### 3.96 $$\int \frac{4+x^2+3 x^4+5 x^6}{(2+3 x^2+x^4)^3} \, dx$$

Optimal. Leaf size=72 $-\frac{x \left (12 x^2+11\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (217 x^2+335\right )}{16 \left (x^4+3 x^2+2\right )}-\frac{257}{8} \tan ^{-1}(x)+\frac{731 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}$

[Out]

-(x*(11 + 12*x^2))/(4*(2 + 3*x^2 + x^4)^2) + (x*(335 + 217*x^2))/(16*(2 + 3*x^2 + x^4)) - (257*ArcTan[x])/8 +
(731*ArcTan[x/Sqrt[2]])/(16*Sqrt[2])

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Rubi [A]  time = 0.037454, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {1678, 1178, 1166, 203} $-\frac{x \left (12 x^2+11\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (217 x^2+335\right )}{16 \left (x^4+3 x^2+2\right )}-\frac{257}{8} \tan ^{-1}(x)+\frac{731 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(x*(11 + 12*x^2))/(4*(2 + 3*x^2 + x^4)^2) + (x*(335 + 217*x^2))/(16*(2 + 3*x^2 + x^4)) - (257*ArcTan[x])/8 +
(731*ArcTan[x/Sqrt[2]])/(16*Sqrt[2])

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(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)*PolynomialQuot
ient[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] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

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{4+x^2+3 x^4+5 x^6}{\left (2+3 x^2+x^4\right )^3} \, dx &=-\frac{x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{1}{8} \int \frac{-38+80 x^2}{\left (2+3 x^2+x^4\right )^2} \, dx\\ &=-\frac{x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \frac{-594+434 x^2}{2+3 x^2+x^4} \, dx\\ &=-\frac{x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac{257}{8} \int \frac{1}{1+x^2} \, dx+\frac{731}{16} \int \frac{1}{2+x^2} \, dx\\ &=-\frac{x \left (11+12 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (335+217 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac{257}{8} \tan ^{-1}(x)+\frac{731 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0595557, size = 56, normalized size = 0.78 $\frac{1}{32} \left (\frac{2 x \left (217 x^6+986 x^4+1391 x^2+626\right )}{\left (x^4+3 x^2+2\right )^2}-1028 \tan ^{-1}(x)+731 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*x*(626 + 1391*x^2 + 986*x^4 + 217*x^6))/(2 + 3*x^2 + x^4)^2 - 1028*ArcTan[x] + 731*Sqrt[2]*ArcTan[x/Sqrt[2
]])/32

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Maple [A]  time = 0.013, size = 53, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ({x}^{2}+2 \right ) ^{2}} \left ({\frac{155\,{x}^{3}}{16}}+{\frac{181\,x}{8}} \right ) }+{\frac{731\,\sqrt{2}}{32}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{1}{ \left ({x}^{2}+1 \right ) ^{2}} \left ( -{\frac{31\,{x}^{3}}{8}}-{\frac{33\,x}{8}} \right ) }-{\frac{257\,\arctan \left ( x \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(155/16*x^3+181/8*x)/(x^2+2)^2+731/32*arctan(1/2*x*2^(1/2))*2^(1/2)-(-31/8*x^3-33/8*x)/(x^2+1)^2-257/8*arctan(
x)

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Maxima [A]  time = 1.51475, size = 81, normalized size = 1.12 \begin{align*} \frac{731}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{217 \, x^{7} + 986 \, x^{5} + 1391 \, x^{3} + 626 \, x}{16 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} - \frac{257}{8} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

731/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(217*x^7 + 986*x^5 + 1391*x^3 + 626*x)/(x^8 + 6*x^6 + 13*x^4 + 12*
x^2 + 4) - 257/8*arctan(x)

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Fricas [A]  time = 1.57343, size = 281, normalized size = 3.9 \begin{align*} \frac{434 \, x^{7} + 1972 \, x^{5} + 2782 \, x^{3} + 731 \, \sqrt{2}{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 1028 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (x\right ) + 1252 \, x}{32 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(434*x^7 + 1972*x^5 + 2782*x^3 + 731*sqrt(2)*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(1/2*sqrt(2)*x) -
1028*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(x) + 1252*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)

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Sympy [A]  time = 0.231106, size = 65, normalized size = 0.9 \begin{align*} \frac{217 x^{7} + 986 x^{5} + 1391 x^{3} + 626 x}{16 x^{8} + 96 x^{6} + 208 x^{4} + 192 x^{2} + 64} - \frac{257 \operatorname{atan}{\left (x \right )}}{8} + \frac{731 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{32} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(217*x**7 + 986*x**5 + 1391*x**3 + 626*x)/(16*x**8 + 96*x**6 + 208*x**4 + 192*x**2 + 64) - 257*atan(x)/8 + 731
*sqrt(2)*atan(sqrt(2)*x/2)/32

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Giac [A]  time = 1.1064, size = 68, normalized size = 0.94 \begin{align*} \frac{731}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{217 \, x^{7} + 986 \, x^{5} + 1391 \, x^{3} + 626 \, x}{16 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} - \frac{257}{8} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

731/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(217*x^7 + 986*x^5 + 1391*x^3 + 626*x)/(x^4 + 3*x^2 + 2)^2 - 257/8
*arctan(x)