3.76.50 \(\int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} (8+2 x+10 x^2)+e^{5 x} (-25 x+e^{-4/x} (20+5 x-25 x^2))}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx\)

Optimal. Leaf size=27 \[ x+\frac {-\frac {2}{5}-e^{5 x}}{1+e^{-4/x} x} \]

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Rubi [F]  time = 3.20, 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 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x + (5*x^3)/E^(8/x) + (8 + 2*x + 10*x^2)/E^(4/x) + E^(5*x)*(-25*x + (20 + 5*x - 25*x^2)/E^(4/x)))/(5*x
+ (10*x^2)/E^(4/x) + (5*x^3)/E^(8/x)),x]

[Out]

x - (8*Defer[Int][(E^(4/x) + x)^(-2), x])/5 + Defer[Int][E^(4/x + 5*x)/(E^(4/x) + x)^2, x] + 4*Defer[Int][E^(4
/x + 5*x)/(x*(E^(4/x) + x)^2), x] - (2*Defer[Int][x/(E^(4/x) + x)^2, x])/5 + (2*Defer[Int][(E^(4/x) + x)^(-1),
x])/5 - 5*Defer[Int][E^(4/x + 5*x)/(E^(4/x) + x), x] + (8*Defer[Int][1/(x*(E^(4/x) + x)), x])/5

Rubi steps

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

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Mathematica [F]  time = 180.09, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(5*x + (5*x^3)/E^(8/x) + (8 + 2*x + 10*x^2)/E^(4/x) + E^(5*x)*(-25*x + (20 + 5*x - 25*x^2)/E^(4/x)))
/(5*x + (10*x^2)/E^(4/x) + (5*x^3)/E^(8/x)),x]

[Out]

$Aborted

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fricas [A]  time = 0.77, size = 36, normalized size = 1.33 \begin {gather*} \frac {5 \, x^{2} e^{\left (-\frac {4}{x}\right )} + 5 \, x - 5 \, e^{\left (5 \, x\right )} - 2}{5 \, {\left (x e^{\left (-\frac {4}{x}\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(10*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*e
xp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.36, size = 42, normalized size = 1.56 \begin {gather*} \frac {5 \, x^{2} + 5 \, x e^{\frac {4}{x}} + 2 \, x - 5 \, e^{\left (5 \, x + \frac {4}{x}\right )}}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(10*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*e
xp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 25, normalized size = 0.93

method	result	size
risch	\(x -\frac {5 \,{\mathrm e}^{5 x}+2}{5 \left (x \,{\mathrm e}^{-\frac {4}{x}}+1\right )}\)	\(25\)
norman	\(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}+\frac {2 x \,{\mathrm e}^{-\frac {4}{x}}}{5}-{\mathrm e}^{5 x}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(10*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*exp(-4/
x)^2+10*x^2*exp(-4/x)+5*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 37, normalized size = 1.37 \begin {gather*} \frac {5 \, x^{2} + 5 \, {\left (x - e^{\left (5 \, x\right )}\right )} e^{\frac {4}{x}} + 2 \, x}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x^3*exp(-4/x)^2+(10*x^2+2*x+8)*exp(-4/x)+5*x)/(5*x^3*e
xp(-4/x)^2+10*x^2*exp(-4/x)+5*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.95, size = 34, normalized size = 1.26 \begin {gather*} x-\frac {\frac {2\,{\mathrm {e}}^{4/x}}{5}+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4/x}}{x+{\mathrm {e}}^{4/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - exp(5*x)*(25*x - exp(-4/x)*(5*x - 25*x^2 + 20)) + exp(-4/x)*(2*x + 10*x^2 + 8) + 5*x^3*exp(-8/x))/(
5*x + 10*x^2*exp(-4/x) + 5*x^3*exp(-8/x)),x)

[Out]

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

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sympy [A]  time = 0.38, size = 26, normalized size = 0.96 \begin {gather*} x - \frac {2}{5 x e^{- \frac {4}{x}} + 5} - \frac {e^{5 x}}{x e^{- \frac {4}{x}} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x**2+5*x+20)*exp(-4/x)-25*x)*exp(5*x)+5*x**3*exp(-4/x)**2+(10*x**2+2*x+8)*exp(-4/x)+5*x)/(5*x
**3*exp(-4/x)**2+10*x**2*exp(-4/x)+5*x),x)

[Out]

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

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