3.72.77 \(\int \frac {-1296 x^2+(-13932-648 x^2) \log (\frac {1}{2} (43+2 x^2)) \log (\log (\frac {1}{2} (43+2 x^2))) \log (\log (\log (\frac {1}{2} (43+2 x^2))))}{e^3 (43 x^5+2 x^7) \log (\frac {1}{2} (43+2 x^2)) \log (\log (\frac {1}{2} (43+2 x^2))) \log ^5(\log (\log (\frac {1}{2} (43+2 x^2))))} \, dx\)

Optimal. Leaf size=20 \[ \frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \]

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Rubi [A]  time = 0.60, antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 1593, 6687} \begin {gather*} \frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1296*x^2 + (-13932 - 648*x^2)*Log[(43 + 2*x^2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]]
)/(E^3*(43*x^5 + 2*x^7)*Log[(43 + 2*x^2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]]^5),x]

[Out]

81/(E^3*x^4*Log[Log[Log[(43 + 2*x^2)/2]]]^4)

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6687

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{\left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx}{e^3}\\ &=\frac {\int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{x^5 \left (43+2 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx}{e^3}\\ &=\frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 20, normalized size = 1.00 \begin {gather*} \frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1296*x^2 + (-13932 - 648*x^2)*Log[(43 + 2*x^2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2
)/2]]])/(E^3*(43*x^5 + 2*x^7)*Log[(43 + 2*x^2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]]^5),x]

[Out]

81/(E^3*x^4*Log[Log[Log[43/2 + x^2]]]^4)

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fricas [A]  time = 0.66, size = 17, normalized size = 0.85 \begin {gather*} \frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/
exp(3)/log(x^2+43/2)/log(log(x^2+43/2))/log(log(log(x^2+43/2)))^5,x, algorithm="fricas")

[Out]

81*e^(-3)/(x^4*log(log(log(x^2 + 43/2)))^4)

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giac [B]  time = 0.66, size = 58, normalized size = 2.90 \begin {gather*} \frac {324 \, e^{\left (-3\right )}}{{\left (2 \, x^{2} + 43\right )}^{2} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} - 86 \, {\left (2 \, x^{2} + 43\right )} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} + 1849 \, \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/
exp(3)/log(x^2+43/2)/log(log(x^2+43/2))/log(log(log(x^2+43/2)))^5,x, algorithm="giac")

[Out]

324*e^(-3)/((2*x^2 + 43)^2*log(log(log(x^2 + 43/2)))^4 - 86*(2*x^2 + 43)*log(log(log(x^2 + 43/2)))^4 + 1849*lo
g(log(log(x^2 + 43/2)))^4)

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maple [A]  time = 0.12, size = 18, normalized size = 0.90

method	result	size
risch	\(\frac {81 \,{\mathrm e}^{-3}}{\ln \left (\ln \left (\ln \left (x^{2}+\frac {43}{2}\right )\right )\right )^{4} x^{4}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-648*x^2-13932)*ln(x^2+43/2)*ln(ln(x^2+43/2))*ln(ln(ln(x^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/exp(3)/ln(x^
2+43/2)/ln(ln(x^2+43/2))/ln(ln(ln(x^2+43/2)))^5,x,method=_RETURNVERBOSE)

[Out]

81/ln(ln(ln(x^2+43/2)))^4/x^4*exp(-3)

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maxima [A]  time = 0.52, size = 24, normalized size = 1.20 \begin {gather*} \frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (-\log \relax (2) + \log \left (2 \, x^{2} + 43\right )\right )\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/
exp(3)/log(x^2+43/2)/log(log(x^2+43/2))/log(log(log(x^2+43/2)))^5,x, algorithm="maxima")

[Out]

81*e^(-3)/(x^4*log(log(-log(2) + log(2*x^2 + 43)))^4)

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mupad [B]  time = 6.89, size = 17, normalized size = 0.85 \begin {gather*} \frac {81\,{\mathrm {e}}^{-3}}{x^4\,{\ln \left (\ln \left (\ln \left (x^2+\frac {43}{2}\right )\right )\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3)*(1296*x^2 + log(log(x^2 + 43/2))*log(log(log(x^2 + 43/2)))*log(x^2 + 43/2)*(648*x^2 + 13932)))/(
log(log(x^2 + 43/2))*log(log(log(x^2 + 43/2)))^5*log(x^2 + 43/2)*(43*x^5 + 2*x^7)),x)

[Out]

(81*exp(-3))/(x^4*log(log(log(x^2 + 43/2)))^4)

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sympy [A]  time = 0.40, size = 20, normalized size = 1.00 \begin {gather*} \frac {81}{x^{4} e^{3} \log {\left (\log {\left (\log {\left (x^{2} + \frac {43}{2} \right )} \right )} \right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x**2-13932)*ln(x**2+43/2)*ln(ln(x**2+43/2))*ln(ln(ln(x**2+43/2)))-1296*x**2)/(2*x**7+43*x**5)
/exp(3)/ln(x**2+43/2)/ln(ln(x**2+43/2))/ln(ln(ln(x**2+43/2)))**5,x)

[Out]

81*exp(-3)/(x**4*log(log(log(x**2 + 43/2)))**4)

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