Integrand size = 136, antiderivative size = 23 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=e^{24} \left (5-x-\left (x+\log (3)+\frac {4}{\log (x)}\right )^2\right ) \] Output:
exp(ln(-x+5-(4/ln(x)+ln(3)+x)^2)+24)
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-e^{24} x-e^{24} x^2-e^{24} x \log (9)-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \] Input:
Integrate[(E^24*(-16 + (-8*x - 8*Log[3])*Log[x] + (5 - x - x^2 - 2*x*Log[3 ] - Log[3]^2)*Log[x]^2)*(-32 + (-8*x - 8*Log[3])*Log[x] + 8*x*Log[x]^2 + ( x + 2*x^2 + 2*x*Log[3])*Log[x]^3))/(Log[x]^2*(16*x*Log[x] + (8*x^2 + 8*x*L og[3])*Log[x]^2 + (-5*x + x^2 + x^3 + 2*x^2*Log[3] + x*Log[3]^2)*Log[x]^3) ),x]
Output:
-(E^24*x) - E^24*x^2 - E^24*x*Log[9] - (16*E^24)/Log[x]^2 - (8*E^24*x)/Log [x] - (8*E^24*Log[3])/Log[x]
Time = 0.75 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {27, 7239, 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.
\(\displaystyle \int \frac {e^{24} \left (\left (-x^2-x-2 x \log (3)+5-\log ^2(3)\right ) \log ^2(x)+(-8 x-8 \log (3)) \log (x)-16\right ) \left (\left (2 x^2+x+2 x \log (3)\right ) \log ^3(x)+8 x \log ^2(x)+(-8 x-8 \log (3)) \log (x)-32\right )}{\log ^2(x) \left (\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (x^3+x^2+2 x^2 \log (3)-5 x+x \log ^2(3)\right ) \log ^3(x)+16 x \log (x)\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{24} \int \frac {\left (-\left (\left (-x^2-2 \log (3) x-x-\log ^2(3)+5\right ) \log ^2(x)\right )+8 (x+\log (3)) \log (x)+16\right ) \left (-\left (\left (2 x^2+2 \log (3) x+x\right ) \log ^3(x)\right )-8 x \log ^2(x)+8 (x+\log (3)) \log (x)+32\right )}{\log ^2(x) \left (-\left (\left (-x^3-2 \log (3) x^2-x^2-\log ^2(3) x+5 x\right ) \log ^3(x)\right )+8 \left (x^2+\log (3) x\right ) \log ^2(x)+16 x \log (x)\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle e^{24} \int \frac {-x (2 x+\log (9)+1) \log ^3(x)-8 x \log ^2(x)+8 (x+\log (3)) \log (x)+32}{x \log ^3(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^{24} \int \left (-2 x-\frac {8}{\log (x)}-\log (9)-1+\frac {8 (x+\log (3))}{\log ^2(x) x}+\frac {32}{\log ^3(x) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{24} \left (-x^2-\frac {16}{\log ^2(x)}-\frac {8 x}{\log (x)}-x (1+\log (9))-\frac {8 \log (3)}{\log (x)}\right )\) |
Input:
Int[(E^24*(-16 + (-8*x - 8*Log[3])*Log[x] + (5 - x - x^2 - 2*x*Log[3] - Lo g[3]^2)*Log[x]^2)*(-32 + (-8*x - 8*Log[3])*Log[x] + 8*x*Log[x]^2 + (x + 2* x^2 + 2*x*Log[3])*Log[x]^3))/(Log[x]^2*(16*x*Log[x] + (8*x^2 + 8*x*Log[3]) *Log[x]^2 + (-5*x + x^2 + x^3 + 2*x^2*Log[3] + x*Log[3]^2)*Log[x]^3)),x]
Output:
E^24*(-x^2 - x*(1 + Log[9]) - 16/Log[x]^2 - (8*x)/Log[x] - (8*Log[3])/Log[ 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 155.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70
method | result | size |
default | \({\mathrm e}^{24} \left (-2 x \ln \left (3\right )-x^{2}-x -\frac {8 \ln \left (3\right )}{\ln \left (x \right )}-\frac {8 x}{\ln \left (x \right )}-\frac {16}{\ln \left (x \right )^{2}}\right )\) | \(39\) |
parallelrisch | \(\text {Expression too large to display}\) | \(1635\) |
Input:
int(((2*x*ln(3)+2*x^2+x)*ln(x)^3+8*x*ln(x)^2+(-8*ln(3)-8*x)*ln(x)-32)*exp( ln(((-ln(3)^2-2*x*ln(3)-x^2-x+5)*ln(x)^2+(-8*ln(3)-8*x)*ln(x)-16)/ln(x)^2) +24)/((x*ln(3)^2+2*x^2*ln(3)+x^3+x^2-5*x)*ln(x)^3+(8*x*ln(3)+8*x^2)*ln(x)^ 2+16*x*ln(x)),x,method=_RETURNVERBOSE)
Output:
exp(24)*(-2*x*ln(3)-x^2-x-8*ln(3)/ln(x)-8*x/ln(x)-16/ln(x)^2)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left (2 \, x e^{24} \log \left (3\right ) + {\left (x^{2} + x\right )} e^{24}\right )} \log \left (x\right )^{2} + 8 \, {\left (x e^{24} + e^{24} \log \left (3\right )\right )} \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \] Input:
integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log( x)-32)*exp(log(((-log(3)^2-2*x*log(3)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*lo g(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(8* x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="fricas")
