\(\int \frac {e^4+16 x^4+e^2 (8 x+8 x^2)}{3 e^4+24 e^2 x^2+48 x^4} \, dx\) [6534]

Optimal result

Integrand size = 44, antiderivative size = 29 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=-7-x+\frac {1}{3} \left (4 x+\frac {x}{\frac {e^2}{4 x}+x}\right ) \]

[Out]

1/3*x+1/3*x/(x+1/4*exp(2)/x)-7

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {28, 1828, 21, 8} \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {x}{3}-\frac {e^2}{3 \left (4 x^2+e^2\right )} \]

[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 1828

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {1}{3} \left (x-\frac {e^2}{e^2+4 x^2}\right ) \]

[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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 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	\(\frac {\frac {4 x^{3}}{3}+\frac {{\mathrm e}^{2} x}{3}-\frac {{\mathrm e}^{2}}{3}}{4 x^{2}+{\mathrm e}^{2}}\)	\(27\)
parallelrisch	\(\frac {16 x^{3}+4 \,{\mathrm e}^{2} x -4 \,{\mathrm e}^{2}}{48 x^{2}+12 \,{\mathrm e}^{2}}\)	\(28\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {4 \, x^{3} + {\left (x - 1\right )} e^{2}}{3 \, {\left (4 \, x^{2} + e^{2}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.52 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {x}{3} - \frac {e^{2}}{12 x^{2} + 3 e^{2}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {1}{3} \, x - \frac {e^{2}}{3 \, {\left (4 \, x^{2} + e^{2}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {1}{3} \, x - \frac {e^{2}}{3 \, {\left (4 \, x^{2} + e^{2}\right )}} \]

[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]

1/3*x - 1/3*e^2/(4*x^2 + e^2)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^4+16 x^4+e^2 \left (8 x+8 x^2\right )}{3 e^4+24 e^2 x^2+48 x^4} \, dx=\frac {x}{3}-\frac {{\mathrm {e}}^2}{3\,\left (4\,x^2+{\mathrm {e}}^2\right )} \]

[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))