Integrand size = 63, antiderivative size = 34 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=x+\frac {3 x \left (i \pi +\log \left (-5+5 e^3\right )\right )}{20 \left (\frac {3}{5}-e^{2 x}\right )} \] Output:
x+3/20*x*ln(-5*exp(3)+5)/(3/5-exp(x)^2)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=\frac {1}{4} x \left (4+\frac {3 i \pi +\log (125)+3 \log \left (-1+e^3\right )}{3-5 e^{2 x}}\right ) \] Input:
Integrate[(36 - 120*E^(2*x) + 100*E^(4*x) + (9 + E^(2*x)*(-15 + 30*x))*(I* Pi + Log[-5 + 5*E^3]))/(36 - 120*E^(2*x) + 100*E^(4*x)),x]
Output:
(x*(4 + ((3*I)*Pi + Log[125] + 3*Log[-1 + E^3])/(3 - 5*E^(2*x))))/4
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(202\) vs. \(2(34)=68\).
Time = 1.03 (sec) , antiderivative size = 202, normalized size of antiderivative = 5.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {7292, 27, 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.
\(\displaystyle \int \frac {-120 e^{2 x}+100 e^{4 x}+\left (e^{2 x} (30 x-15)+9\right ) \left (\log \left (5 e^3-5\right )+i \pi \right )+36}{-120 e^{2 x}+100 e^{4 x}+36} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-120 e^{2 x}+100 e^{4 x}+\left (e^{2 x} (30 x-15)+9\right ) \left (\log \left (5 e^3-5\right )+i \pi \right )+36}{4 \left (3-5 e^{2 x}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \left (3-5 e^{2 x} (1-2 x)\right )-120 e^{2 x}+100 e^{4 x}+36}{\left (3-5 e^{2 x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) (1-2 x)}{3-5 e^{2 x}}+\frac {18 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{\left (3-5 e^{2 x}\right )^2}+4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (x^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )+4 x-\frac {1}{4} (1-2 x)^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )-\frac {1}{2} (1-2 x) \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{2} \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (3-5 e^{2 x}\right )-x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {3 x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )}{3-5 e^{2 x}}-x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )\right )\) |
Input:
Int[(36 - 120*E^(2*x) + 100*E^(4*x) + (9 + E^(2*x)*(-15 + 30*x))*(I*Pi + L og[-5 + 5*E^3]))/(36 - 120*E^(2*x) + 100*E^(4*x)),x]
Output:
(4*x - ((1 - 2*x)^2*(I*Pi + Log[-5*(1 - E^3)]))/4 - x*(I*Pi + Log[-5*(1 - E^3)]) + (3*x*(I*Pi + Log[-5*(1 - E^3)]))/(3 - 5*E^(2*x)) + x^2*(I*Pi + Lo g[-5*(1 - E^3)]) + ((I*Pi + Log[-5*(1 - E^3)])*Log[3 - 5*E^(2*x)])/2 - ((1 - 2*x)*(I*Pi + Log[-5*(1 - E^3)])*Log[1 - (5*E^(2*x))/3])/2 - x*(I*Pi + L og[-5*(1 - E^3)])*Log[1 - (5*E^(2*x))/3])/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76
method | result | size |
risch | \(x -\frac {3 x \left (\ln \left (5\right )+\ln \left (-{\mathrm e}^{3}+1\right )\right )}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}\) | \(26\) |
parallelrisch | \(-\frac {-60 x \,{\mathrm e}^{2 x}+9 x \ln \left (-5 \,{\mathrm e}^{3}+5\right )+36 x}{12 \left (5 \,{\mathrm e}^{2 x}-3\right )}\) | \(34\) |
norman | \(\frac {\left (-3-\frac {3 \ln \left (5\right )}{4}-\frac {3 \ln \left (-{\mathrm e}^{3}+1\right )}{4}\right ) x +5 x \,{\mathrm e}^{2 x}}{5 \,{\mathrm e}^{2 x}-3}\) | \(37\) |
