3.83.29 \(\int \frac {-2-5 x^5 \log (\log (5))}{2 x \log ^2(x^2)+4 x^6 \log (x^2) \log (\log (5))+2 x^{11} \log ^2(\log (5))} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{2 \left (\log \left (x^2\right )+x^5 \log (\log (5))\right )} \]

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Rubi [A]  time = 0.15, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{2 \left (x^5 \log (\log (5))+\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 5*x^5*Log[Log[5]])/(2*x*Log[x^2]^2 + 4*x^6*Log[x^2]*Log[Log[5]] + 2*x^11*Log[Log[5]]^2),x]

[Out]

1/(2*(Log[x^2] + x^5*Log[Log[5]]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-5 x^5 \log (\log (5))}{2 x \left (\log \left (x^2\right )+x^5 \log (\log (5))\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {-2-5 x^5 \log (\log (5))}{x \left (\log \left (x^2\right )+x^5 \log (\log (5))\right )^2} \, dx\\ &=\frac {1}{2 \left (\log \left (x^2\right )+x^5 \log (\log (5))\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{2 \left (\log \left (x^2\right )+x^5 \log (\log (5))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 5*x^5*Log[Log[5]])/(2*x*Log[x^2]^2 + 4*x^6*Log[x^2]*Log[Log[5]] + 2*x^11*Log[Log[5]]^2),x]

[Out]

1/(2*(Log[x^2] + x^5*Log[Log[5]]))

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fricas [A]  time = 0.61, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2 \, {\left (x^{5} \log \left (\log \relax (5)\right ) + \log \left (x^{2}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^5*log(log(5))-2)/(2*x^11*log(log(5))^2+4*x^6*log(x^2)*log(log(5))+2*x*log(x^2)^2),x, algorithm
="fricas")

[Out]

1/2/(x^5*log(log(5)) + log(x^2))

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giac [A]  time = 0.18, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2 \, {\left (x^{5} \log \left (\log \relax (5)\right ) + \log \left (x^{2}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^5*log(log(5))-2)/(2*x^11*log(log(5))^2+4*x^6*log(x^2)*log(log(5))+2*x*log(x^2)^2),x, algorithm
="giac")

[Out]

1/2/(x^5*log(log(5)) + log(x^2))

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maple [A]  time = 0.03, size = 17, normalized size = 0.94

method	result	size
risch	\(\frac {1}{2 x^{5} \ln \left (\ln \relax (5)\right )+2 \ln \left (x^{2}\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*x^5*ln(ln(5))-2)/(2*x^11*ln(ln(5))^2+4*x^6*ln(x^2)*ln(ln(5))+2*x*ln(x^2)^2),x,method=_RETURNVERBOSE)

[Out]

1/2/(x^5*ln(ln(5))+ln(x^2))

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maxima [A]  time = 0.49, size = 16, normalized size = 0.89 \begin {gather*} \frac {1}{2 \, {\left (x^{5} \log \left (\log \relax (5)\right ) + 2 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^5*log(log(5))-2)/(2*x^11*log(log(5))^2+4*x^6*log(x^2)*log(log(5))+2*x*log(x^2)^2),x, algorithm
="maxima")

[Out]

1/2/(x^5*log(log(5)) + 2*log(x))

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mupad [B]  time = 5.07, size = 17, normalized size = 0.94 \begin {gather*} \frac {1}{2\,\ln \left (x^2\right )+2\,x^5\,\ln \left (\ln \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x^5*log(log(5)) + 2)/(2*x^11*log(log(5))^2 + 2*x*log(x^2)^2 + 4*x^6*log(x^2)*log(log(5))),x)

[Out]

1/(2*log(x^2) + 2*x^5*log(log(5)))

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sympy [A]  time = 0.11, size = 17, normalized size = 0.94 \begin {gather*} \frac {1}{2 x^{5} \log {\left (\log {\relax (5 )} \right )} + 2 \log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x**5*ln(ln(5))-2)/(2*x**11*ln(ln(5))**2+4*x**6*ln(x**2)*ln(ln(5))+2*x*ln(x**2)**2),x)

[Out]

1/(2*x**5*log(log(5)) + 2*log(x**2))

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