### 3.98 $$\int \frac{4+x^2+3 x^4+5 x^6}{x^4 (2+3 x^2+x^4)^3} \, dx$$

Optimal. Leaf size=86 $-\frac{x \left (9 x^2+5\right )}{16 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (571 x^2+951\right )}{64 \left (x^4+3 x^2+2\right )}-\frac{1}{6 x^3}+\frac{17}{8 x}-\frac{113}{8} \tan ^{-1}(x)+\frac{1611 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{64 \sqrt{2}}$

[Out]

-1/(6*x^3) + 17/(8*x) - (x*(5 + 9*x^2))/(16*(2 + 3*x^2 + x^4)^2) + (x*(951 + 571*x^2))/(64*(2 + 3*x^2 + x^4))
- (113*ArcTan[x])/8 + (1611*ArcTan[x/Sqrt[2]])/(64*Sqrt[2])

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Rubi [A]  time = 0.118885, antiderivative size = 86, 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} $-\frac{x \left (9 x^2+5\right )}{16 \left (x^4+3 x^2+2\right )^2}+\frac{x \left (571 x^2+951\right )}{64 \left (x^4+3 x^2+2\right )}-\frac{1}{6 x^3}+\frac{17}{8 x}-\frac{113}{8} \tan ^{-1}(x)+\frac{1611 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{64 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(6*x^3) + 17/(8*x) - (x*(5 + 9*x^2))/(16*(2 + 3*x^2 + x^4)^2) + (x*(951 + 571*x^2))/(64*(2 + 3*x^2 + x^4))
- (113*ArcTan[x])/8 + (1611*ArcTan[x/Sqrt[2]])/(64*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^4 \left (2+3 x^2+x^4\right )^3} \, dx &=-\frac{x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}-\frac{1}{8} \int \frac{-16+20 x^2-\frac{73 x^4}{2}+\frac{45 x^6}{2}}{x^4 \left (2+3 x^2+x^4\right )^2} \, dx\\ &=-\frac{x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \frac{32-88 x^2-\frac{573 x^4}{2}+\frac{571 x^6}{2}}{x^4 \left (2+3 x^2+x^4\right )} \, dx\\ &=-\frac{x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}+\frac{1}{32} \int \left (\frac{16}{x^4}-\frac{68}{x^2}-\frac{452}{1+x^2}+\frac{1611}{2 \left (2+x^2\right )}\right ) \, dx\\ &=-\frac{1}{6 x^3}+\frac{17}{8 x}-\frac{x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}-\frac{113}{8} \int \frac{1}{1+x^2} \, dx+\frac{1611}{64} \int \frac{1}{2+x^2} \, dx\\ &=-\frac{1}{6 x^3}+\frac{17}{8 x}-\frac{x \left (5+9 x^2\right )}{16 \left (2+3 x^2+x^4\right )^2}+\frac{x \left (951+571 x^2\right )}{64 \left (2+3 x^2+x^4\right )}-\frac{113}{8} \tan ^{-1}(x)+\frac{1611 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )}{64 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0603144, size = 78, normalized size = 0.91 $\frac{1}{384} \left (-\frac{24 x \left (9 x^2+5\right )}{\left (x^4+3 x^2+2\right )^2}+\frac{6 x \left (571 x^2+951\right )}{x^4+3 x^2+2}-\frac{64}{x^3}+\frac{816}{x}-5424 \tan ^{-1}(x)+4833 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-64/x^3 + 816/x - (24*x*(5 + 9*x^2))/(2 + 3*x^2 + x^4)^2 + (6*x*(951 + 571*x^2))/(2 + 3*x^2 + x^4) - 5424*Arc
Tan[x] + 4833*Sqrt[2]*ArcTan[x/Sqrt[2]])/384

