3.25.73 \(\int \frac {e^{-e^x+x} (-1+5 x+2 x^2-x^3+e^x (-2-3 x+x^2+x^3))}{(20+10 x) (2+5 x-5 x^3+x^5)} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{-e^x+x}}{5 (4+2 x) \left (1+x-x^2\right )} \]

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Rubi [F]  time = 5.96, 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^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-E^x + x)*(-1 + 5*x + 2*x^2 - x^3 + E^x*(-2 - 3*x + x^2 + x^3)))/((20 + 10*x)*(2 + 5*x - 5*x^3 + x^5))
,x]

[Out]

(-2*Defer[Int][E^(-E^x + x)/(1 + Sqrt[5] - 2*x)^2, x])/25 + ((1 + Sqrt[5])*Defer[Int][E^(-E^x + x)/(1 + Sqrt[5
] - 2*x)^2, x])/25 - Defer[Int][E^(-E^x + x)/(1 + Sqrt[5] - 2*x), x]/(25*Sqrt[5]) + Defer[Int][E^(-E^x + x)/(2
+ x)^2, x]/50 - Defer[Int][E^(-E^x + x)/(2 + x), x]/50 + Defer[Int][E^(-E^x + 2*x)/(2 + x), x]/50 + ((15 - 11
*Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 - Sqrt[5] + 2*x), x])/250 + ((5 - 4*Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1
- Sqrt[5] + 2*x), x])/625 - (4*(5 - 3*Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 - Sqrt[5] + 2*x), x])/625 - (2*(5
- Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 - Sqrt[5] + 2*x), x])/625 - ((1 - Sqrt[5])*Defer[Int][E^(-E^x + 2*x)/(-
1 - Sqrt[5] + 2*x), x])/50 - (2*Defer[Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x)^2, x])/25 + ((1 - Sqrt[5])*Defer[
Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x)^2, x])/25 - Defer[Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x), x]/(25*Sqrt[5
]) - (2*(5 + Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x), x])/625 - (4*(5 + 3*Sqrt[5])*Defer[Int][E^
(-E^x + x)/(-1 + Sqrt[5] + 2*x), x])/625 + ((5 + 4*Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x), x])/
625 + ((15 + 11*Sqrt[5])*Defer[Int][E^(-E^x + x)/(-1 + Sqrt[5] + 2*x), x])/250 - ((1 + Sqrt[5])*Defer[Int][E^(
-E^x + 2*x)/(-1 + Sqrt[5] + 2*x), x])/50

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^{-e^x+x}}{10 \left (-2-3 x+x^2+x^3\right )^2}+\frac {e^{-e^x+x} x}{2 \left (-2-3 x+x^2+x^3\right )^2}+\frac {e^{-e^x+x} x^2}{5 \left (-2-3 x+x^2+x^3\right )^2}-\frac {e^{-e^x+x} x^3}{10 \left (-2-3 x+x^2+x^3\right )^2}+\frac {e^{-e^x+2 x}}{10 \left (-2-3 x+x^2+x^3\right )}\right ) \, dx\\ &=-\left (\frac {1}{10} \int \frac {e^{-e^x+x}}{\left (-2-3 x+x^2+x^3\right )^2} \, dx\right )-\frac {1}{10} \int \frac {e^{-e^x+x} x^3}{\left (-2-3 x+x^2+x^3\right )^2} \, dx+\frac {1}{10} \int \frac {e^{-e^x+2 x}}{-2-3 x+x^2+x^3} \, dx+\frac {1}{5} \int \frac {e^{-e^x+x} x^2}{\left (-2-3 x+x^2+x^3\right )^2} \, dx+\frac {1}{2} \int \frac {e^{-e^x+x} x}{\left (-2-3 x+x^2+x^3\right )^2} \, dx\\ &=-\left (\frac {1}{10} \int \left (\frac {e^{-e^x+x}}{25 (2+x)^2}+\frac {2 e^{-e^x+x}}{25 (2+x)}+\frac {e^{-e^x+x} (2-x)}{5 \left (-1-x+x^2\right )^2}+\frac {e^{-e^x+x} (5-2 x)}{25 \left (-1-x+x^2\right )}\right ) \, dx\right )+\frac {1}{10} \int \left (\frac {e^{-e^x+2 x}}{5 (2+x)}+\frac {e^{-e^x+2 x} (3-x)}{5 \left (-1-x+x^2\right )}\right ) \, dx-\frac {1}{10} \int \left (-\frac {8 e^{-e^x+x}}{25 (2+x)^2}-\frac {4 e^{-e^x+x}}{25 (2+x)}+\frac {e^{-e^x+x} x}{5 \left (-1-x+x^2\right )^2}+\frac {4 e^{-e^x+x} (-1+x)}{25 \left (-1-x+x^2\right )}\right ) \, dx+\frac {1}{5} \int \left (\frac {4 e^{-e^x+x}}{25 (2+x)^2}+\frac {4 e^{-e^x+x}}{25 (2+x)}+\frac {e^{-e^x+x}}{5 \left (-1-x+x^2\right )^2}-\frac {4 e^{-e^x+x} (-2+x)}{25 \left (-1-x+x^2\right )}\right ) \, dx+\frac {1}{2} \int \left (-\frac {2 e^{-e^x+x}}{25 (2+x)^2}-\frac {3 e^{-e^x+x}}{25 (2+x)}+\frac {e^{-e^x+x} (-1+x)}{5 \left (-1-x+x^2\right )^2}+\frac {e^{-e^x+x} (-7+3 x)}{25 \left (-1-x+x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{250} \int \frac {e^{-e^x+x}}{(2+x)^2} \, dx\right )-\frac {1}{250} \int \frac {e^{-e^x+x} (5-2 x)}{-1-x+x^2} \, dx-\frac {1}{125} \int \frac {e^{-e^x+x}}{2+x} \, dx+\frac {2}{125} \int \frac {e^{-e^x+x}}{2+x} \, dx-\frac {2}{125} \int \frac {e^{-e^x+x} (-1+x)}{-1-x+x^2} \, dx+\frac {1}{50} \int \frac {e^{-e^x+2 x}}{2+x} \, dx-\frac {1}{50} \int \frac {e^{-e^x+x} (2-x)}{\left (-1-x+x^2\right )^2} \, dx-\frac {1}{50} \int \frac {e^{-e^x+x} x}{\left (-1-x+x^2\right )^2} \, dx+\frac {1}{50} \int \frac {e^{-e^x+2 x} (3-x)}{-1-x+x^2} \, dx+\frac {1}{50} \int \frac {e^{-e^x+x} (-7+3 x)}{-1-x+x^2} \, dx+2 \left (\frac {4}{125} \int \frac {e^{-e^x+x}}{(2+x)^2} \, dx\right )+\frac {4}{125} \int \frac {e^{-e^x+x}}{2+x} \, dx-\frac {4}{125} \int \frac {e^{-e^x+x} (-2+x)}{-1-x+x^2} \, dx-\frac {1}{25} \int \frac {e^{-e^x+x}}{(2+x)^2} \, dx+\frac {1}{25} \int \frac {e^{-e^x+x}}{\left (-1-x+x^2\right )^2} \, dx-\frac {3}{50} \int \frac {e^{-e^x+x}}{2+x} \, dx+\frac {1}{10} \int \frac {e^{-e^x+x} (-1+x)}{\left (-1-x+x^2\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.55, size = 26, normalized size = 0.87 \begin {gather*} -\frac {e^{-e^x+x}}{10 \left (-2-3 x+x^2+x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-E^x + x)*(-1 + 5*x + 2*x^2 - x^3 + E^x*(-2 - 3*x + x^2 + x^3)))/((20 + 10*x)*(2 + 5*x - 5*x^3 +
x^5)),x]

