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

Optimal. Leaf size=53 \[ \frac {x \left (11 x^2+9\right )}{4 \left (x^4+3 x^2+2\right )}-\frac {1}{x}-\frac {19}{2} \tan ^{-1}(x)+\frac {45 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {x \left (11 x^2+9\right )}{4 \left (x^4+3 x^2+2\right )}-\frac {1}{x}-\frac {19}{2} \tan ^{-1}(x)+\frac {45 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + (x*(9 + 11*x^2))/(4*(2 + 3*x^2 + x^4)) - (19*ArcTan[x])/2 + (45*ArcTan[x/Sqrt[2]])/(4*Sqrt[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])

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 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]

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 )^2} \, dx &=\frac {x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac {1}{4} \int \frac {-8+19 x^2-11 x^4}{x^2 \left (2+3 x^2+x^4\right )} \, dx\\ &=\frac {x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac {1}{4} \int \left (-\frac {4}{x^2}+\frac {38}{1+x^2}-\frac {45}{2+x^2}\right ) \, dx\\ &=-\frac {1}{x}+\frac {x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac {19}{2} \int \frac {1}{1+x^2} \, dx+\frac {45}{4} \int \frac {1}{2+x^2} \, dx\\ &=-\frac {1}{x}+\frac {x \left (9+11 x^2\right )}{4 \left (2+3 x^2+x^4\right )}-\frac {19}{2} \tan ^{-1}(x)+\frac {45 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 51, normalized size = 0.96 \begin {gather*} \frac {1}{8} \left (\frac {2 x \left (11 x^2+9\right )}{x^4+3 x^2+2}-\frac {8}{x}-76 \tan ^{-1}(x)+45 \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8/x + (2*x*(9 + 11*x^2))/(2 + 3*x^2 + x^4) - 76*ArcTan[x] + 45*Sqrt[2]*ArcTan[x/Sqrt[2]])/8

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4+x^2+3 x^4+5 x^6}{x^2 \left (2+3 x^2+x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.16, size = 68, normalized size = 1.28 \begin {gather*} \frac {14 \, x^{4} + 45 \, \sqrt {2} {\left (x^{5} + 3 \, x^{3} + 2 \, x\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - 6 \, x^{2} - 76 \, {\left (x^{5} + 3 \, x^{3} + 2 \, x\right )} \arctan \relax (x) - 16}{8 \, {\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} \end {gather*}

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)^2,x, algorithm="fricas")

[Out]

1/8*(14*x^4 + 45*sqrt(2)*(x^5 + 3*x^3 + 2*x)*arctan(1/2*sqrt(2)*x) - 6*x^2 - 76*(x^5 + 3*x^3 + 2*x)*arctan(x)
- 16)/(x^5 + 3*x^3 + 2*x)

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giac [A]  time = 0.38, size = 45, normalized size = 0.85 \begin {gather*} \frac {45}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {7 \, x^{4} - 3 \, x^{2} - 8}{4 \, {\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} - \frac {19}{2} \, \arctan \relax (x) \end {gather*}

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)^2,x, algorithm="giac")

[Out]

45/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(7*x^4 - 3*x^2 - 8)/(x^5 + 3*x^3 + 2*x) - 19/2*arctan(x)

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maple [A]  time = 0.01, size = 43, normalized size = 0.81 \begin {gather*} -\frac {x}{2 \left (x^{2}+1\right )}+\frac {13 x}{4 \left (x^{2}+2\right )}-\frac {19 \arctan \relax (x )}{2}+\frac {45 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{8}-\frac {1}{x} \end {gather*}

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)^2,x)

[Out]

-1/x-1/2/(x^2+1)*x-19/2*arctan(x)+13/4/(x^2+2)*x+45/8*2^(1/2)*arctan(1/2*2^(1/2)*x)

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maxima [A]  time = 1.55, size = 45, normalized size = 0.85 \begin {gather*} \frac {45}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {7 \, x^{4} - 3 \, x^{2} - 8}{4 \, {\left (x^{5} + 3 \, x^{3} + 2 \, x\right )}} - \frac {19}{2} \, \arctan \relax (x) \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

45/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(7*x^4 - 3*x^2 - 8)/(x^5 + 3*x^3 + 2*x) - 19/2*arctan(x)

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mupad [B]  time = 0.07, size = 45, normalized size = 0.85 \begin {gather*} \frac {45\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{8}-\frac {19\,\mathrm {atan}\relax (x)}{2}-\frac {-\frac {7\,x^4}{4}+\frac {3\,x^2}{4}+2}{x^5+3\,x^3+2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(45*2^(1/2)*atan((2^(1/2)*x)/2))/8 - (19*atan(x))/2 - ((3*x^2)/4 - (7*x^4)/4 + 2)/(2*x + 3*x^3 + x^5)

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sympy [A]  time = 0.22, size = 49, normalized size = 0.92 \begin {gather*} \frac {7 x^{4} - 3 x^{2} - 8}{4 x^{5} + 12 x^{3} + 8 x} - \frac {19 \operatorname {atan}{\relax (x )}}{2} + \frac {45 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{8} \end {gather*}

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)**2,x)

[Out]

(7*x**4 - 3*x**2 - 8)/(4*x**5 + 12*x**3 + 8*x) - 19*atan(x)/2 + 45*sqrt(2)*atan(sqrt(2)*x/2)/8

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