3.3.18 \(\int \frac {5+e (-1-2 x)+10 x+e^{e^x} (1+2 x+e^x (3-x-x^2))}{25-10 e+e^2+(10-2 e) e^{e^x}+e^{2 e^x}} \, dx\)

Optimal. Leaf size=26 \[ 1-\frac {3-x-x^2}{5-e+e^{e^x}} \]

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Rubi [F]  time = 1.74, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {5+e (-1-2 x)+10 x+e^{e^x} \left (1+2 x+e^x \left (3-x-x^2\right )\right )}{25-10 e+e^2+(10-2 e) e^{e^x}+e^{2 e^x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5 + E*(-1 - 2*x) + 10*x + E^E^x*(1 + 2*x + E^x*(3 - x - x^2)))/(25 - 10*E + E^2 + (10 - 2*E)*E^E^x + E^(2
*E^x)),x]

[Out]

-3/(5 - E + E^E^x) - 1/(E^x*(5 - E + E^E^x)) + 10*Defer[Int][x/(5*(1 - E/5) + E^E^x)^2, x] - 2*E*Defer[Int][x/
(5*(1 - E/5) + E^E^x)^2, x] + 2*Defer[Int][(E^E^x*x)/(5*(1 - E/5) + E^E^x)^2, x] - Defer[Int][(E^(E^x + x)*x)/
(5*(1 - E/5) + E^E^x)^2, x] - Defer[Int][(E^(E^x + x)*x^2)/(5*(1 - E/5) + E^E^x)^2, x] - Defer[Subst][Defer[In
t][1/((5*(1 - E/5) + E^x)*x^2), x], x, E^x] + 5*Defer[Subst][Defer[Int][1/((5*(1 - E/5) + E^x)^2*x), x], x, E^
x] - E*Defer[Subst][Defer[Int][1/((5*(1 - E/5) + E^x)^2*x), x], x, E^x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+e (-1-2 x)+10 x+e^{e^x} \left (1+2 x+e^x \left (3-x-x^2\right )\right )}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx\\ &=\int \left (\frac {5}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}+\frac {e^{e^x}}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}+\frac {e (-1-2 x)}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}+\frac {10 x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}+\frac {2 e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}+\frac {e^{e^x+x} \left (3-x-x^2\right )}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}\right ) \, dx\\ &=2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+5 \int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+e \int \frac {-1-2 x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+\int \frac {e^{e^x}}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+\int \frac {e^{e^x+x} \left (3-x-x^2\right )}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx\\ &=2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+5 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+e \int \left (-\frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}-\frac {2 x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}\right ) \, dx+\int \left (\frac {3 e^{e^x+x}}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}-\frac {e^{e^x+x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}-\frac {e^{e^x+x} x^2}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2}\right ) \, dx+\operatorname {Subst}\left (\int \frac {e^x}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )\\ &=-\frac {e^{-x}}{5-e+e^{e^x}}+2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+3 \int \frac {e^{e^x+x}}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+5 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-e \int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-(2 e) \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x^2}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right ) x^2} \, dx,x,e^x\right )\\ &=-\frac {e^{-x}}{5-e+e^{e^x}}+2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+3 \operatorname {Subst}\left (\int \frac {e^x}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2} \, dx,x,e^x\right )+5 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-e \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )-(2 e) \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x^2}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right ) x^2} \, dx,x,e^x\right )\\ &=-\frac {e^{-x}}{5-e+e^{e^x}}+2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+3 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+x\right )^2} \, dx,x,e^{e^x}\right )+5 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-e \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )-(2 e) \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x^2}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right ) x^2} \, dx,x,e^x\right )\\ &=-\frac {3}{5-e+e^{e^x}}-\frac {e^{-x}}{5-e+e^{e^x}}+2 \int \frac {e^{e^x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx+5 \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )+10 \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-e \operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right )^2 x} \, dx,x,e^x\right )-(2 e) \int \frac {x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\int \frac {e^{e^x+x} x^2}{\left (5 \left (1-\frac {e}{5}\right )+e^{e^x}\right )^2} \, dx-\operatorname {Subst}\left (\int \frac {1}{\left (5 \left (1-\frac {e}{5}\right )+e^x\right ) x^2} \, dx,x,e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 19, normalized size = 0.73 \begin {gather*} \frac {-3+x+x^2}{5-e+e^{e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E*(-1 - 2*x) + 10*x + E^E^x*(1 + 2*x + E^x*(3 - x - x^2)))/(25 - 10*E + E^2 + (10 - 2*E)*E^E^x
+ E^(2*E^x)),x]