[Out]

-1/10*E^(-E^x + x)/(-2 - 3*x + x^2 + x^3)

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fricas [A]  time = 0.73, size = 31, normalized size = 1.03 \begin {gather*} -\frac {e^{\left (2 \, x - e^{x} - \log \left (10 \, {\left (x + 2\right )} e^{x}\right )\right )}}{x^{2} - x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp(x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x,
algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x^{3} - 2 \, x^{2} - {\left (x^{3} + x^{2} - 3 \, x - 2\right )} e^{x} - 5 \, x + 1\right )} e^{\left (2 \, x - e^{x} - \log \left (10 \, {\left (x + 2\right )} e^{x}\right )\right )}}{x^{5} - 5 \, x^{3} + 5 \, x + 2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp(x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x,
algorithm="giac")

[Out]

integrate(-(x^3 - 2*x^2 - (x^3 + x^2 - 3*x - 2)*e^x - 5*x + 1)*e^(2*x - e^x - log(10*(x + 2)*e^x))/(x^5 - 5*x^
3 + 5*x + 2), x)

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maple [C]  time = 0.11, size = 109, normalized size = 3.63

method	result	size
risch	\(-\frac {{\mathrm e}^{x +\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{3}}{2}-\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )}{2}-\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i \left (2+x \right )\right )}{2}-{\mathrm e}^{x}}}{10 \left (x^{2}-x -1\right ) \left (2+x \right )}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-ln((10*x+20)*exp(x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x,method=
_RETURNVERBOSE)

[Out]

-1/10/(x^2-x-1)/(2+x)*exp(x+1/2*I*Pi*csgn(I*exp(x)*(2+x))^3-1/2*I*Pi*csgn(I*exp(x)*(2+x))^2*csgn(I*exp(x))-1/2
*I*Pi*csgn(I*exp(x)*(2+x))^2*csgn(I*(2+x))+1/2*I*Pi*csgn(I*exp(x)*(2+x))*csgn(I*exp(x))*csgn(I*(2+x))-exp(x))

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maxima [A]  time = 0.77, size = 22, normalized size = 0.73 \begin {gather*} -\frac {e^{\left (x - e^{x}\right )}}{10 \, {\left (x^{3} + x^{2} - 3 \, x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp(x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x,
algorithm="maxima")

[Out]

-1/10*e^(x - e^x)/(x^3 + x^2 - 3*x - 2)

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mupad [B]  time = 1.72, size = 36, normalized size = 1.20 \begin {gather*} \frac {{\mathrm {e}}^{2\,x-{\mathrm {e}}^x}}{20\,{\mathrm {e}}^x-10\,x^2\,{\mathrm {e}}^x-10\,x^3\,{\mathrm {e}}^x+30\,x\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x - exp(x) - log(exp(x)*(10*x + 20)))*(x^3 - 2*x^2 - 5*x + exp(x)*(3*x - x^2 - x^3 + 2) + 1))/(5*x
- 5*x^3 + x^5 + 2),x)

[Out]

exp(2*x - exp(x))/(20*exp(x) - 10*x^2*exp(x) - 10*x^3*exp(x) + 30*x*exp(x))

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sympy [A]  time = 0.26, size = 27, normalized size = 0.90 \begin {gather*} - \frac {e^{- x} e^{2 x - e^{x}}}{10 x^{3} + 10 x^{2} - 30 x - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+x**2-3*x-2)*exp(x)-x**3+2*x**2+5*x-1)*exp(-ln((10*x+20)*exp(x))+2*x-exp(x))/(x**5-5*x**3+5*x+
2),x)

[Out]

-exp(-x)*exp(2*x - exp(x))/(10*x**3 + 10*x**2 - 30*x - 20)

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