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

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

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Rubi [A]  time = 0.07, antiderivative size = 56, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 ) \end {gather*}

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 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])

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 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

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*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 1.02 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.10, size = 69, normalized size = 1.23 \begin {gather*} \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 \relax (x) - 474 \, x}{6 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} \end {gather*}

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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giac [A]  time = 0.31, size = 48, normalized size = 0.86 \begin {gather*} \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 \relax (x) \end {gather*}

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)

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maple [A]  time = 0.01, size = 46, normalized size = 0.82 \begin {gather*} \frac {5 x^{3}}{3}-27 x +\frac {x}{2 x^{2}+2}-\frac {26 x}{x^{2}+2}+\frac {13 \arctan \relax (x )}{2}+33 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right ) \end {gather*}

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

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maxima [A]  time = 1.64, size = 48, normalized size = 0.86 \begin {gather*} \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 \relax (x) \end {gather*}

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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mupad [B]  time = 0.92, size = 48, normalized size = 0.86 \begin {gather*} \frac {13\,\mathrm {atan}\relax (x)}{2}-27\,x+33\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )-\frac {\frac {51\,x^3}{2}+25\,x}{x^4+3\,x^2+2}+\frac {5\,x^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*atan(x))/2 - 27*x + 33*2^(1/2)*atan((2^(1/2)*x)/2) - (25*x + (51*x^3)/2)/(3*x^2 + x^4 + 2) + (5*x^3)/3

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sympy [A]  time = 0.21, size = 54, normalized size = 0.96 \begin {gather*} \frac {5 x^{3}}{3} - 27 x + \frac {- 51 x^{3} - 50 x}{2 x^{4} + 6 x^{2} + 4} + \frac {13 \operatorname {atan}{\relax (x )}}{2} + 33 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )} \end {gather*}

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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