[x]]^2),x]

[Out]

-4*Defer[Int][x/((x + Log[x])*(x*Log[x]^2 - Log[x + Log[x]])^2), x] - 4*Defer[Int][x^2/((x + Log[x])*(x*Log[x]
^2 - Log[x + Log[x]])^2), x] + 8*Defer[Int][(x^3*Log[x])/((x + Log[x])*(x*Log[x]^2 - Log[x + Log[x]])^2), x] +
8*Defer[Int][(x^2*Log[x]^2)/((x + Log[x])*(x*Log[x]^2 - Log[x + Log[x]])^2), x] + 4*Defer[Int][(x^3*Log[x]^2)
/((x + Log[x])*(x*Log[x]^2 - Log[x + Log[x]])^2), x] + 4*Defer[Int][(x^2*Log[x]^3)/((x + Log[x])*(x*Log[x]^2 -
Log[x + Log[x]])^2), x] - 8*Defer[Int][x/(x*Log[x]^2 - Log[x + Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (-1-x-(-2+x) x \log ^2(x)-x \log ^3(x)+2 x \log (x+\log (x))+2 \log (x) \left (x^2+\log (x+\log (x))\right )\right )}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx\\ &=4 \int \frac {x \left (-1-x-(-2+x) x \log ^2(x)-x \log ^3(x)+2 x \log (x+\log (x))+2 \log (x) \left (x^2+\log (x+\log (x))\right )\right )}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx\\ &=4 \int \left (\frac {x \left (-1-x+2 x^2 \log (x)+2 x \log ^2(x)+x^2 \log ^2(x)+x \log ^3(x)\right )}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}-\frac {2 x}{x \log ^2(x)-\log (x+\log (x))}\right ) \, dx\\ &=4 \int \frac {x \left (-1-x+2 x^2 \log (x)+2 x \log ^2(x)+x^2 \log ^2(x)+x \log ^3(x)\right )}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx-8 \int \frac {x}{x \log ^2(x)-\log (x+\log (x))} \, dx\\ &=4 \int \left (-\frac {x}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}-\frac {x^2}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}+\frac {2 x^3 \log (x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}+\frac {2 x^2 \log ^2(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}+\frac {x^3 \log ^2(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}+\frac {x^2 \log ^3(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2}\right ) \, dx-8 \int \frac {x}{x \log ^2(x)-\log (x+\log (x))} \, dx\\ &=-\left (4 \int \frac {x}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx\right )-4 \int \frac {x^2}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx+4 \int \frac {x^3 \log ^2(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx+4 \int \frac {x^2 \log ^3(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx+8 \int \frac {x^3 \log (x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx+8 \int \frac {x^2 \log ^2(x)}{(x+\log (x)) \left (x \log ^2(x)-\log (x+\log (x))\right )^2} \, dx-8 \int \frac {x}{x \log ^2(x)-\log (x+\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.94, size = 20, normalized size = 1.00 \begin {gather*} \frac {4 x^2}{-x \log ^2(x)+\log (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x - 4*x^2 + 8*x^3*Log[x] + (8*x^2 - 4*x^3)*Log[x]^2 - 4*x^2*Log[x]^3 + (8*x^2 + 8*x*Log[x])*Log[
x + Log[x]])/(x^3*Log[x]^4 + x^2*Log[x]^5 + (-2*x^2*Log[x]^2 - 2*x*Log[x]^3)*Log[x + Log[x]] + (x + Log[x])*Lo
g[x + Log[x]]^2),x]

[Out]

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

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fricas [A]  time = 0.74, size = 21, normalized size = 1.05 \begin {gather*} -\frac {4 \, x^{2}}{x \log \relax (x)^{2} - \log \left (x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(x)+8*x^2)*log(x+log(x))-4*x^2*log(x)^3+(-4*x^3+8*x^2)*log(x)^2+8*x^3*log(x)-4*x^2-4*x)/((x
+log(x))*log(x+log(x))^2+(-2*x*log(x)^3-2*x^2*log(x)^2)*log(x+log(x))+x^2*log(x)^5+x^3*log(x)^4),x, algorithm=
"fricas")

[Out]

-4*x^2/(x*log(x)^2 - log(x + log(x)))

