3.88.12 \(\int \frac {\frac {9}{e^4}+85 x^2-36 x^3+4 x^4+\frac {3 (2+18 x-4 x^2)}{e^2}}{\frac {9}{e^4}+81 x^2-36 x^3+4 x^4+\frac {3 (18 x-4 x^2)}{e^2}} \, dx\)

Optimal. Leaf size=25 \[ x-\frac {2}{-7+2 x-\frac {\frac {3}{e^2}+2 x}{x}} \]

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Rubi [A]  time = 0.14, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {1680, 1814, 21, 8} \begin {gather*} \frac {2 e^2 x}{-2 e^2 x^2+9 e^2 x+3}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9/E^4 + 85*x^2 - 36*x^3 + 4*x^4 + (3*(2 + 18*x - 4*x^2))/E^2)/(9/E^4 + 81*x^2 - 36*x^3 + 4*x^4 + (3*(18*x
- 4*x^2))/E^2),x]

[Out]

x + (2*E^2*x)/(3 + 9*E^2*x - 2*E^2*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 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

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} &=\operatorname {Subst}\left (\int \frac {3 \left (192+1424 e^2+2619 e^4\right )+1152 e^4 x-32 e^2 \left (24+73 e^2\right ) x^2+256 e^4 x^4}{\left (3 \left (8+27 e^2\right )-16 e^2 x^2\right )^2} \, dx,x,-\frac {9}{4}+x\right )\\ &=\frac {2 e^2 x}{3+9 e^2 x-2 e^2 x^2}-\frac {\operatorname {Subst}\left (\int \frac {-18 \left (8+27 e^2\right )^2+96 e^2 \left (8+27 e^2\right ) x^2}{3 \left (8+27 e^2\right )-16 e^2 x^2} \, dx,x,-\frac {9}{4}+x\right )}{6 \left (8+27 e^2\right )}\\ &=\frac {2 e^2 x}{3+9 e^2 x-2 e^2 x^2}+\operatorname {Subst}\left (\int 1 \, dx,x,-\frac {9}{4}+x\right )\\ &=x+\frac {2 e^2 x}{3+9 e^2 x-2 e^2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.88 \begin {gather*} x-\frac {2 e^2 x}{-3+e^2 x (-9+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9/E^4 + 85*x^2 - 36*x^3 + 4*x^4 + (3*(2 + 18*x - 4*x^2))/E^2)/(9/E^4 + 81*x^2 - 36*x^3 + 4*x^4 + (3
*(18*x - 4*x^2))/E^2),x]

[Out]

x - (2*E^2*x)/(-3 + E^2*x*(-9 + 2*x))

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fricas [A]  time = 0.49, size = 41, normalized size = 1.64 \begin {gather*} \frac {2 \, x^{3} - 9 \, x^{2} - x e^{\left (\log \relax (3) - 2\right )} - 2 \, x}{2 \, x^{2} - 9 \, x - e^{\left (\log \relax (3) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(3)-2)^2+(-4*x^2+18*x+2)*exp(log(3)-2)+4*x^4-36*x^3+85*x^2)/(exp(log(3)-2)^2+(-4*x^2+18*x)*e
xp(log(3)-2)+4*x^4-36*x^3+81*x^2),x, algorithm="fricas")

[Out]

(2*x^3 - 9*x^2 - x*e^(log(3) - 2) - 2*x)/(2*x^2 - 9*x - e^(log(3) - 2))

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giac [B]  time = 0.17, size = 306, normalized size = 12.24 \begin {gather*} x - \frac {2 \, {\left ({\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 9\right )}^{2} + 8 \, e^{\left (\log \relax (3) - 2\right )}\right )} \log \left (\frac {1}{4} \, \sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + x - \frac {9}{4}\right )}{{\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 9\right )}^{3} + 27 \, {\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 9\right )}^{2} + 162 \, \sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 8 \, {\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 9\right )} e^{\left (\log \relax (3) - 2\right )} - 72 \, e^{\left (\log \relax (3) - 2\right )} - 1458} + \frac {2 \, {\left ({\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + 9\right )}^{2} + 8 \, e^{\left (\log \relax (3) - 2\right )}\right )} \log \left (-\frac {1}{4} \, \sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + x - \frac {9}{4}\right )}{{\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + 9\right )}^{3} - 27 \, {\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + 9\right )}^{2} + 162 \, \sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} - 8 \, {\left (\sqrt {3} \sqrt {27 \, e^{6} + 8 \, e^{4}} e^{\left (-3\right )} + 9\right )} e^{\left (\log \relax (3) - 2\right )} + 72 \, e^{\left (\log \relax (3) - 2\right )} + 1458} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(3)-2)^2+(-4*x^2+18*x+2)*exp(log(3)-2)+4*x^4-36*x^3+85*x^2)/(exp(log(3)-2)^2+(-4*x^2+18*x)*e
xp(log(3)-2)+4*x^4-36*x^3+81*x^2),x, algorithm="giac")

