3.5 \(\int \frac {e^{\frac {1}{-1+x^2}} (1-3 x-x^2+x^3)}{1-x-x^2+x^3} \, dx\)

Optimal. Leaf size=13 \[ e^{\frac {1}{x^2-1}} (x+1) \]

[Out]

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

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Rubi [F]  time = 0.39, 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 = {} \[ \int \frac {e^{\frac {1}{-1+x^2}} \left (1-3 x-x^2+x^3\right )}{1-x-x^2+x^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Defer[Int][E^(-1 + x^2)^(-1), x] + Defer[Int][E^(-1 + x^2)^(-1)/(1 - x), x]/2 - Defer[Int][E^(-1 + x^2)^(-1)/(
-1 + x)^2, x] + Defer[Int][E^(-1 + x^2)^(-1)/(1 + x), x]/2

Rubi steps

\begin {align*} \int \frac {e^{\frac {1}{-1+x^2}} \left (1-3 x-x^2+x^3\right )}{1-x-x^2+x^3} \, dx &=\int \left (e^{\frac {1}{-1+x^2}}-\frac {2 e^{\frac {1}{-1+x^2}} x}{1-x-x^2+x^3}\right ) \, dx\\ &=-\left (2 \int \frac {e^{\frac {1}{-1+x^2}} x}{1-x-x^2+x^3} \, dx\right )+\int e^{\frac {1}{-1+x^2}} \, dx\\ &=-\left (2 \int \left (\frac {e^{\frac {1}{-1+x^2}}}{2 (-1+x)^2}+\frac {e^{\frac {1}{-1+x^2}}}{2 \left (-1+x^2\right )}\right ) \, dx\right )+\int e^{\frac {1}{-1+x^2}} \, dx\\ &=\int e^{\frac {1}{-1+x^2}} \, dx-\int \frac {e^{\frac {1}{-1+x^2}}}{(-1+x)^2} \, dx-\int \frac {e^{\frac {1}{-1+x^2}}}{-1+x^2} \, dx\\ &=\int e^{\frac {1}{-1+x^2}} \, dx-\int \frac {e^{\frac {1}{-1+x^2}}}{(-1+x)^2} \, dx-\int \left (-\frac {e^{\frac {1}{-1+x^2}}}{2 (1-x)}-\frac {e^{\frac {1}{-1+x^2}}}{2 (1+x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {1}{-1+x^2}}}{1-x} \, dx+\frac {1}{2} \int \frac {e^{\frac {1}{-1+x^2}}}{1+x} \, dx+\int e^{\frac {1}{-1+x^2}} \, dx-\int \frac {e^{\frac {1}{-1+x^2}}}{(-1+x)^2} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 13, normalized size = 1.00 \[ e^{\frac {1}{x^2-1}} (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-1 + x^2)^(-1)*(1 + x)

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fricas [A]  time = 0.42, size = 12, normalized size = 0.92 \[ {\left (x + 1\right )} e^{\left (\frac {1}{x^{2} - 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + 1)*e^(1/(x^2 - 1))

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giac [B]  time = 0.94, size = 30, normalized size = 2.31 \[ {\left (x e^{\left (\frac {x^{2}}{x^{2} - 1}\right )} + e^{\left (\frac {x^{2}}{x^{2} - 1}\right )}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \left (x +1\right ) {\mathrm e}^{\frac {1}{x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-x^2-3*x+1)*exp(1/(x^2-1))/(x^3-x^2-x+1),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{3} - x^{2} - 3 \, x + 1\right )} e^{\left (\frac {1}{x^{2} - 1}\right )}}{x^{3} - x^{2} - x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 12, normalized size = 0.92 \[ {\mathrm {e}}^{\frac {1}{x^2-1}}\,\left (x+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/(x^2 - 1))*(3*x + x^2 - x^3 - 1))/(x + x^2 - x^3 - 1),x)

[Out]

exp(1/(x^2 - 1))*(x + 1)

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sympy [A]  time = 0.18, size = 10, normalized size = 0.77 \[ \left (x + 1\right ) e^{\frac {1}{x^{2} - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-x**2-3*x+1)*exp(1/(x**2-1))/(x**3-x**2-x+1),x)

[Out]

(x + 1)*exp(1/(x**2 - 1))

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