3.25.48 \(\int \frac {-5 x+(-30-5 x) \log (6+x)}{54+9 x+(-72-12 x) \log (2)+(24+4 x) \log ^2(2)+(36 x+6 x^2+(-24 x-4 x^2) \log (2)) \log (6+x)+(6 x^2+x^3) \log ^2(6+x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {5}{3-2 \log (2)+x \log (6+x)} \]

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Rubi [A]  time = 0.16, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6688, 12, 6686} \begin {gather*} \frac {5}{x \log (x+6)+3-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x + (-30 - 5*x)*Log[6 + x])/(54 + 9*x + (-72 - 12*x)*Log[2] + (24 + 4*x)*Log[2]^2 + (36*x + 6*x^2 + (-
24*x - 4*x^2)*Log[2])*Log[6 + x] + (6*x^2 + x^3)*Log[6 + x]^2),x]

[Out]

5/(3 - Log[4] + x*Log[6 + 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} &=\int \frac {5 (-x-(6+x) \log (6+x))}{(6+x) \left (3 \left (1-\frac {2 \log (2)}{3}\right )+x \log (6+x)\right )^2} \, dx\\ &=5 \int \frac {-x-(6+x) \log (6+x)}{(6+x) \left (3 \left (1-\frac {2 \log (2)}{3}\right )+x \log (6+x)\right )^2} \, dx\\ &=\frac {5}{3-\log (4)+x \log (6+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 0.94 \begin {gather*} -\frac {5}{-3+\log (4)-x \log (6+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x + (-30 - 5*x)*Log[6 + x])/(54 + 9*x + (-72 - 12*x)*Log[2] + (24 + 4*x)*Log[2]^2 + (36*x + 6*x^
2 + (-24*x - 4*x^2)*Log[2])*Log[6 + x] + (6*x^2 + x^3)*Log[6 + x]^2),x]

[Out]

-5/(-3 + Log[4] - x*Log[6 + x])

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fricas [A]  time = 0.64, size = 16, normalized size = 1.00 \begin {gather*} \frac {5}{x \log \left (x + 6\right ) - 2 \, \log \relax (2) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-30)*log(x+6)-5*x)/((x^3+6*x^2)*log(x+6)^2+((-4*x^2-24*x)*log(2)+6*x^2+36*x)*log(x+6)+(4*x+24)
*log(2)^2+(-12*x-72)*log(2)+9*x+54),x, algorithm="fricas")

[Out]

5/(x*log(x + 6) - 2*log(2) + 3)

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giac [A]  time = 0.35, size = 16, normalized size = 1.00 \begin {gather*} \frac {5}{x \log \left (x + 6\right ) - 2 \, \log \relax (2) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-30)*log(x+6)-5*x)/((x^3+6*x^2)*log(x+6)^2+((-4*x^2-24*x)*log(2)+6*x^2+36*x)*log(x+6)+(4*x+24)
*log(2)^2+(-12*x-72)*log(2)+9*x+54),x, algorithm="giac")

[Out]

5/(x*log(x + 6) - 2*log(2) + 3)

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maple [A]  time = 0.12, size = 18, normalized size = 1.12

method	result	size
norman	\(-\frac {5}{-x \ln \left (x +6\right )+2 \ln \relax (2)-3}\)	\(18\)
risch	\(-\frac {5}{-x \ln \left (x +6\right )+2 \ln \relax (2)-3}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x-30)*ln(x+6)-5*x)/((x^3+6*x^2)*ln(x+6)^2+((-4*x^2-24*x)*ln(2)+6*x^2+36*x)*ln(x+6)+(4*x+24)*ln(2)^2+(
-12*x-72)*ln(2)+9*x+54),x,method=_RETURNVERBOSE)

[Out]

-5/(-x*ln(x+6)+2*ln(2)-3)

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maxima [A]  time = 0.57, size = 16, normalized size = 1.00 \begin {gather*} \frac {5}{x \log \left (x + 6\right ) - 2 \, \log \relax (2) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-30)*log(x+6)-5*x)/((x^3+6*x^2)*log(x+6)^2+((-4*x^2-24*x)*log(2)+6*x^2+36*x)*log(x+6)+(4*x+24)
*log(2)^2+(-12*x-72)*log(2)+9*x+54),x, algorithm="maxima")

[Out]

5/(x*log(x + 6) - 2*log(2) + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int -\frac {5\,x+\ln \left (x+6\right )\,\left (5\,x+30\right )}{\left (x^3+6\,x^2\right )\,{\ln \left (x+6\right )}^2+\left (36\,x-\ln \relax (2)\,\left (4\,x^2+24\,x\right )+6\,x^2\right )\,\ln \left (x+6\right )+9\,x-\ln \relax (2)\,\left (12\,x+72\right )+{\ln \relax (2)}^2\,\left (4\,x+24\right )+54} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + log(x + 6)*(5*x + 30))/(9*x - log(2)*(12*x + 72) + log(x + 6)*(36*x - log(2)*(24*x + 4*x^2) + 6*x^
2) + log(2)^2*(4*x + 24) + log(x + 6)^2*(6*x^2 + x^3) + 54),x)

[Out]

int(-(5*x + log(x + 6)*(5*x + 30))/(9*x - log(2)*(12*x + 72) + log(x + 6)*(36*x - log(2)*(24*x + 4*x^2) + 6*x^
2) + log(2)^2*(4*x + 24) + log(x + 6)^2*(6*x^2 + x^3) + 54), x)

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sympy [A]  time = 0.17, size = 14, normalized size = 0.88 \begin {gather*} \frac {5}{x \log {\left (x + 6 \right )} - 2 \log {\relax (2 )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x-30)*ln(x+6)-5*x)/((x**3+6*x**2)*ln(x+6)**2+((-4*x**2-24*x)*ln(2)+6*x**2+36*x)*ln(x+6)+(4*x+24
)*ln(2)**2+(-12*x-72)*ln(2)+9*x+54),x)

[Out]

5/(x*log(x + 6) - 2*log(2) + 3)

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