3.69.36 \(\int \frac {6+2 e^8-6 e^8 \log (\frac {4}{x})}{(-x+e^8 x \log (\frac {4}{x})) \log ^2(\frac {-x^3+e^8 x^3 \log (\frac {4}{x})}{5 e^8})} \, dx\)

Optimal. Leaf size=24 \[ \frac {2}{\log \left (\frac {1}{5} x^3 \left (-\frac {1}{e^8}+\log \left (\frac {4}{x}\right )\right )\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2561, 6686} \begin {gather*} \frac {2}{\log \left (-\frac {x^3-e^8 x^3 \log \left (\frac {4}{x}\right )}{5 e^8}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + 2*E^8 - 6*E^8*Log[4/x])/((-x + E^8*x*Log[4/x])*Log[(-x^3 + E^8*x^3*Log[4/x])/(5*E^8)]^2),x]

[Out]

2/Log[-1/5*(x^3 - E^8*x^3*Log[4/x])/E^8]

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+2 e^8-6 e^8 \log \left (\frac {4}{x}\right )}{x \left (-1+e^8 \log \left (\frac {4}{x}\right )\right ) \log ^2\left (\frac {-x^3+e^8 x^3 \log \left (\frac {4}{x}\right )}{5 e^8}\right )} \, dx\\ &=\frac {2}{\log \left (-\frac {x^3-e^8 x^3 \log \left (\frac {4}{x}\right )}{5 e^8}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 27, normalized size = 1.12 \begin {gather*} \frac {2}{\log \left (\frac {x^3 \left (-1+e^8 \log \left (\frac {4}{x}\right )\right )}{5 e^8}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 2*E^8 - 6*E^8*Log[4/x])/((-x + E^8*x*Log[4/x])*Log[(-x^3 + E^8*x^3*Log[4/x])/(5*E^8)]^2),x]

[Out]

2/Log[(x^3*(-1 + E^8*Log[4/x]))/(5*E^8)]

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fricas [A]  time = 1.90, size = 27, normalized size = 1.12 \begin {gather*} \frac {2}{\log \left (\frac {1}{5} \, {\left (x^{3} e^{8} \log \left (\frac {4}{x}\right ) - x^{3}\right )} e^{\left (-8\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(8)*log(4/x)+2*exp(8)+6)/(x*exp(8)*log(4/x)-x)/log(1/5*(x^3*exp(8)*log(4/x)-x^3)/exp(8))^2,x,
algorithm="fricas")

[Out]

2/log(1/5*(x^3*e^8*log(4/x) - x^3)*e^(-8))

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giac [B]  time = 0.27, size = 456, normalized size = 19.00 \begin {gather*} -\frac {2 \, {\left (6 \, e^{16} \log \relax (2) \log \left (\frac {4}{x}\right ) - 3 \, e^{16} \log \relax (x) \log \left (\frac {4}{x}\right ) - 6 \, e^{8} \log \relax (2) + 3 \, e^{8} \log \relax (x) - e^{16} \log \left (\frac {4}{x}\right ) - 3 \, e^{8} \log \left (\frac {4}{x}\right ) + e^{8} + 3\right )}}{6 \, e^{16} \log \relax (5) \log \relax (2) \log \left (\frac {4}{x}\right ) - 6 \, e^{16} \log \relax (2) \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) \log \left (\frac {4}{x}\right ) - 3 \, e^{16} \log \relax (5) \log \relax (x) \log \left (\frac {4}{x}\right ) - 18 \, e^{16} \log \relax (2) \log \relax (x) \log \left (\frac {4}{x}\right ) + 3 \, e^{16} \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) \log \relax (x) \log \left (\frac {4}{x}\right ) + 9 \, e^{16} \log \relax (x)^{2} \log \left (\frac {4}{x}\right ) - 2 \, e^{16} \log \relax (5) \log \relax (2) - 6 \, e^{8} \log \relax (5) \log \relax (2) + 2 \, e^{16} \log \relax (2) \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) + 6 \, e^{8} \log \relax (2) \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) + e^{16} \log \relax (5) \log \relax (x) + 3 \, e^{8} \log \relax (5) \log \relax (x) + 6 \, e^{16} \log \relax (2) \log \relax (x) + 18 \, e^{8} \log \relax (2) \log \relax (x) - e^{16} \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) \log \relax (x) - 3 \, e^{8} \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) \log \relax (x) - 3 \, e^{16} \log \relax (x)^{2} - 9 \, e^{8} \log \relax (x)^{2} - 3 \, e^{8} \log \relax (5) \log \left (\frac {4}{x}\right ) + 48 \, e^{16} \log \relax (2) \log \left (\frac {4}{x}\right ) + 3 \, e^{8} \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) \log \left (\frac {4}{x}\right ) - 24 \, e^{16} \log \relax (x) \log \left (\frac {4}{x}\right ) + 9 \, e^{8} \log \relax (x) \log \left (\frac {4}{x}\right ) + e^{8} \log \relax (5) - 16 \, e^{16} \log \relax (2) - 48 \, e^{8} \log \relax (2) - e^{8} \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) + 8 \, e^{16} \log \relax (x) + 21 \, e^{8} \log \relax (x) - 24 \, e^{8} \log \left (\frac {4}{x}\right ) + 8 \, e^{8} + 3 \, \log \relax (5) - 3 \, \log \left (e^{8} \log \left (\frac {4}{x}\right ) - 1\right ) - 9 \, \log \relax (x) + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(8)*log(4/x)+2*exp(8)+6)/(x*exp(8)*log(4/x)-x)/log(1/5*(x^3*exp(8)*log(4/x)-x^3)/exp(8))^2,x,
algorithm="giac")

