3.73.5 \(\int \frac {-4+e^{e^8} (4-4 x^2)+e^{2 e^8} (-1+6 x^2-8 x^3+3 x^4)}{x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {\left (2-e^{e^8} (-1+x)^2\right )^2}{x} \]

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Rubi [B]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 2.80, number of steps used = 2, number of rules used = 1, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {14} \begin {gather*} e^{2 e^8} x^3-4 e^{2 e^8} x^2-2 e^{e^8} \left (2-3 e^{e^8}\right ) x+\frac {\left (2-e^{e^8}\right )^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^E^8*(4 - 4*x^2) + E^(2*E^8)*(-1 + 6*x^2 - 8*x^3 + 3*x^4))/x^2,x]

[Out]

(2 - E^E^8)^2/x - 2*E^E^8*(2 - 3*E^E^8)*x - 4*E^(2*E^8)*x^2 + E^(2*E^8)*x^3

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

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Mathematica [B]  time = 0.03, size = 41, normalized size = 2.05 \begin {gather*} \frac {4-4 e^{e^8} \left (1+x^2\right )+e^{2 e^8} \left (1+6 x^2-4 x^3+x^4\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^E^8*(4 - 4*x^2) + E^(2*E^8)*(-1 + 6*x^2 - 8*x^3 + 3*x^4))/x^2,x]

[Out]

(4 - 4*E^E^8*(1 + x^2) + E^(2*E^8)*(1 + 6*x^2 - 4*x^3 + x^4))/x

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fricas [B]  time = 0.73, size = 37, normalized size = 1.85 \begin {gather*} \frac {{\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} + 1\right )} e^{\left (2 \, e^{8}\right )} - 4 \, {\left (x^{2} + 1\right )} e^{\left (e^{8}\right )} + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^4-8*x^3+6*x^2-1)*exp(exp(2)^4)^2+(-4*x^2+4)*exp(exp(2)^4)-4)/x^2,x, algorithm="fricas")

[Out]

((x^4 - 4*x^3 + 6*x^2 + 1)*e^(2*e^8) - 4*(x^2 + 1)*e^(e^8) + 4)/x

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giac [B]  time = 0.14, size = 50, normalized size = 2.50 \begin {gather*} x^{3} e^{\left (2 \, e^{8}\right )} - 4 \, x^{2} e^{\left (2 \, e^{8}\right )} + 6 \, x e^{\left (2 \, e^{8}\right )} - 4 \, x e^{\left (e^{8}\right )} + \frac {e^{\left (2 \, e^{8}\right )} - 4 \, e^{\left (e^{8}\right )} + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^4-8*x^3+6*x^2-1)*exp(exp(2)^4)^2+(-4*x^2+4)*exp(exp(2)^4)-4)/x^2,x, algorithm="giac")

[Out]

x^3*e^(2*e^8) - 4*x^2*e^(2*e^8) + 6*x*e^(2*e^8) - 4*x*e^(e^8) + (e^(2*e^8) - 4*e^(e^8) + 4)/x

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maple [B]  time = 0.06, size = 54, normalized size = 2.70

method	result	size
default	\({\mathrm e}^{2 \,{\mathrm e}^{8}} x^{3}-4 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{2}+6 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x -4 \,{\mathrm e}^{{\mathrm e}^{8}} x -\frac {-{\mathrm e}^{2 \,{\mathrm e}^{8}}+4 \,{\mathrm e}^{{\mathrm e}^{8}}-4}{x}\)	\(54\)
risch	\({\mathrm e}^{2 \,{\mathrm e}^{8}} x^{3}-4 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{2}+6 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x -4 \,{\mathrm e}^{{\mathrm e}^{8}} x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{8}}}{x}-\frac {4 \,{\mathrm e}^{{\mathrm e}^{8}}}{x}+\frac {4}{x}\)	\(57\)
norman	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{4}-4 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{3}+\left (6 \,{\mathrm e}^{2 \,{\mathrm e}^{8}}-4 \,{\mathrm e}^{{\mathrm e}^{8}}\right ) x^{2}+4+{\mathrm e}^{2 \,{\mathrm e}^{8}}-4 \,{\mathrm e}^{{\mathrm e}^{8}}}{x}\)	\(65\)
gosper	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{4}-4 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{3}+6 \,{\mathrm e}^{2 \,{\mathrm e}^{8}} x^{2}-4 \,{\mathrm e}^{{\mathrm e}^{8}} x^{2}+{\mathrm e}^{2 \,{\mathrm e}^{8}}-4 \,{\mathrm e}^{{\mathrm e}^{8}}+4}{x}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^4-8*x^3+6*x^2-1)*exp(exp(2)^4)^2+(-4*x^2+4)*exp(exp(2)^4)-4)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(2*exp(8))*x^3-4*exp(2*exp(8))*x^2+6*exp(2*exp(8))*x-4*exp(exp(8))*x-(-exp(2*exp(8))+4*exp(exp(8))-4)/x

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maxima [B]  time = 0.37, size = 52, normalized size = 2.60 \begin {gather*} x^{3} e^{\left (2 \, e^{8}\right )} - 4 \, x^{2} e^{\left (2 \, e^{8}\right )} + 2 \, x {\left (3 \, e^{\left (2 \, e^{8}\right )} - 2 \, e^{\left (e^{8}\right )}\right )} + \frac {e^{\left (2 \, e^{8}\right )} - 4 \, e^{\left (e^{8}\right )} + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^4-8*x^3+6*x^2-1)*exp(exp(2)^4)^2+(-4*x^2+4)*exp(exp(2)^4)-4)/x^2,x, algorithm="maxima")

[Out]

x^3*e^(2*e^8) - 4*x^2*e^(2*e^8) + 2*x*(3*e^(2*e^8) - 2*e^(e^8)) + (e^(2*e^8) - 4*e^(e^8) + 4)/x

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mupad [B]  time = 4.34, size = 49, normalized size = 2.45 \begin {gather*} x^3\,{\mathrm {e}}^{2\,{\mathrm {e}}^8}-4\,x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^8}+\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^8}-4\,{\mathrm {e}}^{{\mathrm {e}}^8}+4}{x}+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^8}\,\left (3\,{\mathrm {e}}^{{\mathrm {e}}^8}-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(8))*(4*x^2 - 4) - exp(2*exp(8))*(6*x^2 - 8*x^3 + 3*x^4 - 1) + 4)/x^2,x)

[Out]

x^3*exp(2*exp(8)) - 4*x^2*exp(2*exp(8)) + (exp(2*exp(8)) - 4*exp(exp(8)) + 4)/x + 2*x*exp(exp(8))*(3*exp(exp(8
)) - 2)

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sympy [B]  time = 0.17, size = 54, normalized size = 2.70 \begin {gather*} x^{3} e^{2 e^{8}} - 4 x^{2} e^{2 e^{8}} + x \left (- 4 e^{e^{8}} + 6 e^{2 e^{8}}\right ) + \frac {4 - 4 e^{e^{8}} + e^{2 e^{8}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**4-8*x**3+6*x**2-1)*exp(exp(2)**4)**2+(-4*x**2+4)*exp(exp(2)**4)-4)/x**2,x)

[Out]

x**3*exp(2*exp(8)) - 4*x**2*exp(2*exp(8)) + x*(-4*exp(exp(8)) + 6*exp(2*exp(8))) + (4 - 4*exp(exp(8)) + exp(2*
exp(8)))/x

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