3.97.94 \(\int \frac {e^{-e^2} (-16+81 x^2)}{64 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{4} \left (5+\frac {e^{-e^2} \left (-1+\frac {9 x}{4}\right )^2}{x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 14} \begin {gather*} \frac {81}{64} e^{-e^2} x+\frac {e^{-e^2}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 81*x^2)/(64*E^E^2*x^2),x]

[Out]

1/(4*E^E^2*x) + (81*x)/(64*E^E^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{64} e^{-e^2} \int \frac {-16+81 x^2}{x^2} \, dx\\ &=\frac {1}{64} e^{-e^2} \int \left (81-\frac {16}{x^2}\right ) \, dx\\ &=\frac {e^{-e^2}}{4 x}+\frac {81}{64} e^{-e^2} x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 20, normalized size = 0.77 \begin {gather*} \frac {1}{64} e^{-e^2} \left (\frac {16}{x}+81 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 81*x^2)/(64*E^E^2*x^2),x]

[Out]

(16/x + 81*x)/(64*E^E^2)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 17, normalized size = 0.65 \begin {gather*} \frac {{\left (81 \, x^{2} + 16\right )} e^{\left (-e^{2}\right )}}{64 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(81*x^2-16)/x^2/exp(exp(2)),x, algorithm="fricas")

[Out]

1/64*(81*x^2 + 16)*e^(-e^2)/x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 16, normalized size = 0.62 \begin {gather*} \frac {1}{64} \, {\left (81 \, x + \frac {16}{x}\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(81*x^2-16)/x^2/exp(exp(2)),x, algorithm="giac")

[Out]

1/64*(81*x + 16/x)*e^(-e^2)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 17, normalized size = 0.65

method	result	size
default	\(\frac {{\mathrm e}^{-{\mathrm e}^{2}} \left (81 x +\frac {16}{x}\right )}{64}\)	\(17\)
gosper	\(\frac {\left (81 x^{2}+16\right ) {\mathrm e}^{-{\mathrm e}^{2}}}{64 x}\)	\(18\)
risch	\(\frac {81 x \,{\mathrm e}^{-{\mathrm e}^{2}}}{64}+\frac {{\mathrm e}^{-{\mathrm e}^{2}}}{4 x}\)	\(20\)
norman	\(\frac {\frac {{\mathrm e}^{-{\mathrm e}^{2}}}{4}+\frac {81 \,{\mathrm e}^{-{\mathrm e}^{2}} x^{2}}{64}}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*(81*x^2-16)/x^2/exp(exp(2)),x,method=_RETURNVERBOSE)

[Out]

1/64/exp(exp(2))*(81*x+16/x)

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 16, normalized size = 0.62 \begin {gather*} \frac {1}{64} \, {\left (81 \, x + \frac {16}{x}\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(81*x^2-16)/x^2/exp(exp(2)),x, algorithm="maxima")

[Out]

1/64*(81*x + 16/x)*e^(-e^2)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 17, normalized size = 0.65 \begin {gather*} \frac {{\mathrm {e}}^{-{\mathrm {e}}^2}\,\left (81\,x^2+16\right )}{64\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(2))*((81*x^2)/64 - 1/4))/x^2,x)

[Out]

(exp(-exp(2))*(81*x^2 + 16))/(64*x)

________________________________________________________________________________________

sympy [A]  time = 0.07, size = 12, normalized size = 0.46 \begin {gather*} \frac {81 x + \frac {16}{x}}{64 e^{e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(81*x**2-16)/x**2/exp(exp(2)),x)

[Out]

(81*x + 16/x)*exp(-exp(2))/64

________________________________________________________________________________________