[Out]

(-3 + x + x^2)/(5 - E + E^E^x)

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fricas [A]  time = 0.91, size = 19, normalized size = 0.73 \begin {gather*} -\frac {x^{2} + x - 3}{e - e^{\left (e^{x}\right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-x+3)*exp(x)+2*x+1)*exp(exp(x))+(-2*x-1)*exp(1)+10*x+5)/(exp(exp(x))^2+(-2*exp(1)+10)*exp(exp
(x))+exp(1)^2-10*exp(1)+25),x, algorithm="fricas")

[Out]

-(x^2 + x - 3)/(e - e^(e^x) - 5)

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giac [A]  time = 0.26, size = 19, normalized size = 0.73 \begin {gather*} -\frac {x^{2} + x - 3}{e - e^{\left (e^{x}\right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-x+3)*exp(x)+2*x+1)*exp(exp(x))+(-2*x-1)*exp(1)+10*x+5)/(exp(exp(x))^2+(-2*exp(1)+10)*exp(exp
(x))+exp(1)^2-10*exp(1)+25),x, algorithm="giac")

[Out]

-(x^2 + x - 3)/(e - e^(e^x) - 5)

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maple [A]  time = 0.15, size = 20, normalized size = 0.77

method	result	size
risch	\(-\frac {x^{2}+x -3}{-{\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}-5}\)	\(20\)
norman	\(\frac {-x^{2}-x +3}{-{\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}-5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2-x+3)*exp(x)+2*x+1)*exp(exp(x))+(-2*x-1)*exp(1)+10*x+5)/(exp(exp(x))^2+(-2*exp(1)+10)*exp(exp(x))+e
xp(1)^2-10*exp(1)+25),x,method=_RETURNVERBOSE)

[Out]

-(x^2+x-3)/(-exp(exp(x))+exp(1)-5)

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maxima [A]  time = 0.58, size = 19, normalized size = 0.73 \begin {gather*} -\frac {x^{2} + x - 3}{e - e^{\left (e^{x}\right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-x+3)*exp(x)+2*x+1)*exp(exp(x))+(-2*x-1)*exp(1)+10*x+5)/(exp(exp(x))^2+(-2*exp(1)+10)*exp(exp
(x))+exp(1)^2-10*exp(1)+25),x, algorithm="maxima")

[Out]

-(x^2 + x - 3)/(e - e^(e^x) - 5)

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mupad [B]  time = 0.46, size = 18, normalized size = 0.69 \begin {gather*} \frac {x^2+x-3}{{\mathrm {e}}^{{\mathrm {e}}^x}-\mathrm {e}+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + exp(exp(x))*(2*x - exp(x)*(x + x^2 - 3) + 1) - exp(1)*(2*x + 1) + 5)/(exp(2) - 10*exp(1) + exp(2*e
xp(x)) - exp(exp(x))*(2*exp(1) - 10) + 25),x)

[Out]

(x + x^2 - 3)/(exp(exp(x)) - exp(1) + 5)

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sympy [A]  time = 0.15, size = 15, normalized size = 0.58 \begin {gather*} \frac {x^{2} + x - 3}{e^{e^{x}} - e + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2-x+3)*exp(x)+2*x+1)*exp(exp(x))+(-2*x-1)*exp(1)+10*x+5)/(exp(exp(x))**2+(-2*exp(1)+10)*exp(e
xp(x))+exp(1)**2-10*exp(1)+25),x)

[Out]

(x**2 + x - 3)/(exp(exp(x)) - E + 5)

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