[Out]

x - 2*((sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 9)^2 + 8*e^(log(3) - 2))*log(1/4*sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^
(-3) + x - 9/4)/((sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 9)^3 + 27*(sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 9)^2
+ 162*sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 8*(sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 9)*e^(log(3) - 2) - 72*e^
(log(3) - 2) - 1458) + 2*((sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) + 9)^2 + 8*e^(log(3) - 2))*log(-1/4*sqrt(3)*sqr
t(27*e^6 + 8*e^4)*e^(-3) + x - 9/4)/((sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) + 9)^3 - 27*(sqrt(3)*sqrt(27*e^6 + 8
*e^4)*e^(-3) + 9)^2 + 162*sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) - 8*(sqrt(3)*sqrt(27*e^6 + 8*e^4)*e^(-3) + 9)*e^
(log(3) - 2) + 72*e^(log(3) - 2) + 1458)

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maple [A]  time = 0.12, size = 19, normalized size = 0.76

method	result	size
risch	\(x +\frac {2 x}{3 \left (-\frac {2 x^{2}}{3}+{\mathrm e}^{-2}+3 x \right )}\)	\(19\)
gosper	\(\frac {-4 x^{3}+2 \,{\mathrm e}^{\ln \relax (3)-2} x +9 \,{\mathrm e}^{\ln \relax (3)-2}+85 x}{-4 x^{2}+2 \,{\mathrm e}^{\ln \relax (3)-2}+18 x}\)	\(43\)
norman	\(\frac {\left (-\frac {{\mathrm e}^{2} \left (85 \,{\mathrm e}^{2}+6\right ) x}{2}+2 x^{3} {\mathrm e}^{4}-\frac {27 \,{\mathrm e}^{2}}{2}\right ) {\mathrm e}^{-2}}{2 x^{2} {\mathrm e}^{2}-9 \,{\mathrm e}^{2} x -3}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(ln(3)-2)^2+(-4*x^2+18*x+2)*exp(ln(3)-2)+4*x^4-36*x^3+85*x^2)/(exp(ln(3)-2)^2+(-4*x^2+18*x)*exp(ln(3)-
2)+4*x^4-36*x^3+81*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 23, normalized size = 0.92 \begin {gather*} x - \frac {2 \, x e^{2}}{2 \, x^{2} e^{2} - 9 \, x e^{2} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(3)-2)^2+(-4*x^2+18*x+2)*exp(log(3)-2)+4*x^4-36*x^3+85*x^2)/(exp(log(3)-2)^2+(-4*x^2+18*x)*e
xp(log(3)-2)+4*x^4-36*x^3+81*x^2),x, algorithm="maxima")

[Out]

x - 2*x*e^2/(2*x^2*e^2 - 9*x*e^2 - 3)

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mupad [B]  time = 5.45, size = 23, normalized size = 0.92 \begin {gather*} x+\frac {2\,x\,{\mathrm {e}}^2}{-2\,{\mathrm {e}}^2\,x^2+9\,{\mathrm {e}}^2\,x+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*log(3) - 4) + 85*x^2 - 36*x^3 + 4*x^4 + exp(log(3) - 2)*(18*x - 4*x^2 + 2))/(exp(2*log(3) - 4) + 81
*x^2 - 36*x^3 + 4*x^4 + exp(log(3) - 2)*(18*x - 4*x^2)),x)

[Out]

x + (2*x*exp(2))/(9*x*exp(2) - 2*x^2*exp(2) + 3)

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sympy [A]  time = 0.43, size = 24, normalized size = 0.96 \begin {gather*} x - \frac {2 x e^{2}}{2 x^{2} e^{2} - 9 x e^{2} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(ln(3)-2)**2+(-4*x**2+18*x+2)*exp(ln(3)-2)+4*x**4-36*x**3+85*x**2)/(exp(ln(3)-2)**2+(-4*x**2+18*
x)*exp(ln(3)-2)+4*x**4-36*x**3+81*x**2),x)

[Out]

x - 2*x*exp(2)/(2*x**2*exp(2) - 9*x*exp(2) - 3)

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