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giac [A]  time = 0.28, size = 21, normalized size = 1.05 \begin {gather*} -\frac {4 \, x^{2}}{x \log \relax (x)^{2} - \log \left (x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(x)+8*x^2)*log(x+log(x))-4*x^2*log(x)^3+(-4*x^3+8*x^2)*log(x)^2+8*x^3*log(x)-4*x^2-4*x)/((x
+log(x))*log(x+log(x))^2+(-2*x*log(x)^3-2*x^2*log(x)^2)*log(x+log(x))+x^2*log(x)^5+x^3*log(x)^4),x, algorithm=
"giac")

[Out]

-4*x^2/(x*log(x)^2 - log(x + log(x)))

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maple [A]  time = 0.03, size = 22, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x*ln(x)+8*x^2)*ln(x+ln(x))-4*x^2*ln(x)^3+(-4*x^3+8*x^2)*ln(x)^2+8*x^3*ln(x)-4*x^2-4*x)/((x+ln(x))*ln(x
+ln(x))^2+(-2*x*ln(x)^3-2*x^2*ln(x)^2)*ln(x+ln(x))+x^2*ln(x)^5+x^3*ln(x)^4),x,method=_RETURNVERBOSE)

[Out]

-4*x^2/(x*ln(x)^2-ln(x+ln(x)))

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maxima [A]  time = 0.41, size = 21, normalized size = 1.05 \begin {gather*} -\frac {4 \, x^{2}}{x \log \relax (x)^{2} - \log \left (x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*log(x)+8*x^2)*log(x+log(x))-4*x^2*log(x)^3+(-4*x^3+8*x^2)*log(x)^2+8*x^3*log(x)-4*x^2-4*x)/((x
+log(x))*log(x+log(x))^2+(-2*x*log(x)^3-2*x^2*log(x)^2)*log(x+log(x))+x^2*log(x)^5+x^3*log(x)^4),x, algorithm=
"maxima")

[Out]

-4*x^2/(x*log(x)^2 - log(x + log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {4\,x-8\,x^3\,\ln \relax (x)-{\ln \relax (x)}^2\,\left (8\,x^2-4\,x^3\right )+4\,x^2\,{\ln \relax (x)}^3+4\,x^2-\ln \left (x+\ln \relax (x)\right )\,\left (8\,x\,\ln \relax (x)+8\,x^2\right )}{x^2\,{\ln \relax (x)}^5+x^3\,{\ln \relax (x)}^4-\ln \left (x+\ln \relax (x)\right )\,\left (2\,x^2\,{\ln \relax (x)}^2+2\,x\,{\ln \relax (x)}^3\right )+{\ln \left (x+\ln \relax (x)\right )}^2\,\left (x+\ln \relax (x)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 8*x^3*log(x) - log(x)^2*(8*x^2 - 4*x^3) + 4*x^2*log(x)^3 + 4*x^2 - log(x + log(x))*(8*x*log(x) + 8
*x^2))/(x^2*log(x)^5 + x^3*log(x)^4 - log(x + log(x))*(2*x*log(x)^3 + 2*x^2*log(x)^2) + log(x + log(x))^2*(x +
log(x))),x)

[Out]

int(-(4*x - 8*x^3*log(x) - log(x)^2*(8*x^2 - 4*x^3) + 4*x^2*log(x)^3 + 4*x^2 - log(x + log(x))*(8*x*log(x) + 8
*x^2))/(x^2*log(x)^5 + x^3*log(x)^4 - log(x + log(x))*(2*x*log(x)^3 + 2*x^2*log(x)^2) + log(x + log(x))^2*(x +
log(x))), x)

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sympy [A]  time = 0.30, size = 17, normalized size = 0.85 \begin {gather*} \frac {4 x^{2}}{- x \log {\relax (x )}^{2} + \log {\left (x + \log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x*ln(x)+8*x**2)*ln(x+ln(x))-4*x**2*ln(x)**3+(-4*x**3+8*x**2)*ln(x)**2+8*x**3*ln(x)-4*x**2-4*x)/(
(x+ln(x))*ln(x+ln(x))**2+(-2*x*ln(x)**3-2*x**2*ln(x)**2)*ln(x+ln(x))+x**2*ln(x)**5+x**3*ln(x)**4),x)

[Out]

4*x**2/(-x*log(x)**2 + log(x + log(x)))

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