3.13.83 \(\int \frac {x^2+2 x^2 \log (x)+\log (2500-100 \log (3)+\log ^2(3))}{x^3 \log (x)+x \log (x) \log (2500-100 \log (3)+\log ^2(3))} \, dx\)

Optimal. Leaf size=15 \[ \log \left (\log (x) \left (x^2+\log \left ((-50+\log (3))^2\right )\right )\right ) \]

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Rubi [A]  time = 0.35, antiderivative size = 16, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6741, 6725, 2302, 29, 260} \begin {gather*} \log \left (x^2+\log \left ((\log (3)-50)^2\right )\right )+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + 2*x^2*Log[x] + Log[2500 - 100*Log[3] + Log[3]^2])/(x^3*Log[x] + x*Log[x]*Log[2500 - 100*Log[3] + Lo
g[3]^2]),x]

[Out]

Log[Log[x]] + Log[x^2 + Log[(-50 + Log[3])^2]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2+2 x^2 \log (x)+\log \left (2500-100 \log (3)+\log ^2(3)\right )}{x \log (x) \left (x^2+\log \left ((-50+\log (3))^2\right )\right )} \, dx\\ &=\int \left (\frac {1}{x \log (x)}+\frac {2 x}{x^2+\log \left ((-50+\log (3))^2\right )}\right ) \, dx\\ &=2 \int \frac {x}{x^2+\log \left ((-50+\log (3))^2\right )} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\log \left (x^2+\log \left ((-50+\log (3))^2\right )\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\log (\log (x))+\log \left (x^2+\log \left ((-50+\log (3))^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 16, normalized size = 1.07 \begin {gather*} \log (\log (x))+\log \left (x^2+\log \left ((-50+\log (3))^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 2*x^2*Log[x] + Log[2500 - 100*Log[3] + Log[3]^2])/(x^3*Log[x] + x*Log[x]*Log[2500 - 100*Log[3
] + Log[3]^2]),x]

[Out]

Log[Log[x]] + Log[x^2 + Log[(-50 + Log[3])^2]]

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fricas [A]  time = 0.55, size = 20, normalized size = 1.33 \begin {gather*} \log \left (x^{2} + \log \left (\log \relax (3)^{2} - 100 \, \log \relax (3) + 2500\right )\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3)^2-100*log(3)+2500)+2*x^2*log(x)+x^2)/(x*log(x)*log(log(3)^2-100*log(3)+2500)+x^3*log(x))
,x, algorithm="fricas")

[Out]

log(x^2 + log(log(3)^2 - 100*log(3) + 2500)) + log(log(x))

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giac [A]  time = 0.23, size = 20, normalized size = 1.33 \begin {gather*} \log \left (x^{2} + \log \left (\log \relax (3)^{2} - 100 \, \log \relax (3) + 2500\right )\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3)^2-100*log(3)+2500)+2*x^2*log(x)+x^2)/(x*log(x)*log(log(3)^2-100*log(3)+2500)+x^3*log(x))
,x, algorithm="giac")

[Out]

log(x^2 + log(log(3)^2 - 100*log(3) + 2500)) + log(log(x))

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maple [A]  time = 0.05, size = 19, normalized size = 1.27

method	result	size
risch	\(\ln \left (x^{2}+2 \ln \left (-\ln \relax (3)+50\right )\right )+\ln \left (\ln \relax (x )\right )\)	\(19\)
default	\(\ln \left (\ln \relax (x )\right )+\ln \left (x^{2}+\ln \left (\ln \relax (3)^{2}-100 \ln \relax (3)+2500\right )\right )\)	\(21\)
norman	\(\ln \left (\ln \relax (x )\right )+\ln \left (x^{2}+\ln \left (\ln \relax (3)^{2}-100 \ln \relax (3)+2500\right )\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(ln(3)^2-100*ln(3)+2500)+2*x^2*ln(x)+x^2)/(x*ln(x)*ln(ln(3)^2-100*ln(3)+2500)+x^3*ln(x)),x,method=_RETU
RNVERBOSE)

[Out]

ln(x^2+2*ln(-ln(3)+50))+ln(ln(x))

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maxima [A]  time = 0.57, size = 16, normalized size = 1.07 \begin {gather*} \log \left (x^{2} + 2 \, \log \left (\log \relax (3) - 50\right )\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3)^2-100*log(3)+2500)+2*x^2*log(x)+x^2)/(x*log(x)*log(log(3)^2-100*log(3)+2500)+x^3*log(x))
,x, algorithm="maxima")

[Out]

log(x^2 + 2*log(log(3) - 50)) + log(log(x))

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mupad [B]  time = 1.06, size = 16, normalized size = 1.07 \begin {gather*} \ln \left (x^2+\ln \left ({\left (\ln \relax (3)-50\right )}^2\right )\right )+\ln \left (\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(3)^2 - 100*log(3) + 2500) + 2*x^2*log(x) + x^2)/(x^3*log(x) + x*log(log(3)^2 - 100*log(3) + 2500)
*log(x)),x)

[Out]

log(log((log(3) - 50)^2) + x^2) + log(log(x))

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sympy [A]  time = 0.16, size = 22, normalized size = 1.47 \begin {gather*} \log {\left (x^{2} + \log {\left (- 100 \log {\relax (3 )} + \log {\relax (3 )}^{2} + 2500 \right )} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(ln(3)**2-100*ln(3)+2500)+2*x**2*ln(x)+x**2)/(x*ln(x)*ln(ln(3)**2-100*ln(3)+2500)+x**3*ln(x)),x)

[Out]

log(x**2 + log(-100*log(3) + log(3)**2 + 2500)) + log(log(x))

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