3.41.3 \(\int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} (-500-200 x-20 x^2+e^4 (-50 x-20 x^2-2 x^3)+e^{\frac {x^2}{5+x}} (200 x+20 x^2+e^4 (20 x+22 x^2+2 x^3)))}{25+10 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ 2 e^{e^{\frac {x^2}{5+x}}-x} \left (1+\frac {10}{e^4}+x\right ) \]

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Rubi [F]  time = 3.64, 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^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-4 + E^(x^2/(5 + x)) - x)*(-500 - 200*x - 20*x^2 + E^4*(-50*x - 20*x^2 - 2*x^3) + E^(x^2/(5 + x))*(200
*x + 20*x^2 + E^4*(20*x + 22*x^2 + 2*x^3))))/(25 + 10*x + x^2),x]

[Out]

-20*Defer[Int][E^(-4 + E^(x^2/(5 + x)) - x), x] + 2*(10 + E^4)*Defer[Int][E^(-4 + E^(x^2/(5 + x)) - x + x^2/(5
+ x)), x] - 2*Defer[Int][E^(E^(x^2/(5 + x)) - x)*x, x] + 2*Defer[Int][E^(E^(x^2/(5 + x)) - x + x^2/(5 + x))*x
, x] - 100*(5 - 2*E^4)*Defer[Int][E^(-4 + E^(x^2/(5 + x)) - x + x^2/(5 + x))/(5 + x)^2, x] - 50*Defer[Int][E^(
E^(x^2/(5 + x)) - x + x^2/(5 + x))/(5 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{(5+x)^2} \, dx\\ &=\int \left (-2 e^{-4+e^{\frac {x^2}{5+x}}-x} \left (10+e^4 x\right )+\frac {2 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x (10+x) \left (10+e^4+e^4 x\right )}{(5+x)^2}\right ) \, dx\\ &=-\left (2 \int e^{-4+e^{\frac {x^2}{5+x}}-x} \left (10+e^4 x\right ) \, dx\right )+2 \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x (10+x) \left (10+e^4+e^4 x\right )}{(5+x)^2} \, dx\\ &=-\left (2 \int \left (10 e^{-4+e^{\frac {x^2}{5+x}}-x}+e^{e^{\frac {x^2}{5+x}}-x} x\right ) \, dx\right )+2 \int \left (10 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \left (1+\frac {e^4}{10}\right )+e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x+\frac {50 e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \left (-5+2 e^4\right )}{(5+x)^2}-\frac {25 e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{5+x}\right ) \, dx\\ &=-\left (2 \int e^{e^{\frac {x^2}{5+x}}-x} x \, dx\right )+2 \int e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} x \, dx-20 \int e^{-4+e^{\frac {x^2}{5+x}}-x} \, dx-50 \int \frac {e^{e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{5+x} \, dx-\left (100 \left (5-2 e^4\right )\right ) \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}}}{(5+x)^2} \, dx+\left (2 \left (10+e^4\right )\right ) \int e^{-4+e^{\frac {x^2}{5+x}}-x+\frac {x^2}{5+x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.53, size = 29, normalized size = 1.07 \begin {gather*} e^{-4+e^{\frac {x^2}{5+x}}-x} \left (20+2 e^4 (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + E^(x^2/(5 + x)) - x)*(-500 - 200*x - 20*x^2 + E^4*(-50*x - 20*x^2 - 2*x^3) + E^(x^2/(5 + x)
)*(200*x + 20*x^2 + E^4*(20*x + 22*x^2 + 2*x^3))))/(25 + 10*x + x^2),x]

[Out]

E^(-4 + E^(x^2/(5 + x)) - x)*(20 + 2*E^4*(1 + x))

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fricas [A]  time = 0.70, size = 26, normalized size = 0.96 \begin {gather*} 2 \, {\left ({\left (x + 1\right )} e^{4} + 10\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x^2-50*x)*exp(4)-20*x^2-200*x-5
00)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x))+x),x, algorithm="fricas")

[Out]

2*((x + 1)*e^4 + 10)*e^(-x + e^(x^2/(x + 5)) - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (10 \, x^{2} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{4} - {\left (10 \, x^{2} + {\left (x^{3} + 11 \, x^{2} + 10 \, x\right )} e^{4} + 100 \, x\right )} e^{\left (\frac {x^{2}}{x + 5}\right )} + 100 \, x + 250\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )}}{x^{2} + 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x^2-50*x)*exp(4)-20*x^2-200*x-5
00)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x))+x),x, algorithm="giac")

[Out]

integrate(-2*(10*x^2 + (x^3 + 10*x^2 + 25*x)*e^4 - (10*x^2 + (x^3 + 11*x^2 + 10*x)*e^4 + 100*x)*e^(x^2/(x + 5)
) + 100*x + 250)*e^(-x + e^(x^2/(x + 5)) - 4)/(x^2 + 10*x + 25), x)

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maple [A]  time = 0.13, size = 29, normalized size = 1.07

method	result	size
risch	\(\left (2 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}+20\right ) {\mathrm e}^{-4+{\mathrm e}^{\frac {x^{2}}{5+x}}-x}\)	\(29\)
norman	\(\frac {\left (2 x^{2}+10 \left (10+{\mathrm e}^{4}\right ) {\mathrm e}^{-4}+4 \left (5+3 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4} x \right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{5+x}}-x}}{5+x}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x^2-50*x)*exp(4)-20*x^2-200*x-500)/(x
^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x))+x),x,method=_RETURNVERBOSE)

[Out]

(2*x*exp(4)+2*exp(4)+20)*exp(-4+exp(x^2/(5+x))-x)

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maxima [A]  time = 0.45, size = 27, normalized size = 1.00 \begin {gather*} 2 \, {\left (x e^{4} + e^{4} + 10\right )} e^{\left (-x + e^{\left (x + \frac {25}{x + 5} - 5\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x^2-50*x)*exp(4)-20*x^2-200*x-5
00)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x))+x),x, algorithm="maxima")

[Out]

2*(x*e^4 + e^4 + 10)*e^(-x + e^(x + 25/(x + 5) - 5) - 4)

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mupad [B]  time = 2.89, size = 29, normalized size = 1.07 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{\frac {x^2}{x+5}}-x}\,\left (2\,x+{\mathrm {e}}^{-4}\,\left (2\,{\mathrm {e}}^4+20\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-4)*exp(exp(x^2/(x + 5)) - x)*(200*x - exp(x^2/(x + 5))*(200*x + exp(4)*(20*x + 22*x^2 + 2*x^3) + 20
*x^2) + exp(4)*(50*x + 20*x^2 + 2*x^3) + 20*x^2 + 500))/(10*x + x^2 + 25),x)

[Out]

exp(exp(x^2/(x + 5)) - x)*(2*x + exp(-4)*(2*exp(4) + 20))

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sympy [A]  time = 17.47, size = 27, normalized size = 1.00 \begin {gather*} \frac {\left (2 x e^{4} + 20 + 2 e^{4}\right ) e^{- x + e^{\frac {x^{2}}{x + 5}}}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3+22*x**2+20*x)*exp(4)+20*x**2+200*x)*exp(x**2/(5+x))+(-2*x**3-20*x**2-50*x)*exp(4)-20*x**2-
200*x-500)/(x**2+10*x+25)/exp(4)/exp(-exp(x**2/(5+x))+x),x)

[Out]

(2*x*exp(4) + 20 + 2*exp(4))*exp(-4)*exp(-x + exp(x**2/(x + 5)))

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