\(\int \frac {x^9 (4+x^2+3 x^4+5 x^6)}{(3+2 x^2+x^4)^2} \, dx\) [100]

Optimal result

Integrand size = 31, antiderivative size = 86 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=19 x^2+\frac {19 x^4}{4}-\frac {17 x^6}{6}+\frac {5 x^8}{8}-\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {201 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {183}{4} \log \left (3+2 x^2+x^4\right ) \]

[Out]

19*x^2+19/4*x^4-17/6*x^6+5/8*x^8-25/8*(7*x^2+15)/(x^4+2*x^2+3)-183/4*ln(x^4+2*x^2+3)+201/16*arctan(1/2*(x^2+1)
*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1677, 1674, 1671, 648, 632, 210, 642} \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {201 \arctan \left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5 x^8}{8}-\frac {17 x^6}{6}+\frac {19 x^4}{4}+19 x^2-\frac {25 \left (7 x^2+15\right )}{8 \left (x^4+2 x^2+3\right )}-\frac {183}{4} \log \left (x^4+2 x^2+3\right ) \]

[In]

Int[(x^9*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

19*x^2 + (19*x^4)/4 - (17*x^6)/6 + (5*x^8)/8 - (25*(15 + 7*x^2))/(8*(3 + 2*x^2 + x^4)) + (201*ArcTan[(1 + x^2)
/Sqrt[2]])/(8*Sqrt[2]) - (183*Log[3 + 2*x^2 + x^4])/4

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

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

Rule 1671

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

Rule 1674

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1677

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4 \left (4+x+3 x^2+5 x^3\right )}{\left (3+2 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = -\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \text {Subst}\left (\int \frac {-150-400 x+200 x^2-56 x^4+40 x^5}{3+2 x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \text {Subst}\left (\int \left (304+152 x-136 x^2+40 x^3-\frac {6 (177+244 x)}{3+2 x+x^2}\right ) \, dx,x,x^2\right ) \\ & = 19 x^2+\frac {19 x^4}{4}-\frac {17 x^6}{6}+\frac {5 x^8}{8}-\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {3}{8} \text {Subst}\left (\int \frac {177+244 x}{3+2 x+x^2} \, dx,x,x^2\right ) \\ & = 19 x^2+\frac {19 x^4}{4}-\frac {17 x^6}{6}+\frac {5 x^8}{8}-\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {201}{8} \text {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right )-\frac {183}{4} \text {Subst}\left (\int \frac {2+2 x}{3+2 x+x^2} \, dx,x,x^2\right ) \\ & = 19 x^2+\frac {19 x^4}{4}-\frac {17 x^6}{6}+\frac {5 x^8}{8}-\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {183}{4} \log \left (3+2 x^2+x^4\right )-\frac {201}{4} \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right ) \\ & = 19 x^2+\frac {19 x^4}{4}-\frac {17 x^6}{6}+\frac {5 x^8}{8}-\frac {25 \left (15+7 x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {201 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {183}{4} \log \left (3+2 x^2+x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {1}{48} \left (912 x^2+228 x^4-136 x^6+30 x^8-\frac {150 \left (15+7 x^2\right )}{3+2 x^2+x^4}+603 \sqrt {2} \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )-2196 \log \left (3+2 x^2+x^4\right )\right ) \]

[In]

Integrate[(x^9*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

(912*x^2 + 228*x^4 - 136*x^6 + 30*x^8 - (150*(15 + 7*x^2))/(3 + 2*x^2 + x^4) + 603*Sqrt[2]*ArcTan[(1 + x^2)/Sq
rt[2]] - 2196*Log[3 + 2*x^2 + x^4])/48

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {5 x^{8}}{8}-\frac {17 x^{6}}{6}+\frac {19 x^{4}}{4}+19 x^{2}+\frac {-\frac {175 x^{2}}{8}-\frac {375}{8}}{x^{4}+2 x^{2}+3}-\frac {183 \ln \left (x^{4}+2 x^{2}+3\right )}{4}+\frac {201 \arctan \left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2}\right ) \sqrt {2}}{16}\)	\(71\)
default	\(\frac {5 x^{8}}{8}-\frac {17 x^{6}}{6}+\frac {19 x^{4}}{4}+19 x^{2}-\frac {\frac {175 x^{2}}{4}+\frac {375}{4}}{2 \left (x^{4}+2 x^{2}+3\right )}-\frac {183 \ln \left (x^{4}+2 x^{2}+3\right )}{4}+\frac {201 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}\)	\(74\)