default | \(\ln \left ({\mathrm e}^{x}\right )+\frac {9 \ln \left (5\right ) \left (-\frac {\ln \left (5 \,{\mathrm e}^{2 x}-3\right )}{18}-\frac {1}{6 \left (5 \,{\mathrm e}^{2 x}-3\right )}+\frac {\ln \left ({\mathrm e}^{x}\right )}{9}\right )}{4}+\frac {9 \ln \left (-{\mathrm e}^{3}+1\right ) \left (-\frac {\ln \left (5 \,{\mathrm e}^{2 x}-3\right )}{18}-\frac {1}{6 \left (5 \,{\mathrm e}^{2 x}-3\right )}+\frac {\ln \left ({\mathrm e}^{x}\right )}{9}\right )}{4}+\frac {3 \ln \left (5\right )}{8 \left (5 \,{\mathrm e}^{2 x}-3\right )}+\frac {3 \ln \left (-{\mathrm e}^{3}+1\right )}{8 \left (5 \,{\mathrm e}^{2 x}-3\right )}+\frac {15 \ln \left (-{\mathrm e}^{3}+1\right ) \left (\frac {\ln \left (5 \,{\mathrm e}^{2 x}-3\right )}{60}-\frac {x \,{\mathrm e}^{2 x}}{6 \left (5 \,{\mathrm e}^{2 x}-3\right )}\right )}{2}+\frac {15 \ln \left (5\right ) \left (\frac {\ln \left (5 \,{\mathrm e}^{2 x}-3\right )}{60}-\frac {x \,{\mathrm e}^{2 x}}{6 \left (5 \,{\mathrm e}^{2 x}-3\right )}\right )}{2}\) | \(180\) |
Input:
int((((30*x-15)*exp(x)^2+9)*ln(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36)/ (100*exp(x)^4-120*exp(x)^2+36),x,method=_RETURNVERBOSE)
Output:
x-3/4*x*(ln(5)+ln(-exp(3)+1))/(5*exp(2*x)-3)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=\frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \] Input:
integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x) ^2+36)/(100*exp(x)^4-120*exp(x)^2+36),x, algorithm="fricas")
Output:
1/4*(20*x*e^(2*x) - 3*x*log(-5*e^3 + 5) - 12*x)/(5*e^(2*x) - 3)
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=x + \frac {- 3 x \log {\left (-1 + e^{3} \right )} - 3 x \log {\left (5 \right )} - 3 i \pi x}{20 e^{2 x} - 12} \] Input:
integrate((((30*x-15)*exp(x)**2+9)*ln(-5*exp(3)+5)+100*exp(x)**4-120*exp(x )**2+36)/(100*exp(x)**4-120*exp(x)**2+36),x)
Output:
x + (-3*x*log(-1 + exp(3)) - 3*x*log(5) - 3*I*pi*x)/(20*exp(2*x) - 12)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (22) = 44\).
Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.74 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=\frac {1}{8} \, {\left (2 \, x - \frac {3}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (5 \, e^{\left (2 \, x\right )} - 3\right )\right )} \log \left (-5 \, e^{3} + 5\right ) - \frac {1}{8} \, {\left (\frac {10 \, x e^{\left (2 \, x\right )}}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (e^{\left (2 \, x\right )} - \frac {3}{5}\right )\right )} \log \left (-5 \, e^{3} + 5\right ) + x + \frac {3 \, \log \left (-5 \, e^{3} + 5\right )}{8 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \] Input:
integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x) ^2+36)/(100*exp(x)^4-120*exp(x)^2+36),x, algorithm="maxima")
Output:
1/8*(2*x - 3/(5*e^(2*x) - 3) - log(5*e^(2*x) - 3))*log(-5*e^3 + 5) - 1/8*( 10*x*e^(2*x)/(5*e^(2*x) - 3) - log(e^(2*x) - 3/5))*log(-5*e^3 + 5) + x + 3 /8*log(-5*e^3 + 5)/(5*e^(2*x) - 3)
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=\frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \] Input:
integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x) ^2+36)/(100*exp(x)^4-120*exp(x)^2+36),x, algorithm="giac")
Output:
1/4*(20*x*e^(2*x) - 3*x*log(-5*e^3 + 5) - 12*x)/(5*e^(2*x) - 3)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=x-\frac {3\,x\,\ln \left (5-5\,{\mathrm {e}}^3\right )}{20\,{\mathrm {e}}^{2\,x}-12} \] Input:
int((100*exp(4*x) - 120*exp(2*x) + log(5 - 5*exp(3))*(exp(2*x)*(30*x - 15) + 9) + 36)/(100*exp(4*x) - 120*exp(2*x) + 36),x)
Output:
x - (3*x*log(5 - 5*exp(3)))/(20*exp(2*x) - 12)
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{36-120 e^{2 x}+100 e^{4 x}} \, dx=\frac {x \left (20 e^{2 x}-3 \,\mathrm {log}\left (-5 e^{3}+5\right )-12\right )}{20 e^{2 x}-12} \] Input:
int((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36) /(100*exp(x)^4-120*exp(x)^2+36),x)
Output:
(x*(20*e**(2*x) - 3*log( - 5*e**3 + 5) - 12))/(4*(5*e**(2*x) - 3))