\(\int \frac {-40+40 x+e^{2 x} (-16+80 x-40 x^2)}{e^{4 x} (2-10 x+5 x^2)+e^{2 x} (-4+20 x-10 x^2) \log (\frac {1}{5} (-2+10 x-5 x^2))+(2-10 x+5 x^2) \log ^2(\frac {1}{5} (-2+10 x-5 x^2))} \, dx\) [9606]

Optimal result

Integrand size = 100, antiderivative size = 31 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=-2+e^4+\log (5)+\frac {4}{e^{2 x}-\log \left (-\frac {2}{5}+(2-x) x\right )} \]

[Out]

ln(5)-2+exp(4)+4/(exp(2*x)-ln((2-x)*x-2/5))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6820, 12, 6818} \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x}-\log \left (-x^2+2 x-\frac {2}{5}\right )} \]

[In]

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)*Lo
g[(-2 + 10*x - 5*x^2)/5] + (2 - 10*x + 5*x^2)*Log[(-2 + 10*x - 5*x^2)/5]^2),x]

[Out]

4/(E^(2*x) - Log[-2/5 + 2*x - x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {8 \left (-5+5 x-e^{2 x} \left (2-10 x+5 x^2\right )\right )}{\left (2-10 x+5 x^2\right ) \left (e^{2 x}-\log \left (-\frac {2}{5}+2 x-x^2\right )\right )^2} \, dx \\ & = 8 \int \frac {-5+5 x-e^{2 x} \left (2-10 x+5 x^2\right )}{\left (2-10 x+5 x^2\right ) \left (e^{2 x}-\log \left (-\frac {2}{5}+2 x-x^2\right )\right )^2} \, dx \\ & = \frac {4}{e^{2 x}-\log \left (-\frac {2}{5}+2 x-x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x}-\log \left (-\frac {2}{5}+2 x-x^2\right )} \]

[In]

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]

[Out]

4/(E^(2*x) - Log[-2/5 + 2*x - x^2])

Maple [A] (verified)

Time = 0.27 (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	\(\frac {4}{{\mathrm e}^{2 x}-\ln \left (-x^{2}+2 x -\frac {2}{5}\right )}\)	\(23\)

[In]

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)

[Out]

4/(exp(2*x)-ln(-x^2+2*x-2/5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} - \log \left (-x^{2} + 2 \, x - \frac {2}{5}\right )} \]

[In]

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)*l
og(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2*x)^2),x, algorithm="fricas")

[Out]

4/(e^(2*x) - log(-x^2 + 2*x - 2/5))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{2 x} - \log {\left (- x^{2} + 2 x - \frac {2}{5} \right )}} \]

[In]

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)*exp(2*x)**2),x)

[Out]

4/(exp(2*x) - log(-x**2 + 2*x - 2/5))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} + \log \left (5\right ) - \log \left (-5 \, x^{2} + 10 \, x - 2\right )} \]

[In]

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)*l
og(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2*x)^2),x, algorithm="maxima")

[Out]

4/(e^(2*x) + log(5) - log(-5*x^2 + 10*x - 2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{e^{\left (2 \, x\right )} + \log \left (5\right ) - \log \left (-5 \, x^{2} + 10 \, x - 2\right )} \]

[In]

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)*l
og(-x^2+2*x-2/5)+(5*x^2-10*x+2)*exp(2*x)^2),x, algorithm="giac")

[Out]

4/(e^(2*x) + log(5) - log(-5*x^2 + 10*x - 2))

Mupad [B] (verification not implemented)

Time = 16.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-40+40 x+e^{2 x} \left (-16+80 x-40 x^2\right )}{e^{4 x} \left (2-10 x+5 x^2\right )+e^{2 x} \left (-4+20 x-10 x^2\right ) \log \left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )+\left (2-10 x+5 x^2\right ) \log ^2\left (\frac {1}{5} \left (-2+10 x-5 x^2\right )\right )} \, dx=\frac {4}{{\mathrm {e}}^{2\,x}-\ln \left (-x^2+2\,x-\frac {2}{5}\right )} \]

[In]

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)

[Out]

4/(exp(2*x) - log(2*x - x^2 - 2/5))