\(\int \frac {x^6 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^3} \, dx\) [93]

Optimal result

Integrand size = 31, antiderivative size = 75 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=5 x+\frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {413 \arctan (x)}{8}-\frac {191 \arctan \left (\frac {x}{\sqrt {2}}\right )}{2 \sqrt {2}} \]

[Out]

5*x+1/4*x*(103*x^2+102)/(x^4+3*x^2+2)^2-1/8*x*(15*x^2+244)/(x^4+3*x^2+2)+413/8*arctan(x)-191/4*arctan(1/2*x*2^
(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1682, 1692, 1690, 1180, 209} \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=\frac {413 \arctan (x)}{8}-\frac {191 \arctan \left (\frac {x}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\left (15 x^2+244\right ) x}{8 \left (x^4+3 x^2+2\right )}+\frac {\left (103 x^2+102\right ) x}{4 \left (x^4+3 x^2+2\right )^2}+5 x \]

[In]

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

[Out]

5*x + (x*(102 + 103*x^2))/(4*(2 + 3*x^2 + x^4)^2) - (x*(244 + 15*x^2))/(8*(2 + 3*x^2 + x^4)) + (413*ArcTan[x])
/8 - (191*ArcTan[x/Sqrt[2]])/(2*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 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1682

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[(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] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p +
7)*(b*d - 2*a*e)*x^2, 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] && IGtQ[m/2, 0]

Rule 1690

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rule 1692

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[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[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {1}{8} \int \frac {204-606 x^2-216 x^4+96 x^6-40 x^8}{\left (2+3 x^2+x^4\right )^2} \, dx \\ & = \frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {1}{32} \int \frac {568-924 x^2+160 x^4}{2+3 x^2+x^4} \, dx \\ & = \frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {1}{32} \int \left (160+\frac {4 \left (62-351 x^2\right )}{2+3 x^2+x^4}\right ) \, dx \\ & = 5 x+\frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {1}{8} \int \frac {62-351 x^2}{2+3 x^2+x^4} \, dx \\ & = 5 x+\frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {413}{8} \int \frac {1}{1+x^2} \, dx-\frac {191}{2} \int \frac {1}{2+x^2} \, dx \\ & = 5 x+\frac {x \left (102+103 x^2\right )}{4 \left (2+3 x^2+x^4\right )^2}-\frac {x \left (244+15 x^2\right )}{8 \left (2+3 x^2+x^4\right )}+\frac {413}{8} \tan ^{-1}(x)-\frac {191 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{2 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=\frac {1}{8} \left (\frac {x \left (-124-76 x^2+231 x^4+225 x^6+40 x^8\right )}{\left (2+3 x^2+x^4\right )^2}+413 \arctan (x)-382 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )\right ) \]

[In]

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

[Out]

((x*(-124 - 76*x^2 + 231*x^4 + 225*x^6 + 40*x^8))/(2 + 3*x^2 + x^4)^2 + 413*ArcTan[x] - 382*Sqrt[2]*ArcTan[x/S
qrt[2]])/8

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.71

method	result	size
risch	\(5 x +\frac {-\frac {15}{8} x^{7}-\frac {289}{8} x^{5}-\frac {139}{2} x^{3}-\frac {71}{2} x}{\left (x^{4}+3 x^{2}+2\right )^{2}}+\frac {413 \arctan \left (x \right )}{8}-\frac {191 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{4}\)	\(53\)
default	\(5 x -\frac {16 \left (-\frac {1}{32} x^{3}+\frac {25}{16} x \right )}{\left (x^{2}+2\right )^{2}}-\frac {191 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{4}+\frac {-\frac {19}{8} x^{3}-\frac {21}{8} x}{\left (x^{2}+1\right )^{2}}+\frac {413 \arctan \left (x \right )}{8}\)	\(56\)

[In]

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

[Out]

5*x+(-15/8*x^7-289/8*x^5-139/2*x^3-71/2*x)/(x^4+3*x^2+2)^2+413/8*arctan(x)-191/4*arctan(1/2*x*2^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=\frac {40 \, x^{9} + 225 \, x^{7} + 231 \, x^{5} - 76 \, x^{3} - 382 \, \sqrt {2} {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 413 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )} \arctan \left (x\right ) - 124 \, x}{8 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} \]

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^3,x, algorithm="fricas")

[Out]

1/8*(40*x^9 + 225*x^7 + 231*x^5 - 76*x^3 - 382*sqrt(2)*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(1/2*sqrt(2)*
x) + 413*(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)*arctan(x) - 124*x)/(x^8 + 6*x^6 + 13*x^4 + 12*x^2 + 4)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=5 x + \frac {- 15 x^{7} - 289 x^{5} - 556 x^{3} - 284 x}{8 x^{8} + 48 x^{6} + 104 x^{4} + 96 x^{2} + 32} + \frac {413 \operatorname {atan}{\left (x \right )}}{8} - \frac {191 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{4} \]

[In]

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

[Out]

5*x + (-15*x**7 - 289*x**5 - 556*x**3 - 284*x)/(8*x**8 + 48*x**6 + 104*x**4 + 96*x**2 + 32) + 413*atan(x)/8 -
191*sqrt(2)*atan(sqrt(2)*x/2)/4

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=-\frac {191}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 5 \, x - \frac {15 \, x^{7} + 289 \, x^{5} + 556 \, x^{3} + 284 \, x}{8 \, {\left (x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4\right )}} + \frac {413}{8} \, \arctan \left (x\right ) \]

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^3,x, algorithm="maxima")

[Out]

-191/4*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x - 1/8*(15*x^7 + 289*x^5 + 556*x^3 + 284*x)/(x^8 + 6*x^6 + 13*x^4 +
12*x^2 + 4) + 413/8*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.71 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=-\frac {191}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 5 \, x - \frac {15 \, x^{7} + 289 \, x^{5} + 556 \, x^{3} + 284 \, x}{8 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}^{2}} + \frac {413}{8} \, \arctan \left (x\right ) \]

[In]

integrate(x^6*(5*x^6+3*x^4+x^2+4)/(x^4+3*x^2+2)^3,x, algorithm="giac")

[Out]

-191/4*sqrt(2)*arctan(1/2*sqrt(2)*x) + 5*x - 1/8*(15*x^7 + 289*x^5 + 556*x^3 + 284*x)/(x^4 + 3*x^2 + 2)^2 + 41
3/8*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {x^6 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^3} \, dx=5\,x+\frac {413\,\mathrm {atan}\left (x\right )}{8}-\frac {191\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{4}-\frac {\frac {15\,x^7}{8}+\frac {289\,x^5}{8}+\frac {139\,x^3}{2}+\frac {71\,x}{2}}{x^8+6\,x^6+13\,x^4+12\,x^2+4} \]

[In]

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

[Out]

5*x + (413*atan(x))/8 - (191*2^(1/2)*atan((2^(1/2)*x)/2))/4 - ((71*x)/2 + (139*x^3)/2 + (289*x^5)/8 + (15*x^7)
/8)/(12*x^2 + 13*x^4 + 6*x^6 + x^8 + 4)