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

Optimal. Leaf size=10 \[ -3+\frac {5}{\log (-4+x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 8, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {12, 2390, 2302, 30} \begin {gather*} \frac {5}{\log (x-4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5/((-4 + x)*Log[-4 + x]^2),x]

[Out]

5/Log[-4 + 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} &=-\left (5 \int \frac {1}{(-4+x) \log ^2(-4+x)} \, dx\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-4+x\right )\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-4+x)\right )\right )\\ &=\frac {5}{\log (-4+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 8, normalized size = 0.80 \begin {gather*} \frac {5}{\log (-4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5/Log[-4 + x]

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fricas [A]  time = 0.83, size = 8, normalized size = 0.80 \begin {gather*} \frac {5}{\log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/log(x - 4)

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giac [A]  time = 0.30, size = 8, normalized size = 0.80 \begin {gather*} \frac {5}{\log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/log(x - 4)

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maple [A]  time = 0.42, size = 9, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {5}{\ln \left (x -4\right )}\)	\(9\)
default	\(\frac {5}{\ln \left (x -4\right )}\)	\(9\)
norman	\(\frac {5}{\ln \left (x -4\right )}\)	\(9\)
risch	\(\frac {5}{\ln \left (x -4\right )}\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(x-4)/ln(x-4)^2,x,method=_RETURNVERBOSE)

[Out]

5/ln(x-4)

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maxima [A]  time = 0.47, size = 8, normalized size = 0.80 \begin {gather*} \frac {5}{\log \left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/log(x - 4)

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mupad [B]  time = 0.11, size = 8, normalized size = 0.80 \begin {gather*} \frac {5}{\ln \left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(log(x - 4)^2*(x - 4)),x)

[Out]

5/log(x - 4)

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sympy [A]  time = 0.08, size = 5, normalized size = 0.50 \begin {gather*} \frac {5}{\log {\left (x - 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/log(x - 4)

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