Integrand size = 51, antiderivative size = 26 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=4+\frac {10 x^2}{-x+4 x \left (-e^{2 x}+\log (3)\right )} \] Output:
10/(x*(4*ln(3)-4*exp(x)^2)-x)*x^2+4
Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10 x}{1+4 e^{2 x}-\log (81)} \] Input:
Integrate[(-10 + E^(2*x)*(-40 + 80*x) + 40*Log[3])/(1 + 16*E^(4*x) + E^(2* x)*(8 - 32*Log[3]) - 8*Log[3] + 16*Log[3]^2),x]
Output:
(-10*x)/(1 + 4*E^(2*x) - Log[81])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.05 (sec) , antiderivative size = 210, normalized size of antiderivative = 8.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {7292, 7292, 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.
\(\displaystyle \int \frac {e^{2 x} (80 x-40)-10+40 \log (3)}{16 e^{4 x}+e^{2 x} (8-32 \log (3))+1+16 \log ^2(3)-8 \log (3)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} (80 x-40)-10 (1-4 \log (3))}{\left (4 e^{2 x}+1-4 \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 \left (8 e^{2 x} x-4 e^{2 x}-1+\log (81)\right )}{\left (4 e^{2 x}+1-4 \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 \int -\frac {-8 e^{2 x} x+4 e^{2 x}-\log (81)+1}{\left (1+4 e^{2 x}-4 \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -10 \int \frac {-8 e^{2 x} x+4 e^{2 x}-\log (81)+1}{\left (1+4 e^{2 x}-4 \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -10 \int \left (\frac {2 x (1-\log (81))}{\left (1+4 e^{2 x}-4 \log (3)\right )^2}-\frac {2 x-1}{1+4 e^{2 x}-4 \log (3)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-\log (81))}+\frac {\operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-4 \log (3))}+\frac {x^2}{1-\log (81)}-\frac {(1-2 x)^2}{4 (1-4 \log (3))}-\frac {(1-2 x) \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{2 (1-4 \log (3))}-\frac {x \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{1-\log (81)}+\frac {\log \left (4 e^{2 x}+1-4 \log (3)\right )}{2 (1-\log (81))}+\frac {x}{4 e^{2 x}+1-4 \log (3)}-\frac {x}{1-\log (81)}\right )\) |
Input:
Int[(-10 + E^(2*x)*(-40 + 80*x) + 40*Log[3])/(1 + 16*E^(4*x) + E^(2*x)*(8 - 32*Log[3]) - 8*Log[3] + 16*Log[3]^2),x]
Output:
-10*(-1/4*(1 - 2*x)^2/(1 - 4*Log[3]) + x/(1 + 4*E^(2*x) - 4*Log[3]) - x/(1 - Log[81]) + x^2/(1 - Log[81]) - ((1 - 2*x)*Log[1 + (4*E^(2*x))/(1 - 4*Lo g[3])])/(2*(1 - 4*Log[3])) - (x*Log[1 + (4*E^(2*x))/(1 - 4*Log[3])])/(1 - Log[81]) + Log[1 + 4*E^(2*x) - 4*Log[3]]/(2*(1 - Log[81])) + PolyLog[2, (- 4*E^(2*x))/(1 - 4*Log[3])]/(2*(1 - 4*Log[3])) - PolyLog[2, (-4*E^(2*x))/(1 - 4*Log[3])]/(2*(1 - Log[81])))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69
method | result | size |
norman | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
risch | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
parallelrisch | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
default | \(-\frac {10 \ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}-\frac {40 \left (-\frac {\frac {\ln \left (3\right )}{2}-\frac {1}{8}}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}-\frac {\ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1\right )}{8}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}+\frac {5}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}+40 \ln \left (3\right ) \left (\frac {\ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}+\frac {-\frac {4 \left (\frac {\ln \left (3\right )}{2}-\frac {1}{8}\right )}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}-\frac {\ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1\right )}{2}}{\left (-1+4 \ln \left (3\right )\right )^{2}}\right )+\frac {5 \ln \left (-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1\right )}{-1+4 \ln \left (3\right )}+\frac {40 x \,{\mathrm e}^{2 x}}{\left (-1+4 \ln \left (3\right )\right ) \left (-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1\right )}\) | \(196\) |
Input:
int(((80*x-40)*exp(x)^2+40*ln(3)-10)/(16*exp(x)^4+(-32*ln(3)+8)*exp(x)^2+1 6*ln(3)^2-8*ln(3)+1),x,method=_RETURNVERBOSE)
Output:
