3.84 $$\int \frac{x^4 (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx$$

Optimal. Leaf size=56 $\frac{5 x^3}{3}-\frac{\left (51 x^2+50\right ) x}{2 \left (x^4+3 x^2+2\right )}-27 x+\frac{13}{2} \tan ^{-1}(x)+33 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )$

[Out]

-27*x + (5*x^3)/3 - (x*(50 + 51*x^2))/(2*(2 + 3*x^2 + x^4)) + (13*ArcTan[x])/2 + 33*Sqrt[2]*ArcTan[x/Sqrt[2]]

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Rubi [A]  time = 0.0727921, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.129, Rules used = {1668, 1676, 1166, 203} $\frac{5 x^3}{3}-\frac{\left (51 x^2+50\right ) x}{2 \left (x^4+3 x^2+2\right )}-27 x+\frac{13}{2} \tan ^{-1}(x)+33 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-27*x + (5*x^3)/3 - (x*(50 + 51*x^2))/(2*(2 + 3*x^2 + x^4)) + (13*ArcTan[x])/2 + 33*Sqrt[2]*ArcTan[x/Sqrt[2]]

Rule 1668

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 1676

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 1166

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 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{x^4 \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx &=-\frac{x \left (50+51 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \frac{-100-6 x^2+48 x^4-20 x^6}{2+3 x^2+x^4} \, dx\\ &=-\frac{x \left (50+51 x^2\right )}{2 \left (2+3 x^2+x^4\right )}-\frac{1}{4} \int \left (108-20 x^2-\frac{2 \left (158+145 x^2\right )}{2+3 x^2+x^4}\right ) \, dx\\ &=-27 x+\frac{5 x^3}{3}-\frac{x \left (50+51 x^2\right )}{2 \left (2+3 x^2+x^4\right )}+\frac{1}{2} \int \frac{158+145 x^2}{2+3 x^2+x^4} \, dx\\ &=-27 x+\frac{5 x^3}{3}-\frac{x \left (50+51 x^2\right )}{2 \left (2+3 x^2+x^4\right )}+\frac{13}{2} \int \frac{1}{1+x^2} \, dx+66 \int \frac{1}{2+x^2} \, dx\\ &=-27 x+\frac{5 x^3}{3}-\frac{x \left (50+51 x^2\right )}{2 \left (2+3 x^2+x^4\right )}+\frac{13}{2} \tan ^{-1}(x)+33 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0427317, size = 57, normalized size = 1.02 $\frac{5 x^3}{3}+\frac{-51 x^3-50 x}{2 \left (x^4+3 x^2+2\right )}-27 x+\frac{13}{2} \tan ^{-1}(x)+33 \sqrt{2} \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-27*x + (5*x^3)/3 + (-50*x - 51*x^3)/(2*(2 + 3*x^2 + x^4)) + (13*ArcTan[x])/2 + 33*Sqrt[2]*ArcTan[x/Sqrt[2]]

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Maple [A]  time = 0.01, size = 46, normalized size = 0.8 \begin{align*}{\frac{5\,{x}^{3}}{3}}-27\,x-26\,{\frac{x}{{x}^{2}+2}}+33\,\arctan \left ( 1/2\,x\sqrt{2} \right ) \sqrt{2}+{\frac{x}{2\,{x}^{2}+2}}+{\frac{13\,\arctan \left ( x \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*x^3-27*x-26*x/(x^2+2)+33*arctan(1/2*x*2^(1/2))*2^(1/2)+1/2*x/(x^2+1)+13/2*arctan(x)

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Maxima [A]  time = 1.48765, size = 65, normalized size = 1.16 \begin{align*} \frac{5}{3} \, x^{3} + 33 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 27 \, x - \frac{51 \, x^{3} + 50 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + \frac{13}{2} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*x^3 + 33*sqrt(2)*arctan(1/2*sqrt(2)*x) - 27*x - 1/2*(51*x^3 + 50*x)/(x^4 + 3*x^2 + 2) + 13/2*arctan(x)

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Fricas [A]  time = 1.90295, size = 198, normalized size = 3.54 \begin{align*} \frac{10 \, x^{7} - 132 \, x^{5} - 619 \, x^{3} + 198 \, \sqrt{2}{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) + 39 \,{\left (x^{4} + 3 \, x^{2} + 2\right )} \arctan \left (x\right ) - 474 \, x}{6 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(10*x^7 - 132*x^5 - 619*x^3 + 198*sqrt(2)*(x^4 + 3*x^2 + 2)*arctan(1/2*sqrt(2)*x) + 39*(x^4 + 3*x^2 + 2)*a
rctan(x) - 474*x)/(x^4 + 3*x^2 + 2)

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Sympy [A]  time = 0.189747, size = 53, normalized size = 0.95 \begin{align*} \frac{5 x^{3}}{3} - 27 x - \frac{51 x^{3} + 50 x}{2 x^{4} + 6 x^{2} + 4} + \frac{13 \operatorname{atan}{\left (x \right )}}{2} + 33 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x**3/3 - 27*x - (51*x**3 + 50*x)/(2*x**4 + 6*x**2 + 4) + 13*atan(x)/2 + 33*sqrt(2)*atan(sqrt(2)*x/2)

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Giac [A]  time = 1.14651, size = 65, normalized size = 1.16 \begin{align*} \frac{5}{3} \, x^{3} + 33 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} x\right ) - 27 \, x - \frac{51 \, x^{3} + 50 \, x}{2 \,{\left (x^{4} + 3 \, x^{2} + 2\right )}} + \frac{13}{2} \, \arctan \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*x^3 + 33*sqrt(2)*arctan(1/2*sqrt(2)*x) - 27*x - 1/2*(51*x^3 + 50*x)/(x^4 + 3*x^2 + 2) + 13/2*arctan(x)