3.15.9 \(\int \frac {(744-48 x) \log (3)}{-256-240 x-48 x^2+4 x^3+(-128-56 x+4 x^2) \log (-16+x)+(-16+x) \log ^2(-16+x)} \, dx\)

Optimal. Leaf size=15 \[ \frac {24 \log (3)}{4+2 x+\log (-16+x)} \]

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Rubi [A]  time = 0.16, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 6688, 6686} \begin {gather*} \frac {24 \log (3)}{2 x+\log (x-16)+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((744 - 48*x)*Log[3])/(-256 - 240*x - 48*x^2 + 4*x^3 + (-128 - 56*x + 4*x^2)*Log[-16 + x] + (-16 + x)*Log[
-16 + x]^2),x]

[Out]

(24*Log[3])/(4 + 2*x + Log[-16 + 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 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]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \int \frac {744-48 x}{-256-240 x-48 x^2+4 x^3+\left (-128-56 x+4 x^2\right ) \log (-16+x)+(-16+x) \log ^2(-16+x)} \, dx\\ &=\log (3) \int \frac {-744+48 x}{(16-x) (4+2 x+\log (-16+x))^2} \, dx\\ &=\frac {24 \log (3)}{4+2 x+\log (-16+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 1.00 \begin {gather*} \frac {24 \log (3)}{4+2 x+\log (-16+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((744 - 48*x)*Log[3])/(-256 - 240*x - 48*x^2 + 4*x^3 + (-128 - 56*x + 4*x^2)*Log[-16 + x] + (-16 + x
)*Log[-16 + x]^2),x]

[Out]

(24*Log[3])/(4 + 2*x + Log[-16 + x])

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fricas [A]  time = 0.60, size = 15, normalized size = 1.00 \begin {gather*} \frac {24 \, \log \relax (3)}{2 \, x + \log \left (x - 16\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x+744)*log(3)/((x-16)*log(x-16)^2+(4*x^2-56*x-128)*log(x-16)+4*x^3-48*x^2-240*x-256),x, algorit
hm="fricas")

[Out]

24*log(3)/(2*x + log(x - 16) + 4)

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giac [A]  time = 0.35, size = 15, normalized size = 1.00 \begin {gather*} \frac {24 \, \log \relax (3)}{2 \, x + \log \left (x - 16\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x+744)*log(3)/((x-16)*log(x-16)^2+(4*x^2-56*x-128)*log(x-16)+4*x^3-48*x^2-240*x-256),x, algorit
hm="giac")

[Out]

24*log(3)/(2*x + log(x - 16) + 4)

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maple [A]  time = 0.07, size = 16, normalized size = 1.07

method	result	size
norman	\(\frac {24 \ln \relax (3)}{\ln \left (x -16\right )+2 x +4}\)	\(16\)
risch	\(\frac {24 \ln \relax (3)}{\ln \left (x -16\right )+2 x +4}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*x+744)*ln(3)/((x-16)*ln(x-16)^2+(4*x^2-56*x-128)*ln(x-16)+4*x^3-48*x^2-240*x-256),x,method=_RETURNVER
BOSE)

[Out]

24*ln(3)/(ln(x-16)+2*x+4)

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maxima [A]  time = 0.72, size = 15, normalized size = 1.00 \begin {gather*} \frac {24 \, \log \relax (3)}{2 \, x + \log \left (x - 16\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x+744)*log(3)/((x-16)*log(x-16)^2+(4*x^2-56*x-128)*log(x-16)+4*x^3-48*x^2-240*x-256),x, algorit
hm="maxima")

[Out]

24*log(3)/(2*x + log(x - 16) + 4)

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mupad [B]  time = 1.12, size = 23, normalized size = 1.53 \begin {gather*} -\frac {6\,\ln \relax (3)\,\left (2\,x+\ln \left (x-16\right )\right )}{2\,x+\ln \left (x-16\right )+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)*(48*x - 744))/(240*x + log(x - 16)*(56*x - 4*x^2 + 128) + 48*x^2 - 4*x^3 - log(x - 16)^2*(x - 16)
+ 256),x)

[Out]

-(6*log(3)*(2*x + log(x - 16)))/(2*x + log(x - 16) + 4)

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sympy [A]  time = 0.13, size = 14, normalized size = 0.93 \begin {gather*} \frac {24 \log {\relax (3 )}}{2 x + \log {\left (x - 16 \right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x+744)*ln(3)/((x-16)*ln(x-16)**2+(4*x**2-56*x-128)*ln(x-16)+4*x**3-48*x**2-240*x-256),x)

[Out]

24*log(3)/(2*x + log(x - 16) + 4)

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