\(\int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 (20-2 x+4 x^2)+e^{2 x} (40+4 e^8-4 x+24 x^2)+e^x (-80 x-8 e^8 x+8 x^2-16 x^3)} \, dx\) [5952]

Optimal result

Integrand size = 121, antiderivative size = 28 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=1+\frac {4}{5+\frac {1}{2} \left (e^8-x\right )+\left (e^x-x\right )^2} \]

[Out]

4/(5-1/2*x+1/2*exp(8)+(exp(x)-x)^2)+1

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 12, 6818} \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 x^2-4 e^x x-x+2 e^{2 x}+e^8+10} \]

[In]

Int[(8 - 32*E^(2*x) - 32*x + E^x*(32 + 32*x))/(100 + E^16 + 4*E^(4*x) - 20*x - 16*E^(3*x)*x + 41*x^2 - 4*x^3 +
4*x^4 + E^8*(20 - 2*x + 4*x^2) + E^(2*x)*(40 + 4*E^8 - 4*x + 24*x^2) + E^x*(-80*x - 8*E^8*x + 8*x^2 - 16*x^3)
),x]

[Out]

8/(10 + E^8 + 2*E^(2*x) - x - 4*E^x*x + 2*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 (1-4 e^{2 x}-4 x+4 e^x (1+x)\right )}{\left (2 e^{2 x}+10 \left (1+\frac {e^8}{10}\right )-x-4 e^x x+2 x^2\right )^2} \, dx \\ & = 8 \int \frac {1-4 e^{2 x}-4 x+4 e^x (1+x)}{\left (2 e^{2 x}+10 \left (1+\frac {e^8}{10}\right )-x-4 e^x x+2 x^2\right )^2} \, dx \\ & = \frac {8}{10+e^8+2 e^{2 x}-x-4 e^x x+2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{10+e^8+2 e^{2 x}-x-4 e^x x+2 x^2} \]

[In]

Integrate[(8 - 32*E^(2*x) - 32*x + E^x*(32 + 32*x))/(100 + E^16 + 4*E^(4*x) - 20*x - 16*E^(3*x)*x + 41*x^2 - 4
*x^3 + 4*x^4 + E^8*(20 - 2*x + 4*x^2) + E^(2*x)*(40 + 4*E^8 - 4*x + 24*x^2) + E^x*(-80*x - 8*E^8*x + 8*x^2 - 1
6*x^3)),x]

[Out]

8/(10 + E^8 + 2*E^(2*x) - x - 4*E^x*x + 2*x^2)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\)	\(28\)
risch	\(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\)	\(28\)
parallelrisch	\(\frac {8}{2 \,{\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} x +2 x^{2}+{\mathrm e}^{8}-x +10}\)	\(28\)

[In]

int((-32*exp(x)^2+(32*x+32)*exp(x)-32*x+8)/(4*exp(x)^4-16*x*exp(x)^3+(4*exp(8)+24*x^2-4*x+40)*exp(x)^2+(-8*x*e
xp(8)-16*x^3+8*x^2-80*x)*exp(x)+exp(8)^2+(4*x^2-2*x+20)*exp(8)+4*x^4-4*x^3+41*x^2-20*x+100),x,method=_RETURNVE
RBOSE)

[Out]

8/(2*exp(x)^2-4*exp(x)*x+2*x^2+exp(8)-x+10)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]

[In]

integrate((-32*exp(x)^2+(32*x+32)*exp(x)-32*x+8)/(4*exp(x)^4-16*x*exp(x)^3+(4*exp(8)+24*x^2-4*x+40)*exp(x)^2+(
-8*x*exp(8)-16*x^3+8*x^2-80*x)*exp(x)+exp(8)^2+(4*x^2-2*x+20)*exp(8)+4*x^4-4*x^3+41*x^2-20*x+100),x, algorithm
="fricas")

[Out]

8/(2*x^2 - 4*x*e^x - x + e^8 + 2*e^(2*x) + 10)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {4}{x^{2} - 2 x e^{x} - \frac {x}{2} + e^{2 x} + 5 + \frac {e^{8}}{2}} \]

