3.86.96 \(\int \frac {-3 x-3 x \log (3)+(12+3 x+(12+3 x) \log (3)) \log (4+x)}{4 x^2+x^3+(8 x+2 x^2) \log (4+x)+(4+x) \log ^2(4+x)} \, dx\)

Optimal. Leaf size=15 \[ \frac {3 x (1+\log (3))}{x+\log (4+x)} \]

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Rubi [A]  time = 0.15, antiderivative size = 18, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 5, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 6688, 12, 6711, 32} \begin {gather*} -\frac {3 (1+\log (3))}{\frac {x}{\log (x+4)}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x - 3*x*Log[3] + (12 + 3*x + (12 + 3*x)*Log[3])*Log[4 + x])/(4*x^2 + x^3 + (8*x + 2*x^2)*Log[4 + x] +
(4 + x)*Log[4 + x]^2),x]

[Out]

(-3*(1 + Log[3]))/(1 + x/Log[4 + 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((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*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (-3-3 \log (3))+(12+3 x+(12+3 x) \log (3)) \log (4+x)}{4 x^2+x^3+\left (8 x+2 x^2\right ) \log (4+x)+(4+x) \log ^2(4+x)} \, dx\\ &=\int \frac {3 (1+\log (3)) (-x+(4+x) \log (4+x))}{(4+x) (x+\log (4+x))^2} \, dx\\ &=(3 (1+\log (3))) \int \frac {-x+(4+x) \log (4+x)}{(4+x) (x+\log (4+x))^2} \, dx\\ &=(3 (1+\log (3))) \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log (4+x)}\right )\\ &=-\frac {3 (1+\log (3))}{1+\frac {x}{\log (4+x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 15, normalized size = 1.00 \begin {gather*} \frac {3 x (1+\log (3))}{x+\log (4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x - 3*x*Log[3] + (12 + 3*x + (12 + 3*x)*Log[3])*Log[4 + x])/(4*x^2 + x^3 + (8*x + 2*x^2)*Log[4 +
x] + (4 + x)*Log[4 + x]^2),x]

[Out]

(3*x*(1 + Log[3]))/(x + Log[4 + x])

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fricas [A]  time = 0.78, size = 16, normalized size = 1.07 \begin {gather*} \frac {3 \, {\left (x \log \relax (3) + x\right )}}{x + \log \left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+12)*log(3)+3*x+12)*log(4+x)-3*x*log(3)-3*x)/((4+x)*log(4+x)^2+(2*x^2+8*x)*log(4+x)+x^3+4*x^2)
,x, algorithm="fricas")

[Out]

3*(x*log(3) + x)/(x + log(x + 4))

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giac [A]  time = 0.16, size = 16, normalized size = 1.07 \begin {gather*} \frac {3 \, {\left (x \log \relax (3) + x\right )}}{x + \log \left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+12)*log(3)+3*x+12)*log(4+x)-3*x*log(3)-3*x)/((4+x)*log(4+x)^2+(2*x^2+8*x)*log(4+x)+x^3+4*x^2)
,x, algorithm="giac")

[Out]

3*(x*log(3) + x)/(x + log(x + 4))

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maple [A]  time = 0.13, size = 16, normalized size = 1.07

method	result	size
risch	\(\frac {3 x \left (\ln \relax (3)+1\right )}{\ln \left (4+x \right )+x}\)	\(16\)
norman	\(\frac {\left (-3 \ln \relax (3)-3\right ) \ln \left (4+x \right )}{\ln \left (4+x \right )+x}\)	\(20\)
derivativedivides	\(\frac {3 x}{\ln \left (4+x \right )+x}-\frac {3 \ln \relax (3) \ln \left (4+x \right )}{\ln \left (4+x \right )+x}\)	\(29\)
default	\(\frac {3 x}{\ln \left (4+x \right )+x}-\frac {3 \ln \relax (3) \ln \left (4+x \right )}{\ln \left (4+x \right )+x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x+12)*ln(3)+3*x+12)*ln(4+x)-3*x*ln(3)-3*x)/((4+x)*ln(4+x)^2+(2*x^2+8*x)*ln(4+x)+x^3+4*x^2),x,method=_
RETURNVERBOSE)

[Out]

3*x/(ln(4+x)+x)*(ln(3)+1)

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maxima [A]  time = 0.48, size = 15, normalized size = 1.00 \begin {gather*} \frac {3 \, x {\left (\log \relax (3) + 1\right )}}{x + \log \left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+12)*log(3)+3*x+12)*log(4+x)-3*x*log(3)-3*x)/((4+x)*log(4+x)^2+(2*x^2+8*x)*log(4+x)+x^3+4*x^2)
,x, algorithm="maxima")

[Out]

3*x*(log(3) + 1)/(x + log(x + 4))

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mupad [B]  time = 0.25, size = 63, normalized size = 4.20 \begin {gather*} \frac {\left (\ln \left (27\right )+3\right )\,x^3+\left (24\,\ln \relax (3)+24\right )\,x^2+\left (48\,\ln \relax (3)+48\right )\,x}{16\,x+16\,\ln \left (x+4\right )+8\,x\,\ln \left (x+4\right )+x^2\,\ln \left (x+4\right )+8\,x^2+x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 3*x*log(3) - log(x + 4)*(3*x + log(3)*(3*x + 12) + 12))/(log(x + 4)*(8*x + 2*x^2) + 4*x^2 + x^3 +
log(x + 4)^2*(x + 4)),x)

[Out]

(x*(48*log(3) + 48) + x^3*(log(27) + 3) + x^2*(24*log(3) + 24))/(16*x + 16*log(x + 4) + 8*x*log(x + 4) + x^2*l
og(x + 4) + 8*x^2 + x^3)

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sympy [A]  time = 0.13, size = 15, normalized size = 1.00 \begin {gather*} \frac {3 x + 3 x \log {\relax (3 )}}{x + \log {\left (x + 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x+12)*ln(3)+3*x+12)*ln(4+x)-3*x*ln(3)-3*x)/((4+x)*ln(4+x)**2+(2*x**2+8*x)*ln(4+x)+x**3+4*x**2),
x)

[Out]

(3*x + 3*x*log(3))/(x + log(x + 4))

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