Integrand size = 89, antiderivative size = 20 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \] Output:
x^2/(ln(-x*ln(5)+15)-4)/ln(3)
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\log (3) (-4+\log (15-x \log (5)))} \] Input:
Integrate[(120*x - 9*x^2*Log[5] + (-30*x + 2*x^2*Log[5])*Log[15 - x*Log[5] ])/(-240*Log[3] + 16*x*Log[3]*Log[5] + (120*Log[3] - 8*x*Log[3]*Log[5])*Lo g[15 - x*Log[5]] + (-15*Log[3] + x*Log[3]*Log[5])*Log[15 - x*Log[5]]^2),x]
Output:
x^2/(Log[3]*(-4 + Log[15 - x*Log[5]]))
Leaf count is larger than twice the leaf count of optimal. \(82\) vs. \(2(20)=40\).
Time = 1.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {7239, 27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {-9 x^2 \log (5)+\left (2 x^2 \log (5)-30 x\right ) \log (15-x \log (5))+120 x}{(x \log (3) \log (5)-15 \log (3)) \log ^2(15-x \log (5))+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+16 x \log (3) \log (5)-240 \log (3)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {x (9 x \log (5)-2 (x \log (5)-15) \log (15-x \log (5))-120)}{\log (3) (15-x \log (5)) (4-\log (15-x \log (5)))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {x (-9 \log (5) x-2 (15-x \log (5)) \log (15-x \log (5))+120)}{(15-x \log (5)) (4-\log (15-x \log (5)))^2}dx}{\log (3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {x (-9 \log (5) x-2 (15-x \log (5)) \log (15-x \log (5))+120)}{(15-x \log (5)) (4-\log (15-x \log (5)))^2}dx}{\log (3)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {x^2 \log (5)}{(x \log (5)-15) (\log (15-x \log (5))-4)^2}-\frac {2 x}{\log (15-x \log (5))-4}\right )dx}{\log (3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {(15-x \log (5))^2}{\log ^2(5) (4-\log (15-x \log (5)))}-\frac {30 (15-x \log (5))}{\log ^2(5) (4-\log (15-x \log (5)))}+\frac {225}{\log ^2(5) (4-\log (15-x \log (5)))}}{\log (3)}\) |
Input:
Int[(120*x - 9*x^2*Log[5] + (-30*x + 2*x^2*Log[5])*Log[15 - x*Log[5]])/(-2 40*Log[3] + 16*x*Log[3]*Log[5] + (120*Log[3] - 8*x*Log[3]*Log[5])*Log[15 - x*Log[5]] + (-15*Log[3] + x*Log[3]*Log[5])*Log[15 - x*Log[5]]^2),x]
Output:
-((225/(Log[5]^2*(4 - Log[15 - x*Log[5]])) - (30*(15 - x*Log[5]))/(Log[5]^ 2*(4 - Log[15 - x*Log[5]])) + (15 - x*Log[5])^2/(Log[5]^2*(4 - Log[15 - x* Log[5]])))/Log[3])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.43 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
risch | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
parallelrisch | \(\frac {x^{2}}{\left (\ln \left (-x \ln \left (5\right )+15\right )-4\right ) \ln \left (3\right )}\) | \(21\) |
derivativedivides | \(\frac {\frac {225}{\ln \left (-x \ln \left (5\right )+15\right )-4}-\frac {30 \left (-x \ln \left (5\right )+15\right )}{\ln \left (-x \ln \left (5\right )+15\right )-4}+\frac {\left (-x \ln \left (5\right )+15\right )^{2}}{\ln \left (-x \ln \left (5\right )+15\right )-4}}{\ln \left (5\right )^{2} \ln \left (3\right )}\) | \(68\) |
