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

Optimal. Leaf size=25 \[ 8+\frac {1+e^x-x}{5 \left (e^{5+x}-x\right )} \]

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Rubi [F]  time = 0.70, 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 {1+e^x (1-x)+e^{5+x} (-2+x)}{5 e^{10+2 x}-10 e^{5+x} x+5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 29, normalized size = 1.16 \begin {gather*} \frac {e^5+x-e^5 x}{5 e^5 \left (e^{5+x}-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^5 + x - E^5*x)/(5*E^5*(E^(5 + x) - x))

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fricas [A]  time = 1.10, size = 25, normalized size = 1.00 \begin {gather*} \frac {{\left (x - 1\right )} e^{5} - x}{5 \, {\left (x e^{5} - e^{\left (x + 10\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 27, normalized size = 1.08 \begin {gather*} \frac {x e^{5} - x - e^{5}}{5 \, {\left (x e^{5} - e^{\left (x + 10\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 24, normalized size = 0.96

method	result	size
norman	\(\frac {\frac {1}{5}+\left (\frac {1}{5}-\frac {{\mathrm e}^{5}}{5}\right ) {\mathrm e}^{x}}{{\mathrm e}^{5} {\mathrm e}^{x}-x}\)	\(24\)
risch	\(-\frac {\left (x \,{\mathrm e}^{5}-{\mathrm e}^{5}-x \right ) {\mathrm e}^{-5}}{5 \left ({\mathrm e}^{5+x}-x \right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 26, normalized size = 1.04 \begin {gather*} \frac {x {\left (e^{5} - 1\right )} - e^{5}}{5 \, {\left (x e^{5} - e^{\left (x + 10\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 23, normalized size = 0.92 \begin {gather*} \frac {x\,{\mathrm {e}}^{-5}\,\left ({\mathrm {e}}^5-1\right )-1}{5\,x-5\,{\mathrm {e}}^{x+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 24, normalized size = 0.96 \begin {gather*} \frac {- x e^{5} + x + e^{5}}{- 5 x e^{5} + 5 e^{10} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x*exp(5) + x + exp(5))/(-5*x*exp(5) + 5*exp(10)*exp(x))

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