3.1.16 \(\int \frac {648+8 e^8-8 x^2}{6561+e^{16}+4 e^{12} x+162 x^2+x^4+e^8 (162+6 x^2)+e^4 (324 x+4 x^3)} \, dx\)

Optimal. Leaf size=14 \[ \frac {8 x}{81+\left (e^4+x\right )^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {8 x}{\left (x+e^4\right )^2+81} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(648 + 8*E^8 - 8*x^2)/(6561 + E^16 + 4*E^12*x + 162*x^2 + x^4 + E^8*(162 + 6*x^2) + E^4*(324*x + 4*x^3)),x
]

[Out]

(8*x)/(81 + (E^4 + x)^2)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 {8 \left (81+2 e^4 x-x^2\right )}{\left (81+x^2\right )^2} \, dx,x,e^4+x\right )\\ &=8 \operatorname {Subst}\left (\int \frac {81+2 e^4 x-x^2}{\left (81+x^2\right )^2} \, dx,x,e^4+x\right )\\ &=\frac {8 x}{81+\left (e^4+x\right )^2}-\frac {4}{81} \operatorname {Subst}\left (\int 0 \, dx,x,e^4+x\right )\\ &=\frac {8 x}{81+\left (e^4+x\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.36 \begin {gather*} \frac {8 x}{81+e^8+2 e^4 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(648 + 8*E^8 - 8*x^2)/(6561 + E^16 + 4*E^12*x + 162*x^2 + x^4 + E^8*(162 + 6*x^2) + E^4*(324*x + 4*x
^3)),x]

[Out]

(8*x)/(81 + E^8 + 2*E^4*x + x^2)

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fricas [A]  time = 0.62, size = 17, normalized size = 1.21 \begin {gather*} \frac {8 \, x}{x^{2} + 2 \, x e^{4} + e^{8} + 81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(4)^2-8*x^2+648)/(exp(4)^4+4*x*exp(4)^3+(6*x^2+162)*exp(4)^2+(4*x^3+324*x)*exp(4)+x^4+162*x^2+
6561),x, algorithm="fricas")

[Out]

8*x/(x^2 + 2*x*e^4 + e^8 + 81)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {8 \, {\left (x^{2} - e^{8} - 81\right )}}{x^{4} + 162 \, x^{2} + 4 \, x e^{12} + 6 \, {\left (x^{2} + 27\right )} e^{8} + 4 \, {\left (x^{3} + 81 \, x\right )} e^{4} + e^{16} + 6561}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(4)^2-8*x^2+648)/(exp(4)^4+4*x*exp(4)^3+(6*x^2+162)*exp(4)^2+(4*x^3+324*x)*exp(4)+x^4+162*x^2+
6561),x, algorithm="giac")

[Out]

integrate(-8*(x^2 - e^8 - 81)/(x^4 + 162*x^2 + 4*x*e^12 + 6*(x^2 + 27)*e^8 + 4*(x^3 + 81*x)*e^4 + e^16 + 6561)
, x)

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maple [A]  time = 0.14, size = 18, normalized size = 1.29

method	result	size
risch	\(\frac {8 x}{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}+81}\)	\(18\)
gosper	\(\frac {8 x}{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}+81}\)	\(20\)
norman	\(\frac {8 x}{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}+81}\)	\(20\)
default	\(2 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+4 \textit {\_Z}^{3} {\mathrm e}^{4}+\left (6 \,{\mathrm e}^{8}+162\right ) \textit {\_Z}^{2}+\left (4 \,{\mathrm e}^{12}+324 \,{\mathrm e}^{4}\right ) \textit {\_Z} +{\mathrm e}^{16}+6561+162 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left ({\mathrm e}^{8}-\textit {\_R}^{2}+81\right ) \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{12}+3 \textit {\_R} \,{\mathrm e}^{8}+3 \textit {\_R}^{2} {\mathrm e}^{4}+\textit {\_R}^{3}+81 \,{\mathrm e}^{4}+81 \textit {\_R}}\right )\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(4)^2-8*x^2+648)/(exp(4)^4+4*x*exp(4)^3+(6*x^2+162)*exp(4)^2+(4*x^3+324*x)*exp(4)+x^4+162*x^2+6561),
x,method=_RETURNVERBOSE)

[Out]

8*x/(exp(8)+2*x*exp(4)+x^2+81)

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maxima [A]  time = 0.53, size = 17, normalized size = 1.21 \begin {gather*} \frac {8 \, x}{x^{2} + 2 \, x e^{4} + e^{8} + 81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(4)^2-8*x^2+648)/(exp(4)^4+4*x*exp(4)^3+(6*x^2+162)*exp(4)^2+(4*x^3+324*x)*exp(4)+x^4+162*x^2+
6561),x, algorithm="maxima")

[Out]

8*x/(x^2 + 2*x*e^4 + e^8 + 81)

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mupad [B]  time = 0.12, size = 17, normalized size = 1.21 \begin {gather*} \frac {8\,x}{x^2+2\,{\mathrm {e}}^4\,x+{\mathrm {e}}^8+81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(8) - 8*x^2 + 648)/(exp(16) + exp(4)*(324*x + 4*x^3) + 4*x*exp(12) + exp(8)*(6*x^2 + 162) + 162*x^2
+ x^4 + 6561),x)

[Out]

(8*x)/(exp(8) + 2*x*exp(4) + x^2 + 81)

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sympy [A]  time = 0.29, size = 17, normalized size = 1.21 \begin {gather*} \frac {8 x}{x^{2} + 2 x e^{4} + 81 + e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(4)**2-8*x**2+648)/(exp(4)**4+4*x*exp(4)**3+(6*x**2+162)*exp(4)**2+(4*x**3+324*x)*exp(4)+x**4+
162*x**2+6561),x)

[Out]

8*x/(x**2 + 2*x*exp(4) + 81 + exp(8))

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