3.70.90 \(\int \frac {-4 x+(-20+4 x) \log (-5+x)}{(-5+x) \log ^2(-5+x)} \, dx\)

Optimal. Leaf size=9 \[ \frac {4 x}{\log (-5+x)} \]

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Rubi [B]  time = 0.18, antiderivative size = 22, normalized size of antiderivative = 2.44, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6742, 2411, 2353, 2297, 2298, 2302, 30, 2389} \begin {gather*} \frac {20}{\log (x-5)}-\frac {4 (5-x)}{\log (x-5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x + (-20 + 4*x)*Log[-5 + x])/((-5 + x)*Log[-5 + x]^2),x]

[Out]

20/Log[-5 + x] - (4*(5 - x))/Log[-5 + x]

Rule 30

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

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4 x}{(-5+x) \log ^2(-5+x)}+\frac {4}{\log (-5+x)}\right ) \, dx\\ &=-\left (4 \int \frac {x}{(-5+x) \log ^2(-5+x)} \, dx\right )+4 \int \frac {1}{\log (-5+x)} \, dx\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {5+x}{x \log ^2(x)} \, dx,x,-5+x\right )\right )+4 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )\\ &=4 \text {li}(-5+x)-4 \operatorname {Subst}\left (\int \left (\frac {1}{\log ^2(x)}+\frac {5}{x \log ^2(x)}\right ) \, dx,x,-5+x\right )\\ &=4 \text {li}(-5+x)-4 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-5+x\right )-20 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-5+x\right )\\ &=-\frac {4 (5-x)}{\log (-5+x)}+4 \text {li}(-5+x)-4 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-5+x\right )-20 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-5+x)\right )\\ &=\frac {20}{\log (-5+x)}-\frac {4 (5-x)}{\log (-5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 9, normalized size = 1.00 \begin {gather*} \frac {4 x}{\log (-5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + (-20 + 4*x)*Log[-5 + x])/((-5 + x)*Log[-5 + x]^2),x]

[Out]

(4*x)/Log[-5 + x]

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fricas [A]  time = 0.56, size = 9, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{\log \left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-20)*log(x-5)-4*x)/(x-5)/log(x-5)^2,x, algorithm="fricas")

[Out]

4*x/log(x - 5)

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giac [A]  time = 0.13, size = 9, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{\log \left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-20)*log(x-5)-4*x)/(x-5)/log(x-5)^2,x, algorithm="giac")

[Out]

4*x/log(x - 5)

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maple [A]  time = 0.09, size = 10, normalized size = 1.11

method	result	size
norman	\(\frac {4 x}{\ln \left (x -5\right )}\)	\(10\)
risch	\(\frac {4 x}{\ln \left (x -5\right )}\)	\(10\)
derivativedivides	\(\frac {20}{\ln \left (x -5\right )}+\frac {4 x -20}{\ln \left (x -5\right )}\)	\(21\)
default	\(\frac {20}{\ln \left (x -5\right )}+\frac {4 x -20}{\ln \left (x -5\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x-20)*ln(x-5)-4*x)/(x-5)/ln(x-5)^2,x,method=_RETURNVERBOSE)

[Out]

4*x/ln(x-5)

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maxima [A]  time = 0.41, size = 11, normalized size = 1.22 \begin {gather*} \frac {4 \, x}{\log \left (x - 5\right )} + 20 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-20)*log(x-5)-4*x)/(x-5)/log(x-5)^2,x, algorithm="maxima")

[Out]

4*x/log(x - 5) + 20

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mupad [B]  time = 0.20, size = 9, normalized size = 1.00 \begin {gather*} \frac {4\,x}{\ln \left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - log(x - 5)*(4*x - 20))/(log(x - 5)^2*(x - 5)),x)

[Out]

(4*x)/log(x - 5)

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sympy [A]  time = 0.09, size = 7, normalized size = 0.78 \begin {gather*} \frac {4 x}{\log {\left (x - 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-20)*ln(x-5)-4*x)/(x-5)/ln(x-5)**2,x)

[Out]

4*x/log(x - 5)

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