3.83 \(\int \frac{x^6 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=57 \[ x^5-9 x^3+\frac{\left (103 x^2+102\right ) x}{2 \left (x^4+3 x^2+2\right )}+98 x-\frac{11}{2} \tan ^{-1}(x)-118 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ) \]

[Out]

98*x - 9*x^3 + x^5 + (x*(102 + 103*x^2))/(2*(2 + 3*x^2 + x^4)) - (11*ArcTan[x])/2 - 118*Sqrt[2]*ArcTan[x/Sqrt[
2]]

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Rubi [A]  time = 0.0821456, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1668, 1676, 1166, 203} \[ x^5-9 x^3+\frac{\left (103 x^2+102\right ) x}{2 \left (x^4+3 x^2+2\right )}+98 x-\frac{11}{2} \tan ^{-1}(x)-118 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

98*x - 9*x^3 + x^5 + (x*(102 + 103*x^2))/(2*(2 + 3*x^2 + x^4)) - (11*ArcTan[x])/2 - 118*Sqrt[2]*ArcTan[x/Sqrt[
2]]

Rule 1668

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainde
r[x^m*Pq, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 2]}, S
imp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)
), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a
*c)*PolynomialQuotient[x^m*Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p +
7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && GtQ[Expon[Pq, x^2], 1] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[m/2, 0]

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx &=\frac{x \left (102+103 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{204+6 x^2-108 x^4+48 x^6-20 x^8}{2+3 x^2+x^4} \, dx\\ &=\frac{x \left (102+103 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-392+108 x^2-20 x^4+\frac{2 \left (494+483 x^2\right )}{2+3 x^2+x^4}\right ) \, dx\\ &=98 x-9 x^3+x^5+\frac{x \left (102+103 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{2} \int \frac{494+483 x^2}{2+3 x^2+x^4} \, dx\\ &=98 x-9 x^3+x^5+\frac{x \left (102+103 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{11}{2} \int \frac{1}{1+x^2} \, dx-236 \int \frac{1}{2+x^2} \, dx\\ &=98 x-9 x^3+x^5+\frac{x \left (102+103 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{11}{2} \tan ^{-1}(x)-118 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0479191, size = 58, normalized size = 1.02 \[ x^5-9 x^3+\frac{103 x^3+102 x}{2 \left (x^4+3 x^2+2\right )}+98 x-\frac{11}{2} \tan ^{-1}(x)-118 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(4 + x^2 + 3*x^4 + 5*x^6))/(2 + 3*x^2 + x^4)^2,x]

[Out]

98*x - 9*x^3 + x^5 + (102*x + 103*x^3)/(2*(2 + 3*x^2 + x^4)) - (11*ArcTan[x])/2 - 118*Sqrt[2]*ArcTan[x/Sqrt[2]
]

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Maple [A]  time = 0.013, size = 49, normalized size = 0.9 \begin{align*}{x}^{5}-9\,{x}^{3}+98\,x+52\,{\frac{x}{{x}^{2}+2}}-118\,\arctan \left ( 1/2\,x\sqrt{2} \right ) \sqrt{2}-{\frac{x}{2\,{x}^{2}+2}}-{\frac{11\,\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x)

[Out]

x^5-9*x^3+98*x+52*x/(x^2+2)-118*arctan(1/2*x*2^(1/2))*2^(1/2)-1/2*x/(x^2+1)-11/2*arctan(x)

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Maxima [A]  time = 1.48617, size = 69, normalized size = 1.21 \begin{align*} x^{5} - 9 \, x^{3} - 118 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 98 \, x + \frac{103 \, x^{3} + 102 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{11}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

x^5 - 9*x^3 - 118*sqrt(2)*arctan(1/2*sqrt(2)*x) + 98*x + 1/2*(103*x^3 + 102*x)/(x^4 + 3*x^2 + 2) - 11/2*arctan
(x)

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Fricas [A]  time = 1.75625, size = 209, normalized size = 3.67 \begin{align*} \frac{2 \, x^{9} - 12 \, x^{7} + 146 \, x^{5} + 655 \, x^{3} - 236 \, \sqrt{2}{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 11 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (x\right ) + 494 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/2*(2*x^9 - 12*x^7 + 146*x^5 + 655*x^3 - 236*sqrt(2)*(x^4 + 3*x^2 + 2)*arctan(1/2*sqrt(2)*x) - 11*(x^4 + 3*x^
2 + 2)*arctan(x) + 494*x)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.184926, size = 54, normalized size = 0.95 \begin{align*} x^{5} - 9 x^{3} + 98 x + \frac{103 x^{3} + 102 x}{2 x^{4} + 6 x^{2} + 4} - \frac{11 \operatorname{atan}{\left (x \right )}}{2} - 118 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(5*x**6+3*x**4+x**2+4)/(x**4+3*x**2+2)**2,x)

[Out]

x**5 - 9*x**3 + 98*x + (103*x**3 + 102*x)/(2*x**4 + 6*x**2 + 4) - 11*atan(x)/2 - 118*sqrt(2)*atan(sqrt(2)*x/2)

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Giac [A]  time = 1.08387, size = 69, normalized size = 1.21 \begin{align*} x^{5} - 9 \, x^{3} - 118 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 98 \, x + \frac{103 \, x^{3} + 102 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{11}{2} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

x^5 - 9*x^3 - 118*sqrt(2)*arctan(1/2*sqrt(2)*x) + 98*x + 1/2*(103*x^3 + 102*x)/(x^4 + 3*x^2 + 2) - 11/2*arctan
(x)