3.61.95 \(\int \frac {e^{2 x}+e^{2 e^{25-10 x+x^2}} (4+4 e^x)+e^{e^{25-10 x+x^2}} (-4 e^x+e^{25-8 x+x^2} (20-4 x))}{8 e^{2 e^{25-10 x+x^2}}+2 e^{2 x}-8 e^{e^{25-10 x+x^2}+x}} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^x}{2-e^{-e^{(5-x)^2}+x}}+\frac {x}{2} \]

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Rubi [F]  time = 4.06, 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 {e^{2 x}+e^{2 e^{25-10 x+x^2}} \left (4+4 e^x\right )+e^{e^{25-10 x+x^2}} \left (-4 e^x+e^{25-8 x+x^2} (20-4 x)\right )}{8 e^{2 e^{25-10 x+x^2}}+2 e^{2 x}-8 e^{e^{25-10 x+x^2}+x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x) + E^(2*E^(25 - 10*x + x^2))*(4 + 4*E^x) + E^E^(25 - 10*x + x^2)*(-4*E^x + E^(25 - 8*x + x^2)*(20
- 4*x)))/(8*E^(2*E^(25 - 10*x + x^2)) + 2*E^(2*x) - 8*E^(E^(25 - 10*x + x^2) + x)),x]

[Out]

x/2 + 4*Defer[Int][E^(3*E^(25 - 10*x + x^2))/(2*E^E^(-5 + x)^2 - E^x)^2, x] + 10*Defer[Int][E^(25 + E^(-5 + x)
^2 - 8*x + x^2)/(2*E^E^(-5 + x)^2 - E^x)^2, x] + 2*Defer[Int][E^(2*E^(25 - 10*x + x^2))/(-2*E^E^(-5 + x)^2 + E
^x), x] - 2*Defer[Int][(E^(25 + E^(-5 + x)^2 - 8*x + x^2)*x)/(2*E^E^(-5 + x)^2 - E^x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x}+e^{2 e^{25-10 x+x^2}} \left (4+4 e^x\right )+e^{e^{25-10 x+x^2}} \left (-4 e^x+e^{25-8 x+x^2} (20-4 x)\right )}{2 \left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{2 x}+e^{2 e^{25-10 x+x^2}} \left (4+4 e^x\right )+e^{e^{25-10 x+x^2}} \left (-4 e^x+e^{25-8 x+x^2} (20-4 x)\right )}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {4 e^{2 e^{(-5+x)^2}}+e^{2 x}-4 e^{e^{(-5+x)^2}+x}+4 e^{2 e^{(-5+x)^2}+x}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2}-\frac {4 e^{25+e^{(-5+x)^2}-8 x+x^2} (-5+x)}{\left (-2 e^{e^{(-5+x)^2}}+e^x\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {4 e^{2 e^{(-5+x)^2}}+e^{2 x}-4 e^{e^{(-5+x)^2}+x}+4 e^{2 e^{(-5+x)^2}+x}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx-2 \int \frac {e^{25+e^{(-5+x)^2}-8 x+x^2} (-5+x)}{\left (-2 e^{e^{(-5+x)^2}}+e^x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (1+\frac {8 e^{3 e^{25-10 x+x^2}}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2}+\frac {4 e^{2 e^{25-10 x+x^2}}}{-2 e^{e^{(-5+x)^2}}+e^x}\right ) \, dx-2 \int \left (-\frac {5 e^{25+e^{(-5+x)^2}-8 x+x^2}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2}+\frac {e^{25+e^{(-5+x)^2}-8 x+x^2} x}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2}\right ) \, dx\\ &=\frac {x}{2}+2 \int \frac {e^{2 e^{25-10 x+x^2}}}{-2 e^{e^{(-5+x)^2}}+e^x} \, dx-2 \int \frac {e^{25+e^{(-5+x)^2}-8 x+x^2} x}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx+4 \int \frac {e^{3 e^{25-10 x+x^2}}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx+10 \int \frac {e^{25+e^{(-5+x)^2}-8 x+x^2}}{\left (2 e^{e^{(-5+x)^2}}-e^x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 35, normalized size = 1.13 \begin {gather*} \frac {1}{2} \left (e^x+\frac {e^x}{-1+2 e^{e^{25-10 x+x^2}-x}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x) + E^(2*E^(25 - 10*x + x^2))*(4 + 4*E^x) + E^E^(25 - 10*x + x^2)*(-4*E^x + E^(25 - 8*x + x^2
)*(20 - 4*x)))/(8*E^(2*E^(25 - 10*x + x^2)) + 2*E^(2*x) - 8*E^(E^(25 - 10*x + x^2) + x)),x]

[Out]

