3.48.83 \(\int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} (-1+e^{e^4} (-2 x+2 x^2))}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx\)

Optimal. Leaf size=31 \[ e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{(1-x) x}\right )} \]

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Rubi [F]  time = 16.91, 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 {\exp \left (-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)\right ) (-1+2 x)}{x^2-2 x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + x^2) + 48*Log[5]^2)*(-1 + 2*x))/(x^2
- 2*x^3 + x^4),x]

[Out]

Defer[Int][E^((E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + x^2) - E^4*(1 - (48*Log[5]^2)/E^4))/(
-1 + x)^2, x] - Defer[Int][E^((E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + x^2) - E^4*(1 - (48*L
og[5]^2)/E^4))/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)\right ) (-1+2 x)}{x^2 \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {\exp \left (-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)\right ) (-1+2 x)}{(-1+x)^2 x^2} \, dx\\ &=\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right ) (-1+2 x)}{(1-x)^2 x^2} \, dx\\ &=\int \left (\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(-1+x)^2}-\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2}\right ) \, dx\\ &=\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(-1+x)^2} \, dx-\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.59, size = 30, normalized size = 0.97 \begin {gather*} e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{x-x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + x^2) + 48*Log[5]^2)*(-1 + 2*x)
)/(x^2 - 2*x^3 + x^4),x]

[Out]

E^(E^(48*Log[5]^2)*(2 + 1/(E^E^4*(x - x^2))))

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fricas [B]  time = 0.61, size = 72, normalized size = 2.32 \begin {gather*} e^{\left (-48 \, \log \relax (5)^{2} + \frac {48 \, {\left (x^{2} - x\right )} \log \relax (5)^{2} - {\left (x^{2} - x\right )} e^{4} + {\left (2 \, {\left (x^{2} - x\right )} e^{\left (e^{4}\right )} - 1\right )} e^{\left (48 \, \log \relax (5)^{2} - e^{4}\right )}}{x^{2} - x} + e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-1)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-
2*x^3+x^2)/exp(exp(4)),x, algorithm="fricas")

[Out]

e^(-48*log(5)^2 + (48*(x^2 - x)*log(5)^2 - (x^2 - x)*e^4 + (2*(x^2 - x)*e^(e^4) - 1)*e^(48*log(5)^2 - e^4))/(x
^2 - x) + e^4)

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giac [B]  time = 0.17, size = 136, normalized size = 4.39 \begin {gather*} e^{\left (\frac {48 \, x^{2} \log \relax (5)^{2}}{x^{2} - x} - \frac {x^{2} e^{4}}{x^{2} - x} + \frac {2 \, x^{2} e^{\left (48 \, \log \relax (5)^{2}\right )}}{x^{2} - x} - \frac {48 \, x \log \relax (5)^{2}}{x^{2} - x} - 48 \, \log \relax (5)^{2} + \frac {x e^{4}}{x^{2} - x} - \frac {2 \, x e^{\left (48 \, \log \relax (5)^{2}\right )}}{x^{2} - x} - \frac {e^{\left (48 \, \log \relax (5)^{2} - e^{4}\right )}}{x^{2} - x} + e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-1)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-
2*x^3+x^2)/exp(exp(4)),x, algorithm="giac")

[Out]

e^(48*x^2*log(5)^2/(x^2 - x) - x^2*e^4/(x^2 - x) + 2*x^2*e^(48*log(5)^2)/(x^2 - x) - 48*x*log(5)^2/(x^2 - x) -
48*log(5)^2 + x*e^4/(x^2 - x) - 2*x*e^(48*log(5)^2)/(x^2 - x) - e^(48*log(5)^2 - e^4)/(x^2 - x) + e^4)

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maple [A]  time = 0.28, size = 39, normalized size = 1.26

method	result	size
gosper	\({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \relax (5)^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x \left (x -1\right )}}\)	\(39\)
risch	\({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \relax (5)^{2}-{\mathrm e}^{4}}}{x \left (x -1\right )}}\)	\(39\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \relax (5)^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}-x \,{\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \relax (5)^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}}{x \left (x -1\right )}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x-1)*exp(48*ln(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*ln(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^
2)/exp(exp(4)),x,method=_RETURNVERBOSE)

[Out]

exp((2*x^2*exp(exp(4))-2*x*exp(exp(4))-1)*exp(48*ln(5)^2)/x/(x-1)/exp(exp(4)))

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maxima [A]  time = 0.75, size = 49, normalized size = 1.58 \begin {gather*} e^{\left (-\frac {e^{\left (48 \, \log \relax (5)^{2}\right )}}{x e^{\left (e^{4}\right )} - e^{\left (e^{4}\right )}} + \frac {e^{\left (48 \, \log \relax (5)^{2} - e^{4}\right )}}{x} + 2 \, e^{\left (48 \, \log \relax (5)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-1)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-
2*x^3+x^2)/exp(exp(4)),x, algorithm="maxima")

[Out]

e^(-e^(48*log(5)^2)/(x*e^(e^4) - e^(e^4)) + e^(48*log(5)^2 - e^4)/x + 2*e^(48*log(5)^2))

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mupad [B]  time = 5.21, size = 66, normalized size = 2.13 \begin {gather*} {\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{48\,{\ln \relax (5)}^2}}{x-x^2}}\,{\mathrm {e}}^{-\frac {2\,x^2\,{\mathrm {e}}^{48\,{\ln \relax (5)}^2}}{x-x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^{48\,{\ln \relax (5)}^2}}{x-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(4))*exp(48*log(5)^2)*exp((exp(-exp(4))*exp(48*log(5)^2)*(exp(exp(4))*(2*x - 2*x^2) + 1))/(x - x^
2))*(2*x - 1))/(x^2 - 2*x^3 + x^4),x)

[Out]

exp((2*x*exp(48*log(5)^2))/(x - x^2))*exp(-(2*x^2*exp(48*log(5)^2))/(x - x^2))*exp((exp(-exp(4))*exp(48*log(5)
^2))/(x - x^2))

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sympy [A]  time = 0.36, size = 34, normalized size = 1.10 \begin {gather*} e^{\frac {\left (\left (2 x^{2} - 2 x\right ) e^{e^{4}} - 1\right ) e^{48 \log {\relax (5 )}^{2}}}{\left (x^{2} - x\right ) e^{e^{4}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x-1)*exp(48*ln(5)**2)*exp(((2*x**2-2*x)*exp(exp(4))-1)*exp(48*ln(5)**2)/(x**2-x)/exp(exp(4)))/(x*
*4-2*x**3+x**2)/exp(exp(4)),x)

[Out]

exp(((2*x**2 - 2*x)*exp(exp(4)) - 1)*exp(-exp(4))*exp(48*log(5)**2)/(x**2 - x))

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