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Maple [A]  time = 0.016, size = 64, normalized size = 0.7 \begin{align*}{\frac{1}{8\, \left ({x}^{2}+2 \right ) ^{2}} \left ({\frac{259\,{x}^{3}}{8}}+{\frac{285\,x}{4}} \right ) }+{\frac{1611\,\sqrt{2}}{128}\arctan \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{1}{ \left ({x}^{2}+1 \right ) ^{2}} \left ( -{\frac{39\,{x}^{3}}{8}}-{\frac{41\,x}{8}} \right ) }-{\frac{113\,\arctan \left ( x \right ) }{8}}-{\frac{1}{6\,{x}^{3}}}+{\frac{17}{8\,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)^3,x)

[Out]

1/8*(259/8*x^3+285/4*x)/(x^2+2)^2+1611/128*arctan(1/2*x*2^(1/2))*2^(1/2)-(-39/8*x^3-41/8*x)/(x^2+1)^2-113/8*ar
ctan(x)-1/6/x^3+17/8/x

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Maxima [A]  time = 1.54805, size = 97, normalized size = 1.13 \begin{align*} \frac{1611}{128} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{2121 \, x^{10} + 10408 \, x^{8} + 16989 \, x^{6} + 10126 \, x^{4} + 1248 \, x^{2} - 128}{192 \,{\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )}} - \frac{113}{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^4/(x^4+3*x^2+2)^3,x, algorithm="maxima")

[Out]

1611/128*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/192*(2121*x^10 + 10408*x^8 + 16989*x^6 + 10126*x^4 + 1248*x^2 - 128
)/(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3) - 113/8*arctan(x)

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Fricas [A]  time = 1.54364, size = 336, normalized size = 3.91 \begin{align*} \frac{4242 \, x^{10} + 20816 \, x^{8} + 33978 \, x^{6} + 20252 \, x^{4} + 4833 \, \sqrt{2}{\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 2496 \, x^{2} - 5424 \,{\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, x^{3}\right )} \arctan \left (x\right ) - 256}{384 \,{\left (x^{11} + 6 \, x^{9} + 13 \, x^{7} + 12 \, x^{5} + 4 \, 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)^3,x, algorithm="fricas")

[Out]

1/384*(4242*x^10 + 20816*x^8 + 33978*x^6 + 20252*x^4 + 4833*sqrt(2)*(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3)*a
rctan(1/2*sqrt(2)*x) + 2496*x^2 - 5424*(x^11 + 6*x^9 + 13*x^7 + 12*x^5 + 4*x^3)*arctan(x) - 256)/(x^11 + 6*x^9
+ 13*x^7 + 12*x^5 + 4*x^3)

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Sympy [A]  time = 0.277716, size = 76, normalized size = 0.88 \begin{align*} - \frac{113 \operatorname{atan}{\left (x \right )}}{8} + \frac{1611 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )}}{128} + \frac{2121 x^{10} + 10408 x^{8} + 16989 x^{6} + 10126 x^{4} + 1248 x^{2} - 128}{192 x^{11} + 1152 x^{9} + 2496 x^{7} + 2304 x^{5} + 768 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)**3,x)

[Out]

-113*atan(x)/8 + 1611*sqrt(2)*atan(sqrt(2)*x/2)/128 + (2121*x**10 + 10408*x**8 + 16989*x**6 + 10126*x**4 + 124
8*x**2 - 128)/(192*x**11 + 1152*x**9 + 2496*x**7 + 2304*x**5 + 768*x**3)

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Giac [A]  time = 1.11011, size = 84, normalized size = 0.98 \begin{align*} \frac{1611}{128} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + \frac{571 \, x^{7} + 2664 \, x^{5} + 3959 \, x^{3} + 1882 \, x}{64 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} + \frac{51 \, x^{2} - 4}{24 \, x^{3}} - \frac{113}{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^4/(x^4+3*x^2+2)^3,x, algorithm="giac")

[Out]

1611/128*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/64*(571*x^7 + 2664*x^5 + 3959*x^3 + 1882*x)/(x^4 + 3*x^2 + 2)^2 + 1
/24*(51*x^2 - 4)/x^3 - 113/8*arctan(x)