Output:
-((2*x*e^24*log(3) + (x^2 + x)*e^24)*log(x)^2 + 8*(x*e^24 + e^24*log(3))*l og(x) + 16*e^24)/log(x)^2
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=- x^{2} e^{24} + x \left (- 2 e^{24} \log {\left (3 \right )} - e^{24}\right ) + \frac {\left (- 8 x e^{24} - 8 e^{24} \log {\left (3 \right )}\right ) \log {\left (x \right )} - 16 e^{24}}{\log {\left (x \right )}^{2}} \] Input:
integrate(((2*x*ln(3)+2*x**2+x)*ln(x)**3+8*x*ln(x)**2+(-8*ln(3)-8*x)*ln(x) -32)*exp(ln(((-ln(3)**2-2*x*ln(3)-x**2-x+5)*ln(x)**2+(-8*ln(3)-8*x)*ln(x)- 16)/ln(x)**2)+24)/((x*ln(3)**2+2*x**2*ln(3)+x**3+x**2-5*x)*ln(x)**3+(8*x*l n(3)+8*x**2)*ln(x)**2+16*x*ln(x)),x)
Output:
-x**2*exp(24) + x*(-2*exp(24)*log(3) - exp(24)) + ((-8*x*exp(24) - 8*exp(2 4)*log(3))*log(x) - 16*exp(24))/log(x)**2
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left ({\left (x^{2} + x {\left (2 \, \log \left (3\right ) + 1\right )}\right )} \log \left (x\right )^{2} + 8 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right ) + 16\right )} e^{24}}{\log \left (x\right )^{2}} \] Input:
integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log( x)-32)*exp(log(((-log(3)^2-2*x*log(3)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*lo g(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(8* x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="maxima")
Output:
-((x^2 + x*(2*log(3) + 1))*log(x)^2 + 8*(x + log(3))*log(x) + 16)*e^24/log (x)^2
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.39 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {x^{2} e^{24} \log \left (x\right )^{2} + 2 \, x e^{24} \log \left (3\right ) \log \left (x\right )^{2} + x e^{24} \log \left (x\right )^{2} + 8 \, x e^{24} \log \left (x\right ) + 8 \, e^{24} \log \left (3\right ) \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \] Input:
integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log( x)-32)*exp(log(((-log(3)^2-2*x*log(3)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*lo g(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(8* x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="giac")
Output:
-(x^2*e^24*log(x)^2 + 2*x*e^24*log(3)*log(x)^2 + x*e^24*log(x)^2 + 8*x*e^2 4*log(x) + 8*e^24*log(3)*log(x) + 16*e^24)/log(x)^2
Time = 2.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-x^2\,{\mathrm {e}}^{24}-\frac {16\,{\mathrm {e}}^{24}+8\,{\mathrm {e}}^{24}\,\ln \left (3\right )\,\ln \left (x\right )}{{\ln \left (x\right )}^2}-\frac {x\,\left ({\mathrm {e}}^{24}\,\left (2\,\ln \left (3\right )+1\right )\,{\ln \left (x\right )}^2+8\,{\mathrm {e}}^{24}\,\ln \left (x\right )\right )}{{\ln \left (x\right )}^2} \] Input:
int((exp(log(-(log(x)*(8*x + 8*log(3)) + log(x)^2*(x + 2*x*log(3) + log(3) ^2 + x^2 - 5) + 16)/log(x)^2) + 24)*(8*x*log(x)^2 - log(x)*(8*x + 8*log(3) ) + log(x)^3*(x + 2*x*log(3) + 2*x^2) - 32))/(log(x)^2*(8*x*log(3) + 8*x^2 ) + 16*x*log(x) + log(x)^3*(x*log(3)^2 - 5*x + 2*x^2*log(3) + x^2 + x^3)), x)
Output:
- x^2*exp(24) - (16*exp(24) + 8*exp(24)*log(3)*log(x))/log(x)^2 - (x*(8*ex p(24)*log(x) + exp(24)*log(x)^2*(2*log(3) + 1)))/log(x)^2
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=\frac {e^{24} \left (-2 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (3\right ) x -\mathrm {log}\left (x \right )^{2} x^{2}-\mathrm {log}\left (x \right )^{2} x -8 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )-8 \,\mathrm {log}\left (x \right ) x -16\right )}{\mathrm {log}\left (x \right )^{2}} \] Input:
int(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log(x)-32) *exp(log(((-log(3)^2-2*x*log(3)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*log(x)-1 6)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(8*x*log( 3)+8*x^2)*log(x)^2+16*x*log(x)),x)
Output:
(e**24*( - 2*log(x)**2*log(3)*x - log(x)**2*x**2 - log(x)**2*x - 8*log(x)* log(3) - 8*log(x)*x - 16))/log(x)**2