3.83.69 \(\int -\frac {9}{(-3 x+e^5 x+x \log (x)) \log ^2(-3+e^5+\log (x))} \, dx\)

Optimal. Leaf size=14 \[ -4+\frac {9}{\log \left (-3+e^5+\log (x)\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6, 12, 2390, 2302, 30} \begin {gather*} \frac {9}{\log \left (\log (x)+e^5-3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-9/((-3*x + E^5*x + x*Log[x])*Log[-3 + E^5 + Log[x]]^2),x]

[Out]

9/Log[-3 + E^5 + Log[x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int -\frac {9}{\left (\left (-3+e^5\right ) x+x \log (x)\right ) \log ^2\left (-3+e^5+\log (x)\right )} \, dx\\ &=-\left (9 \int \frac {1}{\left (\left (-3+e^5\right ) x+x \log (x)\right ) \log ^2\left (-3+e^5+\log (x)\right )} \, dx\right )\\ &=-\left (9 \operatorname {Subst}\left (\int \frac {1}{\left (-3+e^5+x\right ) \log ^2\left (-3+e^5+x\right )} \, dx,x,\log (x)\right )\right )\\ &=-\left (9 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-3+e^5+\log (x)\right )\right )\\ &=-\left (9 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (-3+e^5+\log (x)\right )\right )\right )\\ &=\frac {9}{\log \left (-3+e^5+\log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.86 \begin {gather*} \frac {9}{\log \left (-3+e^5+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-9/((-3*x + E^5*x + x*Log[x])*Log[-3 + E^5 + Log[x]]^2),x]

[Out]

9/Log[-3 + E^5 + Log[x]]

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fricas [A]  time = 0.77, size = 11, normalized size = 0.79 \begin {gather*} \frac {9}{\log \left (e^{5} + \log \relax (x) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/(x*log(x)+x*exp(5)-3*x)/log(log(x)+exp(5)-3)^2,x, algorithm="fricas")

[Out]

9/log(e^5 + log(x) - 3)

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giac [A]  time = 0.13, size = 11, normalized size = 0.79 \begin {gather*} \frac {9}{\log \left (e^{5} + \log \relax (x) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/(x*log(x)+x*exp(5)-3*x)/log(log(x)+exp(5)-3)^2,x, algorithm="giac")

[Out]

9/log(e^5 + log(x) - 3)

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maple [A]  time = 0.10, size = 12, normalized size = 0.86

method	result	size
default	\(\frac {9}{\ln \left (\ln \relax (x )+{\mathrm e}^{5}-3\right )}\)	\(12\)
norman	\(\frac {9}{\ln \left (\ln \relax (x )+{\mathrm e}^{5}-3\right )}\)	\(12\)
risch	\(\frac {9}{\ln \left (\ln \relax (x )+{\mathrm e}^{5}-3\right )}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9/(x*ln(x)+x*exp(5)-3*x)/ln(ln(x)+exp(5)-3)^2,x,method=_RETURNVERBOSE)

[Out]

9/ln(ln(x)+exp(5)-3)

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maxima [A]  time = 0.37, size = 11, normalized size = 0.79 \begin {gather*} \frac {9}{\log \left (e^{5} + \log \relax (x) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/(x*log(x)+x*exp(5)-3*x)/log(log(x)+exp(5)-3)^2,x, algorithm="maxima")

[Out]

9/log(e^5 + log(x) - 3)

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mupad [B]  time = 5.58, size = 11, normalized size = 0.79 \begin {gather*} \frac {9}{\ln \left ({\mathrm {e}}^5+\ln \relax (x)-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9/(log(exp(5) + log(x) - 3)^2*(x*exp(5) - 3*x + x*log(x))),x)

[Out]

9/log(exp(5) + log(x) - 3)

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sympy [A]  time = 0.26, size = 10, normalized size = 0.71 \begin {gather*} \frac {9}{\log {\left (\log {\relax (x )} - 3 + e^{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/(x*ln(x)+x*exp(5)-3*x)/ln(ln(x)+exp(5)-3)**2,x)

[Out]

9/log(log(x) - 3 + exp(5))

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