3.84.82 \(\int \frac {-20+60 x+185 x^2+15 x^3+(120 x+10 x^2) \log (x)}{48 x^6+4 x^7+(96 x^5+8 x^6) \log (x)+(48 x^4+4 x^5) \log ^2(x)+(-384 x^3-32 x^4+(-384 x^2-32 x^3) \log (x)) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {5 x}{4 \left (-x^3 (x+\log (x))+4 x \log (12+x)\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {5}{4 \left (x^3+x^2 \log (x)-4 \log (x+12)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 + 60*x + 185*x^2 + 15*x^3 + (120*x + 10*x^2)*Log[x])/(48*x^6 + 4*x^7 + (96*x^5 + 8*x^6)*Log[x] + (48*
x^4 + 4*x^5)*Log[x]^2 + (-384*x^3 - 32*x^4 + (-384*x^2 - 32*x^3)*Log[x])*Log[12 + x] + (768 + 64*x)*Log[12 + x
]^2),x]

[Out]

-5/(4*(x^3 + x^2*Log[x] - 4*Log[12 + 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 \left (-4+12 x+37 x^2+3 x^3+2 x (12+x) \log (x)\right )}{4 (12+x) \left (x^3+x^2 \log (x)-4 \log (12+x)\right )^2} \, dx\\ &=\frac {5}{4} \int \frac {-4+12 x+37 x^2+3 x^3+2 x (12+x) \log (x)}{(12+x) \left (x^3+x^2 \log (x)-4 \log (12+x)\right )^2} \, dx\\ &=-\frac {5}{4 \left (x^3+x^2 \log (x)-4 \log (12+x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.92 \begin {gather*} -\frac {5}{4 \left (x^3+x^2 \log (x)-4 \log (12+x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + 60*x + 185*x^2 + 15*x^3 + (120*x + 10*x^2)*Log[x])/(48*x^6 + 4*x^7 + (96*x^5 + 8*x^6)*Log[x]
+ (48*x^4 + 4*x^5)*Log[x]^2 + (-384*x^3 - 32*x^4 + (-384*x^2 - 32*x^3)*Log[x])*Log[12 + x] + (768 + 64*x)*Log[
12 + x]^2),x]

[Out]

-5/(4*(x^3 + x^2*Log[x] - 4*Log[12 + x]))

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fricas [A]  time = 0.45, size = 20, normalized size = 0.83 \begin {gather*} -\frac {5}{4 \, {\left (x^{3} + x^{2} \log \relax (x) - 4 \, \log \left (x + 12\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+120*x)*log(x)+15*x^3+185*x^2+60*x-20)/((64*x+768)*log(x+12)^2+((-32*x^3-384*x^2)*log(x)-32*
x^4-384*x^3)*log(x+12)+(4*x^5+48*x^4)*log(x)^2+(8*x^6+96*x^5)*log(x)+4*x^7+48*x^6),x, algorithm="fricas")

[Out]

-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))

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giac [A]  time = 0.15, size = 20, normalized size = 0.83 \begin {gather*} -\frac {5}{4 \, {\left (x^{3} + x^{2} \log \relax (x) - 4 \, \log \left (x + 12\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+120*x)*log(x)+15*x^3+185*x^2+60*x-20)/((64*x+768)*log(x+12)^2+((-32*x^3-384*x^2)*log(x)-32*
x^4-384*x^3)*log(x+12)+(4*x^5+48*x^4)*log(x)^2+(8*x^6+96*x^5)*log(x)+4*x^7+48*x^6),x, algorithm="giac")

[Out]

-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

method	result	size
risch	\(-\frac {5}{4 \left (x^{2} \ln \relax (x )+x^{3}-4 \ln \left (x +12\right )\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+120*x)*ln(x)+15*x^3+185*x^2+60*x-20)/((64*x+768)*ln(x+12)^2+((-32*x^3-384*x^2)*ln(x)-32*x^4-384*x
^3)*ln(x+12)+(4*x^5+48*x^4)*ln(x)^2+(8*x^6+96*x^5)*ln(x)+4*x^7+48*x^6),x,method=_RETURNVERBOSE)

[Out]

-5/4/(x^2*ln(x)+x^3-4*ln(x+12))

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maxima [A]  time = 0.40, size = 20, normalized size = 0.83 \begin {gather*} -\frac {5}{4 \, {\left (x^{3} + x^{2} \log \relax (x) - 4 \, \log \left (x + 12\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+120*x)*log(x)+15*x^3+185*x^2+60*x-20)/((64*x+768)*log(x+12)^2+((-32*x^3-384*x^2)*log(x)-32*
x^4-384*x^3)*log(x+12)+(4*x^5+48*x^4)*log(x)^2+(8*x^6+96*x^5)*log(x)+4*x^7+48*x^6),x, algorithm="maxima")

[Out]

-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {60\,x+\ln \relax (x)\,\left (10\,x^2+120\,x\right )+185\,x^2+15\,x^3-20}{\ln \relax (x)\,\left (8\,x^6+96\,x^5\right )+{\ln \left (x+12\right )}^2\,\left (64\,x+768\right )+{\ln \relax (x)}^2\,\left (4\,x^5+48\,x^4\right )-\ln \left (x+12\right )\,\left (\ln \relax (x)\,\left (32\,x^3+384\,x^2\right )+384\,x^3+32\,x^4\right )+48\,x^6+4\,x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x + log(x)*(120*x + 10*x^2) + 185*x^2 + 15*x^3 - 20)/(log(x)*(96*x^5 + 8*x^6) + log(x + 12)^2*(64*x +
768) + log(x)^2*(48*x^4 + 4*x^5) - log(x + 12)*(log(x)*(384*x^2 + 32*x^3) + 384*x^3 + 32*x^4) + 48*x^6 + 4*x^7
),x)

[Out]

int((60*x + log(x)*(120*x + 10*x^2) + 185*x^2 + 15*x^3 - 20)/(log(x)*(96*x^5 + 8*x^6) + log(x + 12)^2*(64*x +
768) + log(x)^2*(48*x^4 + 4*x^5) - log(x + 12)*(log(x)*(384*x^2 + 32*x^3) + 384*x^3 + 32*x^4) + 48*x^6 + 4*x^7
), x)

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sympy [A]  time = 0.32, size = 20, normalized size = 0.83 \begin {gather*} \frac {5}{- 4 x^{3} - 4 x^{2} \log {\relax (x )} + 16 \log {\left (x + 12 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+120*x)*ln(x)+15*x**3+185*x**2+60*x-20)/((64*x+768)*ln(x+12)**2+((-32*x**3-384*x**2)*ln(x)-
32*x**4-384*x**3)*ln(x+12)+(4*x**5+48*x**4)*ln(x)**2+(8*x**6+96*x**5)*ln(x)+4*x**7+48*x**6),x)

[Out]

5/(-4*x**3 - 4*x**2*log(x) + 16*log(x + 12))

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