### 3.88 $$\int \frac{4+x^2+3 x^4+5 x^6}{x^4 (2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=62 $-\frac{x \left (9 x^2+5\right )}{8 \left (x^4+3 x^2+2\right )}-\frac{1}{3 x^3}+\frac{11}{4 x}+\frac{21}{2} \tan ^{-1}(x)-\frac{71 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{8 \sqrt{2}}$

[Out]

-1/(3*x^3) + 11/(4*x) - (x*(5 + 9*x^2))/(8*(2 + 3*x^2 + x^4)) + (21*ArcTan[x])/2 - (71*ArcTan[x/Sqrt[2]])/(8*S
qrt[2])

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Rubi [A]  time = 0.0842406, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.097, Rules used = {1669, 1664, 203} $-\frac{x \left (9 x^2+5\right )}{8 \left (x^4+3 x^2+2\right )}-\frac{1}{3 x^3}+\frac{11}{4 x}+\frac{21}{2} \tan ^{-1}(x)-\frac{71 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{8 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(3*x^3) + 11/(4*x) - (x*(5 + 9*x^2))/(8*(2 + 3*x^2 + x^4)) + (21*ArcTan[x])/2 - (71*ArcTan[x/Sqrt[2]])/(8*S
qrt[2])

Rule 1669

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[x^m*(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])/x^m + (b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e)
/x^m + c*(4*p + 7)*(b*d - 2*a*e)*x^(2 - m), 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] && ILtQ[m/2, 0]

Rule 1664

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x
)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]

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}{x^4 \left (2+3 x^2+x^4\right )^2} \, dx &=-\frac{x \left (5+9 x^2\right )}{8 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{-8+10 x^2-\frac{39 x^4}{2}+\frac{9 x^6}{2}}{x^4 \left (2+3 x^2+x^4\right )} \, dx\\ &=-\frac{x \left (5+9 x^2\right )}{8 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-\frac{4}{x^4}+\frac{11}{x^2}-\frac{42}{1+x^2}+\frac{71}{2 \left (2+x^2\right )}\right ) \, dx\\ &=-\frac{1}{3 x^3}+\frac{11}{4 x}-\frac{x \left (5+9 x^2\right )}{8 \left (2+3 x^2+x^4\right )}-\frac{71}{8} \int \frac{1}{2+x^2} \, dx+\frac{21}{2} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{3 x^3}+\frac{11}{4 x}-\frac{x \left (5+9 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac{21}{2} \tan ^{-1}(x)-\frac{71 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0533273, size = 56, normalized size = 0.9 $\frac{1}{48} \left (-\frac{6 x \left (9 x^2+5\right )}{x^4+3 x^2+2}-\frac{16}{x^3}+\frac{132}{x}+504 \tan ^{-1}(x)-213 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-16/x^3 + 132/x - (6*x*(5 + 9*x^2))/(2 + 3*x^2 + x^4) + 504*ArcTan[x] - 213*Sqrt[2]*ArcTan[x/Sqrt[2]])/48

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Maple [A]  time = 0.014, size = 48, normalized size = 0.8 \begin{align*} -{\frac{13\,x}{8\,{x}^{2}+16}}-{\frac{71\,\sqrt{2}}{16}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }+{\frac{x}{2\,{x}^{2}+2}}+{\frac{21\,\arctan \left ( x \right ) }{2}}-{\frac{1}{3\,{x}^{3}}}+{\frac{11}{4\,x}} \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/(x^4+3*x^2+2)^2,x)

[Out]

-13/8*x/(x^2+2)-71/16*arctan(1/2*x*2^(1/2))*2^(1/2)+1/2*x/(x^2+1)+21/2*arctan(x)-1/3/x^3+11/4/x

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Maxima [A]  time = 1.47063, size = 70, normalized size = 1.13 \begin{align*} -\frac{71}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{39 \, x^{6} + 175 \, x^{4} + 108 \, x^{2} - 16}{24 \,{\left (x^{7} + 3 \, x^{5} + 2 \, x^{3}\right )}} + \frac{21}{2} \, \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/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

-71/16*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/24*(39*x^6 + 175*x^4 + 108*x^2 - 16)/(x^7 + 3*x^5 + 2*x^3) + 21/2*arc
tan(x)

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Fricas [A]  time = 1.58719, size = 213, normalized size = 3.44 \begin{align*} \frac{78 \, x^{6} + 350 \, x^{4} - 213 \, \sqrt{2}{\left (x^{7} + 3 \, x^{5} + 2 \, x^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 216 \, x^{2} + 504 \,{\left (x^{7} + 3 \, x^{5} + 2 \, x^{3}\right )} \arctan \left (x\right ) - 32}{48 \,{\left (x^{7} + 3 \, x^{5} + 2 \, x^{3}\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/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/48*(78*x^6 + 350*x^4 - 213*sqrt(2)*(x^7 + 3*x^5 + 2*x^3)*arctan(1/2*sqrt(2)*x) + 216*x^2 + 504*(x^7 + 3*x^5
+ 2*x^3)*arctan(x) - 32)/(x^7 + 3*x^5 + 2*x^3)

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Sympy [A]  time = 0.225061, size = 56, normalized size = 0.9 \begin{align*} \frac{21 \operatorname{atan}{\left (x \right )}}{2} - \frac{71 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{16} + \frac{39 x^{6} + 175 x^{4} + 108 x^{2} - 16}{24 x^{7} + 72 x^{5} + 48 x^{3}} \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/(x**4+3*x**2+2)**2,x)

[Out]

21*atan(x)/2 - 71*sqrt(2)*atan(sqrt(2)*x/2)/16 + (39*x**6 + 175*x**4 + 108*x**2 - 16)/(24*x**7 + 72*x**5 + 48*
x**3)

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Giac [A]  time = 1.11528, size = 70, normalized size = 1.13 \begin{align*} -\frac{71}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - \frac{9 \, x^{3} + 5 \, x}{8 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + \frac{33 \, x^{2} - 4}{12 \, x^{3}} + \frac{21}{2} \, \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/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

-71/16*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/8*(9*x^3 + 5*x)/(x^4 + 3*x^2 + 2) + 1/12*(33*x^2 - 4)/x^3 + 21/2*arct
an(x)