3.66.34 \(\int \frac {e^4+16 x^4+e^2 (8 x+8 x^2)}{3 e^4+24 e^2 x^2+48 x^4} \, dx\)

Optimal. Leaf size=29 \[ -7-x+\frac {1}{3} \left (4 x+\frac {x}{\frac {e^2}{4 x}+x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {28, 1814, 21, 8} \begin {gather*} \frac {x}{3}-\frac {e^2}{3 \left (4 x^2+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4 + 16*x^4 + E^2*(8*x + 8*x^2))/(3*E^4 + 24*E^2*x^2 + 48*x^4),x]

[Out]

x/3 - E^2/(3*(E^2 + 4*x^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=48 \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{\left (12 e^2+48 x^2\right )^2} \, dx\\ &=-\frac {e^2}{3 \left (e^2+4 x^2\right )}-\frac {2 \int \frac {-2 e^4-8 e^2 x^2}{12 e^2+48 x^2} \, dx}{e^2}\\ &=-\frac {e^2}{3 \left (e^2+4 x^2\right )}+\frac {\int 1 \, dx}{3}\\ &=\frac {x}{3}-\frac {e^2}{3 \left (e^2+4 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.76 \begin {gather*} \frac {1}{3} \left (x-\frac {e^2}{e^2+4 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4 + 16*x^4 + E^2*(8*x + 8*x^2))/(3*E^4 + 24*E^2*x^2 + 48*x^4),x]

[Out]

(x - E^2/(E^2 + 4*x^2))/3

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fricas [A]  time = 0.70, size = 24, normalized size = 0.83 \begin {gather*} \frac {4 \, x^{3} + {\left (x - 1\right )} e^{2}}{3 \, {\left (4 \, x^{2} + e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)^2+(8*x^2+8*x)*exp(2)+16*x^4)/(3*exp(2)^2+24*x^2*exp(2)+48*x^4),x, algorithm="fricas")

[Out]

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

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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((exp(2)^2+(8*x^2+8*x)*exp(2)+16*x^4)/(3*exp(2)^2+24*x^2*exp(2)+48*x^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/3*(exp(2)*exp(2)*1/2/exp(4)*ln(sageVAR
x^2+(16*exp(4)/256)^(1/4)*(16*exp(4)/256)^(1/4))+exp(2)^2*sqrt(-exp(2)^2+exp(4))/(-exp(2)*exp(2)^2+exp(6))*ata
n(sageVARx/(16*exp(4)

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maple [A]  time = 0.11, size = 19, normalized size = 0.66

method	result	size
risch	\(\frac {x}{3}-\frac {{\mathrm e}^{2}}{3 \left (4 x^{2}+{\mathrm e}^{2}\right )}\)	\(19\)
gosper	\(\frac {4 x^{3}+{\mathrm e}^{2} x -{\mathrm e}^{2}}{12 x^{2}+3 \,{\mathrm e}^{2}}\)	\(27\)
norman	\(\frac {\frac {4 x^{3}}{3}+\frac {{\mathrm e}^{2} x}{3}-\frac {{\mathrm e}^{2}}{3}}{4 x^{2}+{\mathrm e}^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)^2+(8*x^2+8*x)*exp(2)+16*x^4)/(3*exp(2)^2+24*x^2*exp(2)+48*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 18, normalized size = 0.62 \begin {gather*} \frac {1}{3} \, x - \frac {e^{2}}{3 \, {\left (4 \, x^{2} + e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)^2+(8*x^2+8*x)*exp(2)+16*x^4)/(3*exp(2)^2+24*x^2*exp(2)+48*x^4),x, algorithm="maxima")

[Out]

1/3*x - 1/3*e^2/(4*x^2 + e^2)

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mupad [B]  time = 0.08, size = 20, normalized size = 0.69 \begin {gather*} \frac {x}{3}-\frac {{\mathrm {e}}^2}{3\,\left (4\,x^2+{\mathrm {e}}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4) + exp(2)*(8*x + 8*x^2) + 16*x^4)/(3*exp(4) + 24*x^2*exp(2) + 48*x^4),x)

[Out]

x/3 - exp(2)/(3*(exp(2) + 4*x^2))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.52 \begin {gather*} \frac {x}{3} - \frac {e^{2}}{12 x^{2} + 3 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)**2+(8*x**2+8*x)*exp(2)+16*x**4)/(3*exp(2)**2+24*x**2*exp(2)+48*x**4),x)

[Out]

x/3 - exp(2)/(12*x**2 + 3*exp(2))

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