3.44.40 \(\int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} (324-4 e^3+e^{2+x} (-81 x+e^3 x))}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {\left (-81+e^3\right ) x}{-2-e^{\frac {e^{2+x}}{4}}+2 x} \]

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Rubi [F]  time = 0.74, 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 {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(648 - 8*E^3 + E^(E^(2 + x)/4)*(324 - 4*E^3 + E^(2 + x)*(-81*x + E^3*x)))/(16 + 4*E^(E^(2 + x)/2) + E^(E^(
2 + x)/4)*(16 - 16*x) - 32*x + 16*x^2),x]

[Out]

(81 - E^3)*Defer[Int][(2 + E^(E^(2 + x)/4) - 2*x)^(-1), x] + 2*(81 - E^3)*Defer[Int][x/(2 + E^(E^(2 + x)/4) -
2*x)^2, x] - ((81 - E^3)*Defer[Int][(E^((8 + E^(2 + x) + 4*x)/4)*x)/(2 + E^(E^(2 + x)/4) - 2*x)^2, x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (81-e^3\right ) \left (8+4 e^{\frac {e^{2+x}}{4}}-e^{2+\frac {e^{2+x}}{4}+x} x\right )}{4 \left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx\\ &=\frac {1}{4} \left (81-e^3\right ) \int \frac {8+4 e^{\frac {e^{2+x}}{4}}-e^{2+\frac {e^{2+x}}{4}+x} x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx\\ &=\frac {1}{4} \left (81-e^3\right ) \int \left (\frac {4 \left (2+e^{\frac {e^{2+x}}{4}}\right )}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2}-\frac {e^{\frac {1}{4} \left (8+e^{2+x}+4 x\right )} x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2}\right ) \, dx\\ &=\left (81-e^3\right ) \int \frac {2+e^{\frac {e^{2+x}}{4}}}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx+\frac {1}{4} \left (-81+e^3\right ) \int \frac {e^{\frac {1}{4} \left (8+e^{2+x}+4 x\right )} x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx\\ &=\left (81-e^3\right ) \int \left (\frac {1}{2+e^{\frac {e^{2+x}}{4}}-2 x}+\frac {2 x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2}\right ) \, dx+\frac {1}{4} \left (-81+e^3\right ) \int \frac {e^{\frac {1}{4} \left (8+e^{2+x}+4 x\right )} x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx\\ &=\left (81-e^3\right ) \int \frac {1}{2+e^{\frac {e^{2+x}}{4}}-2 x} \, dx+\left (2 \left (81-e^3\right )\right ) \int \frac {x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx+\frac {1}{4} \left (-81+e^3\right ) \int \frac {e^{\frac {1}{4} \left (8+e^{2+x}+4 x\right )} x}{\left (2+e^{\frac {e^{2+x}}{4}}-2 x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 26, normalized size = 0.96 \begin {gather*} -\frac {\left (-81+e^3\right ) x}{2+e^{\frac {e^{2+x}}{4}}-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(648 - 8*E^3 + E^(E^(2 + x)/4)*(324 - 4*E^3 + E^(2 + x)*(-81*x + E^3*x)))/(16 + 4*E^(E^(2 + x)/2) +
E^(E^(2 + x)/4)*(16 - 16*x) - 32*x + 16*x^2),x]

[Out]

-(((-81 + E^3)*x)/(2 + E^(E^(2 + x)/4) - 2*x))

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fricas [A]  time = 0.54, size = 25, normalized size = 0.93 \begin {gather*} \frac {x e^{3} - 81 \, x}{2 \, x - e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16
*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+16),x, algorithm="fricas")

[Out]

(x*e^3 - 81*x)/(2*x - e^(1/4*e^(x + 2)) - 2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x e^{3} - 81 \, x\right )} e^{\left (x + 2\right )} - 4 \, e^{3} + 324\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, e^{3} + 648}{4 \, {\left (4 \, x^{2} - 4 \, {\left (x - 1\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, x + e^{\left (\frac {1}{2} \, e^{\left (x + 2\right )}\right )} + 4\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16
*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+16),x, algorithm="giac")

[Out]

integrate(1/4*(((x*e^3 - 81*x)*e^(x + 2) - 4*e^3 + 324)*e^(1/4*e^(x + 2)) - 8*e^3 + 648)/(4*x^2 - 4*(x - 1)*e^
(1/4*e^(x + 2)) - 8*x + e^(1/2*e^(x + 2)) + 4), x)

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maple [A]  time = 0.22, size = 23, normalized size = 0.85

method	result	size
risch	\(\frac {\left ({\mathrm e}^{3}-81\right ) x}{2 x -2-{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}}\)	\(23\)
norman	\(\frac {\left (-\frac {81}{2}+\frac {{\mathrm e}^{3}}{2}\right ) {\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}-81+{\mathrm e}^{3}}{2 x -2-{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16*x+16)
*exp(1/4*exp(2+x))+16*x^2-32*x+16),x,method=_RETURNVERBOSE)

[Out]

(exp(3)-81)/(2*x-2-exp(1/4*exp(2+x)))*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{4} \, \int \frac {{\left ({\left (x e^{3} - 81 \, x\right )} e^{\left (x + 2\right )} - 4 \, e^{3} + 324\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, e^{3} + 648}{4 \, x^{2} - 4 \, {\left (x - 1\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, x + e^{\left (\frac {1}{2} \, e^{\left (x + 2\right )}\right )} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16
*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+16),x, algorithm="maxima")

[Out]

1/4*integrate((((x*e^3 - 81*x)*e^(x + 2) - 4*e^3 + 324)*e^(1/4*e^(x + 2)) - 8*e^3 + 648)/(4*x^2 - 4*(x - 1)*e^
(1/4*e^(x + 2)) - 8*x + e^(1/2*e^(x + 2)) + 4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {8\,{\mathrm {e}}^3+{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{4}}\,\left (4\,{\mathrm {e}}^3+{\mathrm {e}}^{x+2}\,\left (81\,x-x\,{\mathrm {e}}^3\right )-324\right )-648}{4\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{2}}-32\,x-{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{4}}\,\left (16\,x-16\right )+16\,x^2+16} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*exp(3) + exp(exp(x + 2)/4)*(4*exp(3) + exp(x + 2)*(81*x - x*exp(3)) - 324) - 648)/(4*exp(exp(x + 2)/2)
- 32*x - exp(exp(x + 2)/4)*(16*x - 16) + 16*x^2 + 16),x)

[Out]

-int((8*exp(3) + exp(exp(x + 2)/4)*(4*exp(3) + exp(x + 2)*(81*x - x*exp(3)) - 324) - 648)/(4*exp(exp(x + 2)/2)
- 32*x - exp(exp(x + 2)/4)*(16*x - 16) + 16*x^2 + 16), x)

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sympy [A]  time = 0.18, size = 20, normalized size = 0.74 \begin {gather*} \frac {- x e^{3} + 81 x}{- 2 x + e^{\frac {e^{x + 2}}{4}} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+648)/(4*exp(1/4*exp(2+x))**2+(-1
6*x+16)*exp(1/4*exp(2+x))+16*x**2-32*x+16),x)

[Out]

(-x*exp(3) + 81*x)/(-2*x + exp(exp(x + 2)/4) + 2)

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