10*x/(-4*exp(x)^2+4*ln(3)-1)
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10 \, x}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1} \] Input:
integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*ex p(x)^2+16*log(3)^2-8*log(3)+1),x, algorithm="fricas")
Output:
-10*x/(4*e^(2*x) - 4*log(3) + 1)
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=- \frac {10 x}{4 e^{2 x} - 4 \log {\left (3 \right )} + 1} \] Input:
integrate(((80*x-40)*exp(x)**2+40*ln(3)-10)/(16*exp(x)**4+(-32*ln(3)+8)*ex p(x)**2+16*ln(3)**2-8*ln(3)+1),x)
Output:
-10*x/(4*exp(2*x) - 4*log(3) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 224, normalized size of antiderivative = 8.62 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=20 \, {\left (\frac {2 \, x}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} - \frac {\log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right )}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} - \frac {1}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1}\right )} \log \left (3\right ) - \frac {40 \, x e^{\left (2 \, x\right )}}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1} - \frac {10 \, x}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} + \frac {5 \, \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right )}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} + \frac {5 \, \log \left (e^{\left (2 \, x\right )} - \log \left (3\right ) + \frac {1}{4}\right )}{4 \, \log \left (3\right ) - 1} + \frac {5}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1} + \frac {5}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1} \] Input:
integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*ex p(x)^2+16*log(3)^2-8*log(3)+1),x, algorithm="maxima")
Output:
20*(2*x/(16*log(3)^2 - 8*log(3) + 1) - log(4*e^(2*x) - 4*log(3) + 1)/(16*l og(3)^2 - 8*log(3) + 1) - 1/(4*(4*log(3) - 1)*e^(2*x) - 16*log(3)^2 + 8*lo g(3) - 1))*log(3) - 40*x*e^(2*x)/(4*(4*log(3) - 1)*e^(2*x) - 16*log(3)^2 + 8*log(3) - 1) - 10*x/(16*log(3)^2 - 8*log(3) + 1) + 5*log(4*e^(2*x) - 4*l og(3) + 1)/(16*log(3)^2 - 8*log(3) + 1) + 5*log(e^(2*x) - log(3) + 1/4)/(4 *log(3) - 1) + 5/(4*(4*log(3) - 1)*e^(2*x) - 16*log(3)^2 + 8*log(3) - 1) + 5/(4*e^(2*x) - 4*log(3) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 5.35 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {5 \, {\left (8 \, x \log \left (3\right ) + 4 \, e^{\left (2 \, x\right )} \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - 4 \, \log \left (3\right ) \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - 4 \, e^{\left (2 \, x\right )} \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right ) + 4 \, \log \left (3\right ) \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right ) - 2 \, x + \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right )\right )}}{16 \, e^{\left (2 \, x\right )} \log \left (3\right ) - 16 \, \log \left (3\right )^{2} - 4 \, e^{\left (2 \, x\right )} + 8 \, \log \left (3\right ) - 1} \] Input:
integrate(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*ex p(x)^2+16*log(3)^2-8*log(3)+1),x, algorithm="giac")
Output:
-5*(8*x*log(3) + 4*e^(2*x)*log(4*e^(2*x) - 4*log(3) + 1) - 4*log(3)*log(4* e^(2*x) - 4*log(3) + 1) - 4*e^(2*x)*log(-4*e^(2*x) + 4*log(3) - 1) + 4*log (3)*log(-4*e^(2*x) + 4*log(3) - 1) - 2*x + log(4*e^(2*x) - 4*log(3) + 1) - log(-4*e^(2*x) + 4*log(3) - 1))/(16*e^(2*x)*log(3) - 16*log(3)^2 - 4*e^(2 *x) + 8*log(3) - 1)
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10\,x}{4\,{\mathrm {e}}^{2\,x}-\ln \left (81\right )+1} \] Input:
int((40*log(3) + exp(2*x)*(80*x - 40) - 10)/(16*exp(4*x) - 8*log(3) - exp( 2*x)*(32*log(3) - 8) + 16*log(3)^2 + 1),x)
Output:
-(10*x)/(4*exp(2*x) - log(81) + 1)
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10 x}{4 e^{2 x}-4 \,\mathrm {log}\left (3\right )+1} \] Input:
int(((80*x-40)*exp(x)^2+40*log(3)-10)/(16*exp(x)^4+(-32*log(3)+8)*exp(x)^2 +16*log(3)^2-8*log(3)+1),x)
Output:
( - 10*x)/(4*e**(2*x) - 4*log(3) + 1)