### 3.89 $$\int \frac{4+x^2+3 x^4+5 x^6}{x^6 (2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=69 $-\frac{x \left (3-5 x^2\right )}{16 \left (x^4+3 x^2+2\right )}+\frac{11}{12 x^3}-\frac{1}{5 x^5}-\frac{23}{4 x}-\frac{23}{2} \tan ^{-1}(x)+\frac{97 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}$

[Out]

-1/(5*x^5) + 11/(12*x^3) - 23/(4*x) - (x*(3 - 5*x^2))/(16*(2 + 3*x^2 + x^4)) - (23*ArcTan[x])/2 + (97*ArcTan[x
/Sqrt[2]])/(16*Sqrt[2])

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Rubi [A]  time = 0.0904842, antiderivative size = 69, 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} $-\frac{x \left (3-5 x^2\right )}{16 \left (x^4+3 x^2+2\right )}+\frac{11}{12 x^3}-\frac{1}{5 x^5}-\frac{23}{4 x}-\frac{23}{2} \tan ^{-1}(x)+\frac{97 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(5*x^5) + 11/(12*x^3) - 23/(4*x) - (x*(3 - 5*x^2))/(16*(2 + 3*x^2 + x^4)) - (23*ArcTan[x])/2 + (97*ArcTan[x
/Sqrt[2]])/(16*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^6 \left (2+3 x^2+x^4\right )^2} \, dx &=-\frac{x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{-8+10 x^2-17 x^4+\frac{39 x^6}{4}-\frac{5 x^8}{4}}{x^6 \left (2+3 x^2+x^4\right )} \, dx\\ &=-\frac{x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (-\frac{4}{x^6}+\frac{11}{x^4}-\frac{23}{x^2}+\frac{46}{1+x^2}-\frac{97}{4 \left (2+x^2\right )}\right ) \, dx\\ &=-\frac{1}{5 x^5}+\frac{11}{12 x^3}-\frac{23}{4 x}-\frac{x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}+\frac{97}{16} \int \frac{1}{2+x^2} \, dx-\frac{23}{2} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{5 x^5}+\frac{11}{12 x^3}-\frac{23}{4 x}-\frac{x \left (3-5 x^2\right )}{16 \left (2+3 x^2+x^4\right )}-\frac{23}{2} \tan ^{-1}(x)+\frac{97 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{16 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0595521, size = 61, normalized size = 0.88 $\frac{1}{480} \left (\frac{30 x \left (5 x^2-3\right )}{x^4+3 x^2+2}+\frac{440}{x^3}-\frac{96}{x^5}-\frac{2760}{x}-5520 \tan ^{-1}(x)+1455 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-96/x^5 + 440/x^3 - 2760/x + (30*x*(-3 + 5*x^2))/(2 + 3*x^2 + x^4) - 5520*ArcTan[x] + 1455*Sqrt[2]*ArcTan[x/S
qrt[2]])/480

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Maple [A]  time = 0.016, size = 53, normalized size = 0.8 \begin{align*}{\frac{13\,x}{16\,{x}^{2}+32}}+{\frac{97\,\sqrt{2}}{32}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{x}{2\,{x}^{2}+2}}-{\frac{23\,\arctan \left ( x \right ) }{2}}-{\frac{1}{5\,{x}^{5}}}+{\frac{11}{12\,{x}^{3}}}-{\frac{23}{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^6/(x^4+3*x^2+2)^2,x)

[Out]

13/16*x/(x^2+2)+97/32*arctan(1/2*x*2^(1/2))*2^(1/2)-1/2*x/(x^2+1)-23/2*arctan(x)-1/5/x^5+11/12/x^3-23/4/x

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Maxima [A]  time = 1.46622, size = 77, normalized size = 1.12 \begin{align*} \frac{97}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - \frac{1305 \, x^{8} + 3965 \, x^{6} + 2148 \, x^{4} - 296 \, x^{2} + 96}{240 \,{\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )}} - \frac{23}{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^6/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

97/32*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/240*(1305*x^8 + 3965*x^6 + 2148*x^4 - 296*x^2 + 96)/(x^9 + 3*x^7 + 2*x
^5) - 23/2*arctan(x)

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Fricas [A]  time = 1.53035, size = 239, normalized size = 3.46 \begin{align*} -\frac{2610 \, x^{8} + 7930 \, x^{6} + 4296 \, x^{4} - 1455 \, \sqrt{2}{\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 592 \, x^{2} + 5520 \,{\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\right )} \arctan \left (x\right ) + 192}{480 \,{\left (x^{9} + 3 \, x^{7} + 2 \, x^{5}\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^6/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

-1/480*(2610*x^8 + 7930*x^6 + 4296*x^4 - 1455*sqrt(2)*(x^9 + 3*x^7 + 2*x^5)*arctan(1/2*sqrt(2)*x) - 592*x^2 +
5520*(x^9 + 3*x^7 + 2*x^5)*arctan(x) + 192)/(x^9 + 3*x^7 + 2*x^5)

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Sympy [A]  time = 0.24267, size = 61, normalized size = 0.88 \begin{align*} - \frac{23 \operatorname{atan}{\left (x \right )}}{2} + \frac{97 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{32} - \frac{1305 x^{8} + 3965 x^{6} + 2148 x^{4} - 296 x^{2} + 96}{240 x^{9} + 720 x^{7} + 480 x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-23*atan(x)/2 + 97*sqrt(2)*atan(sqrt(2)*x/2)/32 - (1305*x**8 + 3965*x**6 + 2148*x**4 - 296*x**2 + 96)/(240*x**
9 + 720*x**7 + 480*x**5)

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Giac [A]  time = 1.14244, size = 77, normalized size = 1.12 \begin{align*} \frac{97}{32} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{5 \, x^{3} - 3 \, x}{16 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac{345 \, x^{4} - 55 \, x^{2} + 12}{60 \, x^{5}} - \frac{23}{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^6/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

97/32*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/16*(5*x^3 - 3*x)/(x^4 + 3*x^2 + 2) - 1/60*(345*x^4 - 55*x^2 + 12)/x^5
- 23/2*arctan(x)