[Out]

-2*(6*e^16*log(2)*log(4/x) - 3*e^16*log(x)*log(4/x) - 6*e^8*log(2) + 3*e^8*log(x) - e^16*log(4/x) - 3*e^8*log(
4/x) + e^8 + 3)/(6*e^16*log(5)*log(2)*log(4/x) - 6*e^16*log(2)*log(e^8*log(4/x) - 1)*log(4/x) - 3*e^16*log(5)*
log(x)*log(4/x) - 18*e^16*log(2)*log(x)*log(4/x) + 3*e^16*log(e^8*log(4/x) - 1)*log(x)*log(4/x) + 9*e^16*log(x
)^2*log(4/x) - 2*e^16*log(5)*log(2) - 6*e^8*log(5)*log(2) + 2*e^16*log(2)*log(e^8*log(4/x) - 1) + 6*e^8*log(2)
*log(e^8*log(4/x) - 1) + e^16*log(5)*log(x) + 3*e^8*log(5)*log(x) + 6*e^16*log(2)*log(x) + 18*e^8*log(2)*log(x
) - e^16*log(e^8*log(4/x) - 1)*log(x) - 3*e^8*log(e^8*log(4/x) - 1)*log(x) - 3*e^16*log(x)^2 - 9*e^8*log(x)^2
- 3*e^8*log(5)*log(4/x) + 48*e^16*log(2)*log(4/x) + 3*e^8*log(e^8*log(4/x) - 1)*log(4/x) - 24*e^16*log(x)*log(
4/x) + 9*e^8*log(x)*log(4/x) + e^8*log(5) - 16*e^16*log(2) - 48*e^8*log(2) - e^8*log(e^8*log(4/x) - 1) + 8*e^1
6*log(x) + 21*e^8*log(x) - 24*e^8*log(4/x) + 8*e^8 + 3*log(5) - 3*log(e^8*log(4/x) - 1) - 9*log(x) + 24)

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maple [A]  time = 0.12, size = 30, normalized size = 1.25

method	result	size
norman	\(\frac {2}{\ln \left (\frac {\left (x^{3} {\mathrm e}^{8} \ln \left (\frac {4}{x}\right )-x^{3}\right ) {\mathrm e}^{-8}}{5}\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*exp(8)*ln(4/x)+2*exp(8)+6)/(x*exp(8)*ln(4/x)-x)/ln(1/5*(x^3*exp(8)*ln(4/x)-x^3)/exp(8))^2,x,method=_RE
TURNVERBOSE)

[Out]

2/ln(1/5*(x^3*exp(8)*ln(4/x)-x^3)/exp(8))

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maxima [A]  time = 0.50, size = 29, normalized size = 1.21 \begin {gather*} -\frac {2}{\log \relax (5) - \log \left (2 \, e^{8} \log \relax (2) - e^{8} \log \relax (x) - 1\right ) - 3 \, \log \relax (x) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(8)*log(4/x)+2*exp(8)+6)/(x*exp(8)*log(4/x)-x)/log(1/5*(x^3*exp(8)*log(4/x)-x^3)/exp(8))^2,x,
algorithm="maxima")

[Out]

-2/(log(5) - log(2*e^8*log(2) - e^8*log(x) - 1) - 3*log(x) + 8)

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mupad [B]  time = 4.97, size = 24, normalized size = 1.00 \begin {gather*} \frac {2}{\ln \left (\frac {x^3\,\ln \left (\frac {4}{x}\right )}{5}-\frac {x^3\,{\mathrm {e}}^{-8}}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(8) - 6*exp(8)*log(4/x) + 6)/(log(-exp(-8)*(x^3/5 - (x^3*exp(8)*log(4/x))/5))^2*(x - x*exp(8)*log(4
/x))),x)

[Out]

2/log((x^3*log(4/x))/5 - (x^3*exp(-8))/5)

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sympy [A]  time = 0.44, size = 24, normalized size = 1.00 \begin {gather*} \frac {2}{\log {\left (\frac {\frac {x^{3} e^{8} \log {\left (\frac {4}{x} \right )}}{5} - \frac {x^{3}}{5}}{e^{8}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*exp(8)*ln(4/x)+2*exp(8)+6)/(x*exp(8)*ln(4/x)-x)/ln(1/5*(x**3*exp(8)*ln(4/x)-x**3)/exp(8))**2,x)

[Out]

2/log((x**3*exp(8)*log(4/x)/5 - x**3/5)*exp(-8))

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