3.1.97 \(\int \frac {4+x^2+3 x^4+5 x^6}{x^2 (2+3 x^2+x^4)^3} \, dx\) [97]

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

[Out]

-1/2/x+1/8*x*(11*x^2+9)/(x^4+3*x^2+2)^2-1/32*x*(347*x^2+547)/(x^4+3*x^2+2)+189/8*arctan(x)-1119/64*arctan(1/2*
x*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 = {1683, 1678, 209} \begin {gather*} \frac {189 \text {ArcTan}(x)}{8}-\frac {1119 \text {ArcTan}\left (\frac {x}{\sqrt {2}}\right )}{32 \sqrt {2}}+\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} \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)^3),x]

[Out]

-1/2*1/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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1678

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 1683

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 )^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*}

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Mathematica [A]
time = 0.05, size = 63, normalized size = 0.80 \begin {gather*} \frac {1}{64} \left (-\frac {2 \left (64+1250 x^2+2499 x^4+1684 x^6+363 x^8\right )}{x \left (2+3 x^2+x^4\right )^2}+1512 \tan ^{-1}(x)-1119 \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)^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 = 0.04, size = 58, normalized size = 0.73

method	result	size
risch	\(\frac {-\frac {363}{32} x^{8}-\frac {421}{8} x^{6}-\frac {2499}{32} x^{4}-\frac {625}{16} x^{2}-2}{x \left (x^{4}+3 x^{2}+2\right )^{2}}+\frac {189 \arctan \left (x \right )}{8}-\frac {1119 \arctan \left (\frac {\sqrt {2}\, x}{2}\right ) \sqrt {2}}{64}\)	\(56\)
default	\(\frac {-\frac {35}{8} x^{3}-\frac {37}{8} x}{\left (x^{2}+1\right )^{2}}+\frac {189 \arctan \left (x \right )}{8}-\frac {1}{2 x}-\frac {\frac {207}{16} x^{3}+\frac {233}{8} x}{2 \left (x^{2}+2\right )^{2}}-\frac {1119 \arctan \left (\frac {\sqrt {2}\, x}{2}\right ) \sqrt {2}}{64}\)	\(58\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 65, normalized size = 0.82 \begin {gather*} -\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 {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)^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 = 0.37, size = 108, normalized size = 1.37 \begin {gather*} -\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 {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)^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 = 0.11, size = 71, normalized size = 0.90 \begin {gather*} \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 {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)**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 = 5.46, size = 55, normalized size = 0.70 \begin {gather*} -\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 {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)^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)

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Mupad [B]
time = 0.92, size = 65, normalized size = 0.82 \begin {gather*} \frac {189\,\mathrm {atan}\left (x\right )}{8}-\frac {1119\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{64}-\frac {\frac {363\,x^8}{32}+\frac {421\,x^6}{8}+\frac {2499\,x^4}{32}+\frac {625\,x^2}{16}+2}{x^9+6\,x^7+13\,x^5+12\,x^3+4\,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)^3),x)

[Out]

(189*atan(x))/8 - (1119*2^(1/2)*atan((2^(1/2)*x)/2))/64 - ((625*x^2)/16 + (2499*x^4)/32 + (421*x^6)/8 + (363*x
^8)/32 + 2)/(4*x + 12*x^3 + 13*x^5 + 6*x^7 + x^9)

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