[In]

integrate((-32*exp(x)**2+(32*x+32)*exp(x)-32*x+8)/(4*exp(x)**4-16*x*exp(x)**3+(4*exp(8)+24*x**2-4*x+40)*exp(x)
**2+(-8*x*exp(8)-16*x**3+8*x**2-80*x)*exp(x)+exp(8)**2+(4*x**2-2*x+20)*exp(8)+4*x**4-4*x**3+41*x**2-20*x+100),
x)

[Out]

4/(x**2 - 2*x*exp(x) - x/2 + exp(2*x) + 5 + exp(8)/2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {8}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]

[In]

integrate((-32*exp(x)^2+(32*x+32)*exp(x)-32*x+8)/(4*exp(x)^4-16*x*exp(x)^3+(4*exp(8)+24*x^2-4*x+40)*exp(x)^2+(
-8*x*exp(8)-16*x^3+8*x^2-80*x)*exp(x)+exp(8)^2+(4*x^2-2*x+20)*exp(8)+4*x^4-4*x^3+41*x^2-20*x+100),x, algorithm
="maxima")

[Out]

8/(2*x^2 - 4*x*e^x - x + e^8 + 2*e^(2*x) + 10)

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=\frac {16}{2 \, x^{2} - 4 \, x e^{x} - x + e^{8} + 2 \, e^{\left (2 \, x\right )} + 10} \]

[In]

integrate((-32*exp(x)^2+(32*x+32)*exp(x)-32*x+8)/(4*exp(x)^4-16*x*exp(x)^3+(4*exp(8)+24*x^2-4*x+40)*exp(x)^2+(
-8*x*exp(8)-16*x^3+8*x^2-80*x)*exp(x)+exp(8)^2+(4*x^2-2*x+20)*exp(8)+4*x^4-4*x^3+41*x^2-20*x+100),x, algorithm
="giac")

[Out]

16/(2*x^2 - 4*x*e^x - x + e^8 + 2*e^(2*x) + 10)

Mupad [B] (verification not implemented)

Time = 12.50 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54 \[ \int \frac {8-32 e^{2 x}-32 x+e^x (32+32 x)}{100+e^{16}+4 e^{4 x}-20 x-16 e^{3 x} x+41 x^2-4 x^3+4 x^4+e^8 \left (20-2 x+4 x^2\right )+e^{2 x} \left (40+4 e^8-4 x+24 x^2\right )+e^x \left (-80 x-8 e^8 x+8 x^2-16 x^3\right )} \, dx=-\frac {\frac {16\,x^2}{{\mathrm {e}}^8+10}+\frac {16\,{\mathrm {e}}^{2\,x}}{{\mathrm {e}}^8+10}-\frac {8\,x}{{\mathrm {e}}^8+10}-\frac {32\,x\,{\mathrm {e}}^x}{{\mathrm {e}}^8+10}}{2\,{\mathrm {e}}^{2\,x}-x+{\mathrm {e}}^8-4\,x\,{\mathrm {e}}^x+2\,x^2+10} \]

[In]

int(-(32*x + 32*exp(2*x) - exp(x)*(32*x + 32) - 8)/(4*exp(4*x) - 20*x + exp(16) - 16*x*exp(3*x) - exp(x)*(80*x
+ 8*x*exp(8) - 8*x^2 + 16*x^3) + exp(8)*(4*x^2 - 2*x + 20) + exp(2*x)*(4*exp(8) - 4*x + 24*x^2 + 40) + 41*x^2
- 4*x^3 + 4*x^4 + 100),x)

[Out]

-((16*x^2)/(exp(8) + 10) + (16*exp(2*x))/(exp(8) + 10) - (8*x)/(exp(8) + 10) - (32*x*exp(x))/(exp(8) + 10))/(2
*exp(2*x) - x + exp(8) - 4*x*exp(x) + 2*x^2 + 10)