### 3.94 $$\int \frac{x^4 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^3} \, dx$$

Optimal. Leaf size=72 $-\frac{x \left (51 x^2+50\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (125 x^2+254\right )}{8 \left (x^4+3 x^2+2\right )}-\frac{369}{8} \tan ^{-1}(x)+\frac{267 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}}$

[Out]

-(x*(50 + 51*x^2))/(4*(2 + 3*x^2 + x^4)^2) + (x*(254 + 125*x^2))/(8*(2 + 3*x^2 + x^4)) - (369*ArcTan[x])/8 + (
267*ArcTan[x/Sqrt[2]])/(4*Sqrt[2])

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Rubi [A]  time = 0.0678667, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.129, Rules used = {1668, 1678, 1166, 203} $-\frac{x \left (51 x^2+50\right )}{4 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (125 x^2+254\right )}{8 \left (x^4+3 x^2+2\right )}-\frac{369}{8} \tan ^{-1}(x)+\frac{267 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(x*(50 + 51*x^2))/(4*(2 + 3*x^2 + x^4)^2) + (x*(254 + 125*x^2))/(8*(2 + 3*x^2 + x^4)) - (369*ArcTan[x])/8 + (
267*ArcTan[x/Sqrt[2]])/(4*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 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 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^4 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx &=-\frac{x \left (50+51 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac{1}{8} \int \frac{-100+294 x^2+96 x^4-40 x^6}{\left (2+3 x^2+x^4\right )^2} \, dx\\ &=-\frac{x \left (50+51 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (254+125 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \frac{-816+660 x^2}{2+3 x^2+x^4} \, dx\\ &=-\frac{x \left (50+51 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (254+125 x^2\right )}{8 \left (2+3 x^2+x^4\right )}-\frac{369}{8} \int \frac{1}{1+x^2} \, dx+\frac{267}{4} \int \frac{1}{2+x^2} \, dx\\ &=-\frac{x \left (50+51 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (254+125 x^2\right )}{8 \left (2+3 x^2+x^4\right )}-\frac{369}{8} \tan ^{-1}(x)+\frac{267 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{4 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0605023, size = 55, normalized size = 0.76 $\frac{1}{8} \left (\frac{x \left (125 x^6+629 x^4+910 x^2+408\right )}{\left (x^4+3 x^2+2\right )^2}-369 \tan ^{-1}(x)+267 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((x*(408 + 910*x^2 + 629*x^4 + 125*x^6))/(2 + 3*x^2 + x^4)^2 - 369*ArcTan[x] + 267*Sqrt[2]*ArcTan[x/Sqrt[2]])/
8

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Maple [A]  time = 0.012, size = 54, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ \left ({x}^{2}+2 \right ) ^{2}} \left ({\frac{51\,{x}^{3}}{8}}+{\frac{77\,x}{4}} \right ) }+{\frac{267\,\sqrt{2}}{8}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{1}{ \left ({x}^{2}+1 \right ) ^{2}} \left ( -{\frac{23\,{x}^{3}}{8}}-{\frac{25\,x}{8}} \right ) }-{\frac{369\,\arctan \left ( x \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(51/8*x^3+77/4*x)/(x^2+2)^2+267/8*arctan(1/2*x*2^(1/2))*2^(1/2)-(-23/8*x^3-25/8*x)/(x^2+1)^2-369/8*arctan(x)

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Maxima [A]  time = 1.49012, size = 81, normalized size = 1.12 \begin{align*} \frac{267}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{125 \, x^{7} + 629 \, x^{5} + 910 \, x^{3} + 408 \, x}{8 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} - \frac{369}{8} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

267/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/8*(125*x^7 + 629*x^5 + 910*x^3 + 408*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2
+ 4) - 369/8*arctan(x)

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Fricas [A]  time = 1.64032, size = 274, normalized size = 3.81 \begin{align*} \frac{125 \, x^{7} + 629 \, x^{5} + 910 \, x^{3} + 267 \, \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 ) - 369 \,{\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (x\right ) + 408 \, x}{8 \,{\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(x^4*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^3,x, algorithm="fricas")

[Out]

1/8*(125*x^7 + 629*x^5 + 910*x^3 + 267*sqrt(2)*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(1/2*sqrt(2)*x) - 369
*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(x) + 408*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)

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Sympy [A]  time = 0.236653, size = 65, normalized size = 0.9 \begin{align*} \frac{125 x^{7} + 629 x^{5} + 910 x^{3} + 408 x}{8 x^{8} + 48 x^{6} + 104 x^{4} + 96 x^{2} + 32} - \frac{369 \operatorname{atan}{\left (x \right )}}{8} + \frac{267 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{8} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(125*x**7 + 629*x**5 + 910*x**3 + 408*x)/(8*x**8 + 48*x**6 + 104*x**4 + 96*x**2 + 32) - 369*atan(x)/8 + 267*sq
rt(2)*atan(sqrt(2)*x/2)/8

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Giac [A]  time = 1.10008, size = 68, normalized size = 0.94 \begin{align*} \frac{267}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{125 \, x^{7} + 629 \, x^{5} + 910 \, x^{3} + 408 \, x}{8 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} - \frac{369}{8} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

267/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/8*(125*x^7 + 629*x^5 + 910*x^3 + 408*x)/(x^4 + 3*x^2 + 2)^2 - 369/8*ar
ctan(x)