3.99.41 \(\int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{5 e^7 x^3+2 e^{e^5} x \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (5+\frac {2 e^{-7+e^5} \log (x)}{x^2}\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2561, 6712, 31} \begin {gather*} \log \left (\frac {2 e^{e^5} \log (x)}{x^2}+5 e^7\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^E^5 - 4*E^E^5*Log[x])/(5*E^7*x^3 + 2*E^E^5*x*Log[x]),x]

[Out]

Log[5*E^7 + (2*E^E^5*Log[x])/x^2]

Rule 31

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

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 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{e^5}-4 e^{e^5} \log (x)}{x \left (5 e^7 x^2+2 e^{e^5} \log (x)\right )} \, dx\\ &=\left (2 e^{e^5}\right ) \operatorname {Subst}\left (\int \frac {1}{5 e^7+2 e^{e^5} x} \, dx,x,\frac {\log (x)}{x^2}\right )\\ &=\log \left (5 e^7+\frac {2 e^{e^5} \log (x)}{x^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 1.41 \begin {gather*} -2 \log (x)+\log \left (5 e^7 x^2+2 e^{e^5} \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^E^5 - 4*E^E^5*Log[x])/(5*E^7*x^3 + 2*E^E^5*x*Log[x]),x]

[Out]

-2*Log[x] + Log[5*E^7*x^2 + 2*E^E^5*Log[x]]

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fricas [A]  time = 0.78, size = 21, normalized size = 1.24 \begin {gather*} \log \left (5 \, x^{2} e^{7} + 2 \, e^{\left (e^{5}\right )} \log \relax (x)\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(5*x^2*e^7 + 2*e^(e^5)*log(x)) - 2*log(x)

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giac [A]  time = 0.12, size = 21, normalized size = 1.24 \begin {gather*} \log \left (-5 \, x^{2} e^{7} - 2 \, e^{\left (e^{5}\right )} \log \relax (x)\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-5*x^2*e^7 - 2*e^(e^5)*log(x)) - 2*log(x)

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maple [A]  time = 0.05, size = 22, normalized size = 1.29

method	result	size
risch	\(-2 \ln \relax (x )+\ln \left (\ln \relax (x )+\frac {5 x^{2} {\mathrm e}^{7-{\mathrm e}^{5}}}{2}\right )\)	\(22\)
norman	\(-2 \ln \relax (x )+\ln \left (5 x^{2} {\mathrm e} \,{\mathrm e}^{6}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \ln \relax (x )\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(exp(5))*ln(x)+2*exp(exp(5)))/(2*x*exp(exp(5))*ln(x)+5*x^3*exp(1)*exp(3)^2),x,method=_RETURNVERBOSE
)

[Out]

-2*ln(x)+ln(ln(x)+5/2*x^2*exp(7-exp(5)))

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maxima [A]  time = 0.38, size = 28, normalized size = 1.65 \begin {gather*} \log \left (\frac {1}{2} \, {\left (5 \, x^{2} e^{7} + 2 \, e^{\left (e^{5}\right )} \log \relax (x)\right )} e^{\left (-e^{5}\right )}\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/2*(5*x^2*e^7 + 2*e^(e^5)*log(x))*e^(-e^5)) - 2*log(x)

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mupad [B]  time = 5.71, size = 19, normalized size = 1.12 \begin {gather*} \ln \left (\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^5-7}\,\ln \relax (x)}{5}+x^2\right )-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(exp(5)) - 4*exp(exp(5))*log(x))/(5*x^3*exp(7) + 2*x*exp(exp(5))*log(x)),x)

[Out]

log((2*exp(exp(5) - 7)*log(x))/5 + x^2) - 2*log(x)

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sympy [A]  time = 0.18, size = 24, normalized size = 1.41 \begin {gather*} - 2 \log {\relax (x )} + \log {\left (\frac {5 x^{2} e^{7}}{2 e^{e^{5}}} + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(x) + log(5*x**2*exp(7)*exp(-exp(5))/2 + log(x))

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