(E^x + E^x/(-1 + 2*E^(E^(25 - 10*x + x^2) - x)) + x)/2

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fricas [B]  time = 0.85, size = 71, normalized size = 2.29 \begin {gather*} \frac {2 \, {\left (x + e^{x}\right )} e^{\left ({\left (x e^{\left (2 \, x\right )} + e^{\left (x^{2} - 8 \, x + 25\right )}\right )} e^{\left (-2 \, x\right )}\right )} - x e^{\left (2 \, x\right )}}{2 \, {\left (2 \, e^{\left ({\left (x e^{\left (2 \, x\right )} + e^{\left (x^{2} - 8 \, x + 25\right )}\right )} e^{\left (-2 \, x\right )}\right )} - e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+4)*exp(exp(x^2-10*x+25))^2+((-4*x+20)*exp(x)^2*exp(x^2-10*x+25)-4*exp(x))*exp(exp(x^2-10*
x+25))+exp(x)^2)/(8*exp(exp(x^2-10*x+25))^2-8*exp(x)*exp(exp(x^2-10*x+25))+2*exp(x)^2),x, algorithm="fricas")

[Out]

1/2*(2*(x + e^x)*e^((x*e^(2*x) + e^(x^2 - 8*x + 25))*e^(-2*x)) - x*e^(2*x))/(2*e^((x*e^(2*x) + e^(x^2 - 8*x +
25))*e^(-2*x)) - e^(2*x))

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giac [B]  time = 0.24, size = 936, normalized size = 30.19 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+4)*exp(exp(x^2-10*x+25))^2+((-4*x+20)*exp(x)^2*exp(x^2-10*x+25)-4*exp(x))*exp(exp(x^2-10*
x+25))+exp(x)^2)/(8*exp(exp(x^2-10*x+25))^2-8*exp(x)*exp(exp(x^2-10*x+25))+2*exp(x)^2),x, algorithm="giac")

[Out]

1/2*(4*x^3*e^(2*x^2 - 9*x + 50) - 16*x^3*e^(2*x^2 - 10*x + e^(x^2 - 10*x + 25) + 50) + 16*x^3*e^(2*x^2 - 11*x
+ 2*e^(x^2 - 10*x + 25) + 50) - 8*x^2*e^(2*x^2 - 9*x + e^(x^2 - 10*x + 25) + 50) - 40*x^2*e^(2*x^2 - 9*x + 50)
+ 16*x^2*e^(2*x^2 - 10*x + 2*e^(x^2 - 10*x + 25) + 50) + 160*x^2*e^(2*x^2 - 10*x + e^(x^2 - 10*x + 25) + 50)
- 160*x^2*e^(2*x^2 - 11*x + 2*e^(x^2 - 10*x + 25) + 50) - 4*x^2*e^(x^2 + x + 25) - 16*x^2*e^(x^2 - x + 2*e^(x^
2 - 10*x + 25) + 25) + 16*x^2*e^(x^2 + e^(x^2 - 10*x + 25) + 25) + 80*x*e^(2*x^2 - 9*x + e^(x^2 - 10*x + 25) +
50) + 100*x*e^(2*x^2 - 9*x + 50) - 160*x*e^(2*x^2 - 10*x + 2*e^(x^2 - 10*x + 25) + 50) - 400*x*e^(2*x^2 - 10*
x + e^(x^2 - 10*x + 25) + 50) + 400*x*e^(2*x^2 - 11*x + 2*e^(x^2 - 10*x + 25) + 50) + 8*x*e^(x^2 + x + e^(x^2
- 10*x + 25) + 25) + 20*x*e^(x^2 + x + 25) + 80*x*e^(x^2 - x + 2*e^(x^2 - 10*x + 25) + 25) - 16*x*e^(x^2 + 2*e
^(x^2 - 10*x + 25) + 25) - 80*x*e^(x^2 + e^(x^2 - 10*x + 25) + 25) + x*e^(11*x) - 4*x*e^(10*x + e^(x^2 - 10*x
+ 25)) + 4*x*e^(9*x + 2*e^(x^2 - 10*x + 25)) - 200*e^(2*x^2 - 9*x + e^(x^2 - 10*x + 25) + 50) + 400*e^(2*x^2 -
10*x + 2*e^(x^2 - 10*x + 25) + 50) - 40*e^(x^2 + x + e^(x^2 - 10*x + 25) + 25) + 80*e^(x^2 + 2*e^(x^2 - 10*x
+ 25) + 25) - 2*e^(11*x + e^(x^2 - 10*x + 25)) + 4*e^(10*x + 2*e^(x^2 - 10*x + 25)))/(4*x^2*e^(2*x^2 - 9*x + 5
0) - 16*x^2*e^(2*x^2 - 10*x + e^(x^2 - 10*x + 25) + 50) + 16*x^2*e^(2*x^2 - 11*x + 2*e^(x^2 - 10*x + 25) + 50)
- 40*x*e^(2*x^2 - 9*x + 50) + 160*x*e^(2*x^2 - 10*x + e^(x^2 - 10*x + 25) + 50) - 160*x*e^(2*x^2 - 11*x + 2*e
^(x^2 - 10*x + 25) + 50) - 4*x*e^(x^2 + x + 25) - 16*x*e^(x^2 - x + 2*e^(x^2 - 10*x + 25) + 25) + 16*x*e^(x^2
+ e^(x^2 - 10*x + 25) + 25) + 100*e^(2*x^2 - 9*x + 50) - 400*e^(2*x^2 - 10*x + e^(x^2 - 10*x + 25) + 50) + 400
*e^(2*x^2 - 11*x + 2*e^(x^2 - 10*x + 25) + 50) + 20*e^(x^2 + x + 25) + 80*e^(x^2 - x + 2*e^(x^2 - 10*x + 25) +
25) - 80*e^(x^2 + e^(x^2 - 10*x + 25) + 25) + e^(11*x) - 4*e^(10*x + e^(x^2 - 10*x + 25)) + 4*e^(9*x + 2*e^(x
^2 - 10*x + 25)))