default | \(\frac {\frac {225}{\ln \left (-x \ln \left (5\right )+15\right )-4}-\frac {30 \left (-x \ln \left (5\right )+15\right )}{\ln \left (-x \ln \left (5\right )+15\right )-4}+\frac {\left (-x \ln \left (5\right )+15\right )^{2}}{\ln \left (-x \ln \left (5\right )+15\right )-4}}{\ln \left (5\right )^{2} \ln \left (3\right )}\) | \(68\) |
Input:
int(((2*x^2*ln(5)-30*x)*ln(-x*ln(5)+15)-9*x^2*ln(5)+120*x)/((x*ln(3)*ln(5) -15*ln(3))*ln(-x*ln(5)+15)^2+(-8*x*ln(3)*ln(5)+120*ln(3))*ln(-x*ln(5)+15)+ 16*x*ln(3)*ln(5)-240*ln(3)),x,method=_RETURNVERBOSE)
Output:
x^2/(ln(-x*ln(5)+15)-4)/ln(3)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \] Input:
integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*l og(3)*log(5)-15*log(3))*log(-x*log(5)+15)^2+(-8*x*log(3)*log(5)+120*log(3) )*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="fricas")
Output:
x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log {\left (3 \right )} \log {\left (- x \log {\left (5 \right )} + 15 \right )} - 4 \log {\left (3 \right )}} \] Input:
integrate(((2*x**2*ln(5)-30*x)*ln(-x*ln(5)+15)-9*x**2*ln(5)+120*x)/((x*ln( 3)*ln(5)-15*ln(3))*ln(-x*ln(5)+15)**2+(-8*x*ln(3)*ln(5)+120*ln(3))*ln(-x*l n(5)+15)+16*x*ln(3)*ln(5)-240*ln(3)),x)
Output:
x**2/(log(3)*log(-x*log(5) + 15) - 4*log(3))
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \] Input:
integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*l og(3)*log(5)-15*log(3))*log(-x*log(5)+15)^2+(-8*x*log(3)*log(5)+120*log(3) )*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="maxima")
Output:
x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\log \left (3\right ) \log \left (-x \log \left (5\right ) + 15\right ) - 4 \, \log \left (3\right )} \] Input:
integrate(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*l og(3)*log(5)-15*log(3))*log(-x*log(5)+15)^2+(-8*x*log(3)*log(5)+120*log(3) )*log(-x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x, algorithm="giac")
Output:
x^2/(log(3)*log(-x*log(5) + 15) - 4*log(3))
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^2}{\ln \left (3\right )\,\left (\ln \left (15-x\,\ln \left (5\right )\right )-4\right )} \] Input:
int((log(15 - x*log(5))*(30*x - 2*x^2*log(5)) - 120*x + 9*x^2*log(5))/(240 *log(3) - log(15 - x*log(5))*(120*log(3) - 8*x*log(3)*log(5)) + log(15 - x *log(5))^2*(15*log(3) - x*log(3)*log(5)) - 16*x*log(3)*log(5)),x)
Output:
x^2/(log(3)*(log(15 - x*log(5)) - 4))
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {120 x-9 x^2 \log (5)+\left (-30 x+2 x^2 \log (5)\right ) \log (15-x \log (5))}{-240 \log (3)+16 x \log (3) \log (5)+(120 \log (3)-8 x \log (3) \log (5)) \log (15-x \log (5))+(-15 \log (3)+x \log (3) \log (5)) \log ^2(15-x \log (5))} \, dx=\frac {x^{2}}{\mathrm {log}\left (3\right ) \left (\mathrm {log}\left (-\mathrm {log}\left (5\right ) x +15\right )-4\right )} \] Input:
int(((2*x^2*log(5)-30*x)*log(-x*log(5)+15)-9*x^2*log(5)+120*x)/((x*log(3)* log(5)-15*log(3))*log(-x*log(5)+15)^2+(-8*x*log(3)*log(5)+120*log(3))*log( -x*log(5)+15)+16*x*log(3)*log(5)-240*log(3)),x)
Output:
x**2/(log(3)*(log( - log(5)*x + 15) - 4))