3.40.82 \(\int \frac {-2 e x^2+e^{\frac {1}{2} (28+e^4)} (-2 e^5+4 e x)}{e^8 x^2-2 e^4 x^3+x^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 e \left (e^{\frac {1}{2} \left (28+e^4\right )}-x\right )}{\left (e^4-x\right ) x} \]

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Rubi [A]  time = 0.09, antiderivative size = 44, normalized size of antiderivative = 1.47, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1594, 27, 1820} \begin {gather*} \frac {2 e^{11+\frac {e^4}{2}}}{x}-\frac {2 e \left (1-e^{10+\frac {e^4}{2}}\right )}{e^4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E*x^2 + E^((28 + E^4)/2)*(-2*E^5 + 4*E*x))/(E^8*x^2 - 2*E^4*x^3 + x^4),x]

[Out]

(-2*E*(1 - E^(10 + E^4/2)))/(E^4 - x) + (2*E^(11 + E^4/2))/x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 e x^2+e^{\frac {1}{2} \left (28+e^4\right )} \left (-2 e^5+4 e x\right )}{x^2 \left (e^8-2 e^4 x+x^2\right )} \, dx\\ &=\int \frac {-2 e x^2+e^{\frac {1}{2} \left (28+e^4\right )} \left (-2 e^5+4 e x\right )}{x^2 \left (-e^4+x\right )^2} \, dx\\ &=\int \left (\frac {2 e \left (-1+e^{10+\frac {e^4}{2}}\right )}{\left (e^4-x\right )^2}-\frac {2 e^{11+\frac {e^4}{2}}}{x^2}\right ) \, dx\\ &=-\frac {2 e \left (1-e^{10+\frac {e^4}{2}}\right )}{e^4-x}+\frac {2 e^{11+\frac {e^4}{2}}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.00 \begin {gather*} -\frac {2 e \left (-e^{14+\frac {e^4}{2}}+x\right )}{\left (e^4-x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E*x^2 + E^((28 + E^4)/2)*(-2*E^5 + 4*E*x))/(E^8*x^2 - 2*E^4*x^3 + x^4),x]

[Out]

(-2*E*(-E^(14 + E^4/2) + x))/((E^4 - x)*x)

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fricas [A]  time = 0.75, size = 27, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (x e - e^{\left (\frac {1}{2} \, e^{4} + 15\right )}\right )}}{x^{2} - x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*exp(4)+4*x*exp(1))*exp(1/2*exp(4)+14)-2*x^2*exp(1))/(x^2*exp(4)^2-2*x^3*exp(4)+x^4),x, a
lgorithm="fricas")

[Out]

2*(x*e - e^(1/2*e^4 + 15))/(x^2 - x*e^4)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*exp(4)+4*x*exp(1))*exp(1/2*exp(4)+14)-2*x^2*exp(1))/(x^2*exp(4)^2-2*x^3*exp(4)+x^4),x, a
lgorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -2*((-exp(4)*exp(5)*exp(exp(4)/2+14)+exp
(8)*exp(1)*exp(exp(4)/2+14))/exp(8)^2*ln(sageVARx^2-2*sageVARx*exp(4)+exp(8))+(2*exp(4)^2*exp(5)*exp(exp(4)/2+
14)-2*exp(4)*exp(8)*e

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maple [A]  time = 0.11, size = 27, normalized size = 0.90

method	result	size
gosper	\(\frac {2 \left ({\mathrm e}^{\frac {{\mathrm e}^{4}}{2}+14}-x \right ) {\mathrm e}}{x \left ({\mathrm e}^{4}-x \right )}\)	\(27\)
risch	\(\frac {-2 x \,{\mathrm e}+2 \,{\mathrm e}^{15+\frac {{\mathrm e}^{4}}{2}}}{\left ({\mathrm e}^{4}-x \right ) x}\)	\(28\)
norman	\(\frac {-2 x \,{\mathrm e}+2 \,{\mathrm e} \,{\mathrm e}^{\frac {{\mathrm e}^{4}}{2}} {\mathrm e}^{14}}{x \left ({\mathrm e}^{4}-x \right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(1)*exp(4)+4*x*exp(1))*exp(1/2*exp(4)+14)-2*x^2*exp(1))/(x^2*exp(4)^2-2*x^3*exp(4)+x^4),x,method=_
RETURNVERBOSE)

[Out]

2*(exp(1/2*exp(4)+14)-x)/x/(exp(4)-x)*exp(1)

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maxima [A]  time = 0.46, size = 27, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (x e - e^{\left (\frac {1}{2} \, e^{4} + 15\right )}\right )}}{x^{2} - x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*exp(4)+4*x*exp(1))*exp(1/2*exp(4)+14)-2*x^2*exp(1))/(x^2*exp(4)^2-2*x^3*exp(4)+x^4),x, a
lgorithm="maxima")

[Out]

2*(x*e - e^(1/2*e^4 + 15))/(x^2 - x*e^4)

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mupad [B]  time = 0.20, size = 28, normalized size = 0.93 \begin {gather*} \frac {2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{2}+15}-2\,x\,\mathrm {e}}{x\,{\mathrm {e}}^4-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(4)/2 + 14)*(2*exp(5) - 4*x*exp(1)) + 2*x^2*exp(1))/(x^2*exp(8) - 2*x^3*exp(4) + x^4),x)

[Out]

(2*exp(exp(4)/2 + 15) - 2*x*exp(1))/(x*exp(4) - x^2)

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sympy [A]  time = 0.43, size = 27, normalized size = 0.90 \begin {gather*} - \frac {- 2 e x + 2 e^{15} e^{\frac {e^{4}}{2}}}{x^{2} - x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*exp(4)+4*x*exp(1))*exp(1/2*exp(4)+14)-2*x**2*exp(1))/(x**2*exp(4)**2-2*x**3*exp(4)+x**4)
,x)

[Out]

-(-2*E*x + 2*exp(15)*exp(exp(4)/2))/(x**2 - x*exp(4))

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