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maple [A]  time = 0.26, size = 29, normalized size = 0.94

method	result	size
risch	\(\frac {x}{2}+\frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{2 x}}{2 \left ({\mathrm e}^{x}-2 \,{\mathrm e}^{{\mathrm e}^{\left (x -5\right )^{2}}}\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(x)+4)*exp(exp(x^2-10*x+25))^2+((-4*x+20)*exp(x)^2*exp(x^2-10*x+25)-4*exp(x))*exp(exp(x^2-10*x+25))
+exp(x)^2)/(8*exp(exp(x^2-10*x+25))^2-8*exp(x)*exp(exp(x^2-10*x+25))+2*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 40, normalized size = 1.29 \begin {gather*} \frac {x e^{x} - 2 \, {\left (x + e^{x}\right )} e^{\left (e^{\left (x^{2} - 10 \, x + 25\right )}\right )}}{2 \, {\left (e^{x} - 2 \, e^{\left (e^{\left (x^{2} - 10 \, x + 25\right )}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+4)*exp(exp(x^2-10*x+25))^2+((-4*x+20)*exp(x)^2*exp(x^2-10*x+25)-4*exp(x))*exp(exp(x^2-10*
x+25))+exp(x)^2)/(8*exp(exp(x^2-10*x+25))^2-8*exp(x)*exp(exp(x^2-10*x+25))+2*exp(x)^2),x, algorithm="maxima")

[Out]

1/2*(x*e^x - 2*(x + e^x)*e^(e^(x^2 - 10*x + 25)))/(e^x - 2*e^(e^(x^2 - 10*x + 25)))

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mupad [B]  time = 4.52, size = 87, normalized size = 2.81 \begin {gather*} \frac {x}{2}+\frac {{\mathrm {e}}^x}{2}+\frac {{\mathrm {e}}^{3\,x}+10\,{\mathrm {e}}^{x^2-7\,x+25}-2\,x\,{\mathrm {e}}^{x^2-7\,x+25}}{2\,\left (2\,{\mathrm {e}}^{{\mathrm {e}}^{-10\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{25}}-{\mathrm {e}}^x\right )\,\left (10\,{\mathrm {e}}^{x^2-9\,x+25}+{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{x^2-9\,x+25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - exp(exp(x^2 - 10*x + 25))*(4*exp(x) + exp(2*x)*exp(x^2 - 10*x + 25)*(4*x - 20)) + exp(2*exp(x^
2 - 10*x + 25))*(4*exp(x) + 4))/(8*exp(2*exp(x^2 - 10*x + 25)) + 2*exp(2*x) - 8*exp(exp(x^2 - 10*x + 25))*exp(
x)),x)

[Out]

x/2 + exp(x)/2 + (exp(3*x) + 10*exp(x^2 - 7*x + 25) - 2*x*exp(x^2 - 7*x + 25))/(2*(2*exp(exp(-10*x)*exp(x^2)*e
xp(25)) - exp(x))*(10*exp(x^2 - 9*x + 25) + exp(x) - 2*x*exp(x^2 - 9*x + 25)))

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sympy [A]  time = 0.27, size = 31, normalized size = 1.00 \begin {gather*} \frac {x}{2} + \frac {e^{x}}{2} + \frac {e^{2 x}}{- 2 e^{x} + 4 e^{e^{x^{2} - 10 x + 25}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)+4)*exp(exp(x**2-10*x+25))**2+((-4*x+20)*exp(x)**2*exp(x**2-10*x+25)-4*exp(x))*exp(exp(x**
2-10*x+25))+exp(x)**2)/(8*exp(exp(x**2-10*x+25))**2-8*exp(x)*exp(exp(x**2-10*x+25))+2*exp(x)**2),x)

[Out]

x/2 + exp(x)/2 + exp(2*x)/(-2*exp(x) + 4*exp(exp(x**2 - 10*x + 25)))

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