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

Optimal. Leaf size=79 \[ \frac{x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}-\frac{x \left (347 x^2+547\right )}{32 \left (x^4+3 x^2+2\right )}-\frac{1}{2 x}+\frac{189}{8} \tan ^{-1}(x)-\frac{1119 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}} \]

[Out]

-1/(2*x) + (x*(9 + 11*x^2))/(8*(2 + 3*x^2 + x^4)^2) - (x*(547 + 347*x^2))/(32*(2 + 3*x^2 + x^4)) + (189*ArcTan
[x])/8 - (1119*ArcTan[x/Sqrt[2]])/(32*Sqrt[2])

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Rubi [A]  time = 0.103181, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {1669, 1664, 203} \[ \frac{x \left (11 x^2+9\right )}{8 \left (x^4+3 x^2+2\right )^2}-\frac{x \left (347 x^2+547\right )}{32 \left (x^4+3 x^2+2\right )}-\frac{1}{2 x}+\frac{189}{8} \tan ^{-1}(x)-\frac{1119 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*x) + (x*(9 + 11*x^2))/(8*(2 + 3*x^2 + x^4)^2) - (x*(547 + 347*x^2))/(32*(2 + 3*x^2 + x^4)) + (189*ArcTan
[x])/8 - (1119*ArcTan[x/Sqrt[2]])/(32*Sqrt[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^2 \left (2+3 x^2+x^4\right )^3} \, dx &=\frac{x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac{1}{8} \int \frac{-16+29 x^2-55 x^4}{x^2 \left (2+3 x^2+x^4\right )^2} \, dx\\ &=\frac{x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (547+347 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \frac{32+441 x^2-347 x^4}{x^2 \left (2+3 x^2+x^4\right )} \, dx\\ &=\frac{x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (547+347 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \left (\frac{16}{x^2}+\frac{756}{1+x^2}-\frac{1119}{2+x^2}\right ) \, dx\\ &=-\frac{1}{2 x}+\frac{x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (547+347 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac{189}{8} \int \frac{1}{1+x^2} \, dx-\frac{1119}{32} \int \frac{1}{2+x^2} \, dx\\ &=-\frac{1}{2 x}+\frac{x \left (9+11 x^2\right )}{8 \left (2+3 x^2+x^4\right )^2}-\frac{x \left (547+347 x^2\right )}{32 \left (2+3 x^2+x^4\right )}+\frac{189}{8} \tan ^{-1}(x)-\frac{1119 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{32 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0678454, size = 63, normalized size = 0.8 \[ \frac{1}{64} \left (-\frac{2 \left (363 x^8+1684 x^6+2499 x^4+1250 x^2+64\right )}{x \left (x^4+3 x^2+2\right )^2}+1512 \tan ^{-1}(x)-1119 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(64 + 1250*x^2 + 2499*x^4 + 1684*x^6 + 363*x^8))/(x*(2 + 3*x^2 + x^4)^2) + 1512*ArcTan[x] - 1119*Sqrt[2]*
ArcTan[x/Sqrt[2]])/64

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Maple [A]  time = 0.014, size = 58, normalized size = 0.7 \begin{align*} -{\frac{1}{2\, \left ({x}^{2}+2 \right ) ^{2}} \left ({\frac{207\,{x}^{3}}{16}}+{\frac{233\,x}{8}} \right ) }-{\frac{1119\,\sqrt{2}}{64}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }+{\frac{1}{ \left ({x}^{2}+1 \right ) ^{2}} \left ( -{\frac{35\,{x}^{3}}{8}}-{\frac{37\,x}{8}} \right ) }+{\frac{189\,\arctan \left ( x \right ) }{8}}-{\frac{1}{2\,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^2/(x^4+3*x^2+2)^3,x)

[Out]

-1/2*(207/16*x^3+233/8*x)/(x^2+2)^2-1119/64*arctan(1/2*x*2^(1/2))*2^(1/2)+(-35/8*x^3-37/8*x)/(x^2+1)^2+189/8*a
rctan(x)-1/2/x

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Maxima [A]  time = 1.48621, size = 88, normalized size = 1.11 \begin{align*} -\frac{1119}{64} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - \frac{363 \, x^{8} + 1684 \, x^{6} + 2499 \, x^{4} + 1250 \, x^{2} + 64}{32 \,{\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )}} + \frac{189}{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^2/(x^4+3*x^2+2)^3,x, algorithm="maxima")

[Out]

-1119/64*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/32*(363*x^8 + 1684*x^6 + 2499*x^4 + 1250*x^2 + 64)/(x^9 + 6*x^7 + 1
3*x^5 + 12*x^3 + 4*x) + 189/8*arctan(x)

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Fricas [A]  time = 1.56684, size = 302, normalized size = 3.82 \begin{align*} -\frac{726 \, x^{8} + 3368 \, x^{6} + 4998 \, x^{4} + 1119 \, \sqrt{2}{\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 2500 \, x^{2} - 1512 \,{\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, x\right )} \arctan \left (x\right ) + 128}{64 \,{\left (x^{9} + 6 \, x^{7} + 13 \, x^{5} + 12 \, x^{3} + 4 \, 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^2/(x^4+3*x^2+2)^3,x, algorithm="fricas")

[Out]

-1/64*(726*x^8 + 3368*x^6 + 4998*x^4 + 1119*sqrt(2)*(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*x)*arctan(1/2*sqrt(2)*x
) + 2500*x^2 - 1512*(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*x)*arctan(x) + 128)/(x^9 + 6*x^7 + 13*x^5 + 12*x^3 + 4*
x)

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Sympy [A]  time = 0.260194, size = 70, normalized size = 0.89 \begin{align*} - \frac{363 x^{8} + 1684 x^{6} + 2499 x^{4} + 1250 x^{2} + 64}{32 x^{9} + 192 x^{7} + 416 x^{5} + 384 x^{3} + 128 x} + \frac{189 \operatorname{atan}{\left (x \right )}}{8} - \frac{1119 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{64} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(363*x**8 + 1684*x**6 + 2499*x**4 + 1250*x**2 + 64)/(32*x**9 + 192*x**7 + 416*x**5 + 384*x**3 + 128*x) + 189*
atan(x)/8 - 1119*sqrt(2)*atan(sqrt(2)*x/2)/64

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Giac [A]  time = 1.12745, size = 74, normalized size = 0.94 \begin{align*} -\frac{1119}{64} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - \frac{347 \, x^{7} + 1588 \, x^{5} + 2291 \, x^{3} + 1058 \, x}{32 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} - \frac{1}{2 \, x} + \frac{189}{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^2/(x^4+3*x^2+2)^3,x, algorithm="giac")

[Out]

-1119/64*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/32*(347*x^7 + 1588*x^5 + 2291*x^3 + 1058*x)/(x^4 + 3*x^2 + 2)^2 - 1
/2/x + 189/8*arctan(x)