Integrand size = 100, antiderivative size = 31 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=-2+e^4+\log (5)+\frac {4}{e^{2 x}-\log \left (-\frac {2}{5}+(2-x) x\right )} \] Output:
ln(5)-2+exp(4)+4/(exp(2*x)-ln((2-x)*x-2/5))
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x}-\log \left (-\frac {2}{5}+2 x-x^2\right )} \] Input:
Integrate[(-40 + 40*x + E^(2*x)*(-16 + 80*x - 40*x^2))/(E^(4*x)*(2 - 10*x + 5*x^2) + E^(2*x)*(-4 + 20*x - 10*x^2)*Log[(-2 + 10*x - 5*x^2)/5] + (2 - 10*x + 5*x^2)*Log[(-2 + 10*x - 5*x^2)/5]^2),x]
Output:
4/(E^(2*x) - Log[-2/5 + 2*x - x^2])
Time = 0.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, 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.
\(\displaystyle \int \frac {e^{2 x} \left (-40 x^2+80 x-16\right )+40 x-40}{e^{4 x} \left (5 x^2-10 x+2\right )+\left (5 x^2-10 x+2\right ) \log ^2\left (\frac {1}{5} \left (-5 x^2+10 x-2\right )\right )+e^{2 x} \left (-10 x^2+20 x-4\right ) \log \left (\frac {1}{5} \left (-5 x^2+10 x-2\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {8 \left (-e^{2 x} \left (5 x^2-10 x+2\right )+5 x-5\right )}{\left (5 x^2-10 x+2\right ) \left (e^{2 x}-\log \left (-x^2+2 x-\frac {2}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int -\frac {-5 x+e^{2 x} \left (5 x^2-10 x+2\right )+5}{\left (5 x^2-10 x+2\right ) \left (e^{2 x}-\log \left (-x^2+2 x-\frac {2}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int \frac {-5 x+e^{2 x} \left (5 x^2-10 x+2\right )+5}{\left (5 x^2-10 x+2\right ) \left (e^{2 x}-\log \left (-x^2+2 x-\frac {2}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {4}{e^{2 x}-\log \left (-x^2+2 x-\frac {2}{5}\right )}\) |
Input:
Int[(-40 + 40*x + E^(2*x)*(-16 + 80*x - 40*x^2))/(E^(4*x)*(2 - 10*x + 5*x^ 2) + E^(2*x)*(-4 + 20*x - 10*x^2)*Log[(-2 + 10*x - 5*x^2)/5] + (2 - 10*x + 5*x^2)*Log[(-2 + 10*x - 5*x^2)/5]^2),x]
Output:
4/(E^(2*x) - Log[-2/5 + 2*x - x^2])
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 = 0.45 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {4}{{\mathrm e}^{2 x}-\ln \left (-x^{2}+2 x -\frac {2}{5}\right )}\) | \(23\) |
parallelrisch | \(\frac {4}{{\mathrm e}^{2 x}-\ln \left (-x^{2}+2 x -\frac {2}{5}\right )}\) | \(23\) |
Input:
int(((-40*x^2+80*x-16)*exp(2*x)+40*x-40)/((5*x^2-10*x+2)*ln(-x^2+2*x-2/5)^ 2+(-10*x^2+20*x-4)*exp(2*x)*ln(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2*x)^2),x, method=_RETURNVERBOSE)
Output:
4/(exp(2*x)-ln(-x^2+2*x-2/5))
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} - \log \left (-x^{2} + 2 \, x - \frac {2}{5}\right )} \] Input:
integrate(((-40*x^2+80*x-16)*exp(2*x)+40*x-40)/((5*x^2-10*x+2)*log(-x^2+2* x-2/5)^2+(-10*x^2+20*x-4)*exp(2*x)*log(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2* x)^2),x, algorithm="fricas")
Output:
4/(e^(2*x) - log(-x^2 + 2*x - 2/5))
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x} - \log {\left (- x^{2} + 2 x - \frac {2}{5} \right )}} \] Input:
integrate(((-40*x**2+80*x-16)*exp(2*x)+40*x-40)/((5*x**2-10*x+2)*ln(-x**2+ 2*x-2/5)**2+(-10*x**2+20*x-4)*exp(2*x)*ln(-x**2+2*x-2/5)+(5*x**2-10*x+2)*e xp(2*x)**2),x)
Output:
4/(exp(2*x) - log(-x**2 + 2*x - 2/5))
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} + \log \left (5\right ) - \log \left (-5 \, x^{2} + 10 \, x - 2\right )} \] Input:
integrate(((-40*x^2+80*x-16)*exp(2*x)+40*x-40)/((5*x^2-10*x+2)*log(-x^2+2* x-2/5)^2+(-10*x^2+20*x-4)*exp(2*x)*log(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2* x)^2),x, algorithm="maxima")
Output:
4/(e^(2*x) + log(5) - log(-5*x^2 + 10*x - 2))
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} + \log \left (5\right ) - \log \left (-5 \, x^{2} + 10 \, x - 2\right )} \] Input:
integrate(((-40*x^2+80*x-16)*exp(2*x)+40*x-40)/((5*x^2-10*x+2)*log(-x^2+2* x-2/5)^2+(-10*x^2+20*x-4)*exp(2*x)*log(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2* x)^2),x, algorithm="giac")
Output:
4/(e^(2*x) + log(5) - log(-5*x^2 + 10*x - 2))
Time = 4.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{{\mathrm {e}}^{2\,x}-\ln \left (-x^2+2\,x-\frac {2}{5}\right )} \] Input:
int(-(exp(2*x)*(40*x^2 - 80*x + 16) - 40*x + 40)/(log(2*x - x^2 - 2/5)^2*( 5*x^2 - 10*x + 2) + exp(4*x)*(5*x^2 - 10*x + 2) - exp(2*x)*log(2*x - x^2 - 2/5)*(10*x^2 - 20*x + 4)),x)
Output:
4/(exp(2*x) - log(2*x - x^2 - 2/5))
Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x}-\mathrm {log}\left (-x^{2}+2 x -\frac {2}{5}\right )} \] Input:
int(((-40*x^2+80*x-16)*exp(2*x)+40*x-40)/((5*x^2-10*x+2)*log(-x^2+2*x-2/5) ^2+(-10*x^2+20*x-4)*exp(2*x)*log(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2*x)^2), x)
Output:
4/(e**(2*x) - log(( - 5*x**2 + 10*x - 2)/5))