[In]

int(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x,method=_RETURNVERBOSE)

[Out]

5/8*x^8-17/6*x^6+19/4*x^4+19*x^2+(-175/8*x^2-375/8)/(x^4+2*x^2+3)-183/4*ln(x^4+2*x^2+3)+201/16*arctan(1/2*(x^2
+1)*2^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {30 \, x^{12} - 76 \, x^{10} + 46 \, x^{8} + 960 \, x^{6} + 2508 \, x^{4} + 603 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + 1686 \, x^{2} - 2196 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) - 2250}{48 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

1/48*(30*x^12 - 76*x^10 + 46*x^8 + 960*x^6 + 2508*x^4 + 603*sqrt(2)*(x^4 + 2*x^2 + 3)*arctan(1/2*sqrt(2)*(x^2
+ 1)) + 1686*x^2 - 2196*(x^4 + 2*x^2 + 3)*log(x^4 + 2*x^2 + 3) - 2250)/(x^4 + 2*x^2 + 3)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5 x^{8}}{8} - \frac {17 x^{6}}{6} + \frac {19 x^{4}}{4} + 19 x^{2} + \frac {- 175 x^{2} - 375}{8 x^{4} + 16 x^{2} + 24} - \frac {183 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{4} + \frac {201 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \]

[In]

integrate(x**9*(5*x**6+3*x**4+x**2+4)/(x**4+2*x**2+3)**2,x)

[Out]

5*x**8/8 - 17*x**6/6 + 19*x**4/4 + 19*x**2 + (-175*x**2 - 375)/(8*x**4 + 16*x**2 + 24) - 183*log(x**4 + 2*x**2
+ 3)/4 + 201*sqrt(2)*atan(sqrt(2)*x**2/2 + sqrt(2)/2)/16

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5}{8} \, x^{8} - \frac {17}{6} \, x^{6} + \frac {19}{4} \, x^{4} + 19 \, x^{2} + \frac {201}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - \frac {25 \, {\left (7 \, x^{2} + 15\right )}}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac {183}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

5/8*x^8 - 17/6*x^6 + 19/4*x^4 + 19*x^2 + 201/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) - 25/8*(7*x^2 + 15)/(x^4
+ 2*x^2 + 3) - 183/4*log(x^4 + 2*x^2 + 3)

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5}{8} \, x^{8} - \frac {17}{6} \, x^{6} + \frac {19}{4} \, x^{4} + 19 \, x^{2} + \frac {201}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {366 \, x^{4} + 557 \, x^{2} + 723}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac {183}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]

[In]

integrate(x^9*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

5/8*x^8 - 17/6*x^6 + 19/4*x^4 + 19*x^2 + 201/16*sqrt(2)*arctan(1/2*sqrt(2)*(x^2 + 1)) + 1/8*(366*x^4 + 557*x^2
+ 723)/(x^4 + 2*x^2 + 3) - 183/4*log(x^4 + 2*x^2 + 3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {x^9 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {201\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16}-\frac {\frac {175\,x^2}{8}+\frac {375}{8}}{x^4+2\,x^2+3}-\frac {183\,\ln \left (x^4+2\,x^2+3\right )}{4}+19\,x^2+\frac {19\,x^4}{4}-\frac {17\,x^6}{6}+\frac {5\,x^8}{8} \]

[In]

int((x^9*(x^2 + 3*x^4 + 5*x^6 + 4))/(2*x^2 + x^4 + 3)^2,x)

[Out]

(201*2^(1/2)*atan(2^(1/2)/2 + (2^(1/2)*x^2)/2))/16 - ((175*x^2)/8 + 375/8)/(2*x^2 + x^4 + 3) - (183*log(2*x^2
+ x^4 + 3))/4 + 19*x^2 + (19*x^4)/4 - (17*x^6)/6 + (5*x^8)/8