Integrand size = 113, antiderivative size = 24 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=\frac {5 x}{4 \left (-x^3 (x+\log (x))+4 x \log (12+x)\right )} \] Output:
5/4*x/(4*ln(x+12)*x-x^3*(x+ln(x)))
Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=-\frac {5}{4 \left (x^3+x^2 \log (x)-4 \log (12+x)\right )} \] Input:
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]
Output:
-5/(4*(x^3 + x^2*Log[x] - 4*Log[12 + x]))
Time = 0.52 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7239, 27, 25, 7237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {15 x^3+185 x^2+\left (10 x^2+120 x\right ) \log (x)+60 x-20}{4 x^7+48 x^6+\left (8 x^6+96 x^5\right ) \log (x)+\left (4 x^5+48 x^4\right ) \log ^2(x)+\left (-32 x^4-384 x^3+\left (-32 x^3-384 x^2\right ) \log (x)\right ) \log (x+12)+(64 x+768) \log ^2(x+12)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (3 x^3+37 x^2+12 x+2 (x+12) x \log (x)-4\right )}{4 (x+12) \left (x^3+x^2 \log (x)-4 \log (x+12)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{4} \int -\frac {-3 x^3-37 x^2-2 (x+12) \log (x) x-12 x+4}{(x+12) \left (x^3+\log (x) x^2-4 \log (x+12)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {5}{4} \int \frac {-3 x^3-37 x^2-2 (x+12) \log (x) x-12 x+4}{(x+12) \left (x^3+\log (x) x^2-4 \log (x+12)\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {5}{4 \left (x^3+x^2 \log (x)-4 \log (x+12)\right )}\) |
Input:
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]
Output:
-5/(4*(x^3 + x^2*Log[x] - 4*Log[12 + x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {5}{4 \left (x^{3}+x^{2} \ln \left (x \right )-4 \ln \left (x +12\right )\right )}\) | \(21\) |
risch | \(-\frac {5}{4 \left (x^{3}+x^{2} \ln \left (x \right )-4 \ln \left (x +12\right )\right )}\) | \(21\) |
parallelrisch | \(-\frac {5}{4 \left (x^{3}+x^{2} \ln \left (x \right )-4 \ln \left (x +12\right )\right )}\) | \(21\) |
Input:
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)
Output:
-5/4/(x^3+x^2*ln(x)-4*ln(x+12))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=-\frac {5}{4 \, {\left (x^{3} + x^{2} \log \left (x\right ) - 4 \, \log \left (x + 12\right )\right )}} \] Input:
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")
Output:
-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=\frac {5}{- 4 x^{3} - 4 x^{2} \log {\left (x \right )} + 16 \log {\left (x + 12 \right )}} \] Input:
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)
Output:
5/(-4*x**3 - 4*x**2*log(x) + 16*log(x + 12))
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=-\frac {5}{4 \, {\left (x^{3} + x^{2} \log \left (x\right ) - 4 \, \log \left (x + 12\right )\right )}} \] Input:
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")
Output:
-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=-\frac {5}{4 \, {\left (x^{3} + x^{2} \log \left (x\right ) - 4 \, \log \left (x + 12\right )\right )}} \] Input:
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")
Output:
-5/4/(x^3 + x^2*log(x) - 4*log(x + 12))
Timed out. \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=\int \frac {60\,x+\ln \left (x\right )\,\left (10\,x^2+120\,x\right )+185\,x^2+15\,x^3-20}{\ln \left (x\right )\,\left (8\,x^6+96\,x^5\right )+{\ln \left (x+12\right )}^2\,\left (64\,x+768\right )+{\ln \left (x\right )}^2\,\left (4\,x^5+48\,x^4\right )-\ln \left (x+12\right )\,\left (\ln \left (x\right )\,\left (32\,x^3+384\,x^2\right )+384\,x^3+32\,x^4\right )+48\,x^6+4\,x^7} \,d x \] Input:
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)
Output:
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)
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-20+60 x+185 x^2+15 x^3+\left (120 x+10 x^2\right ) \log (x)}{48 x^6+4 x^7+\left (96 x^5+8 x^6\right ) \log (x)+\left (48 x^4+4 x^5\right ) \log ^2(x)+\left (-384 x^3-32 x^4+\left (-384 x^2-32 x^3\right ) \log (x)\right ) \log (12+x)+(768+64 x) \log ^2(12+x)} \, dx=\frac {5}{16 \,\mathrm {log}\left (x +12\right )-4 \,\mathrm {log}\left (x \right ) x^{2}-4 x^{3}} \] Input:
int(((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)
Output:
5/(4*(4*log(x + 12) - log(x)*x**2 - x**3))