3.4.50 \(\int \frac {(-1+x^4) (1+x^4)}{(1+x^2+x^4)^{5/2}} \, dx\)

Optimal. Leaf size=29 \[ -\frac {x \left (3 x^4+2 x^2+3\right )}{3 \left (x^4+x^2+1\right )^{3/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1678, 1588} \begin {gather*} \frac {x^3}{3 \left (x^4+x^2+1\right )^{3/2}}-\frac {x}{\sqrt {x^4+x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)*(1 + x^4))/(1 + x^2 + x^4)^(5/2),x]

[Out]

x^3/(3*(1 + x^2 + x^4)^(3/2)) - x/Sqrt[1 + x^2 + x^4]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 1678

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*} \int \frac {\left (-1+x^4\right ) \left (1+x^4\right )}{\left (1+x^2+x^4\right )^{5/2}} \, dx &=\frac {x^3}{3 \left (1+x^2+x^4\right )^{3/2}}+\frac {1}{9} \int \frac {-9+9 x^4}{\left (1+x^2+x^4\right )^{3/2}} \, dx\\ &=\frac {x^3}{3 \left (1+x^2+x^4\right )^{3/2}}-\frac {x}{\sqrt {1+x^2+x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 29, normalized size = 1.00 \begin {gather*} -\frac {x \left (3 x^4+2 x^2+3\right )}{3 \left (x^4+x^2+1\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^4)*(1 + x^4))/(1 + x^2 + x^4)^(5/2),x]

[Out]

-1/3*(x*(3 + 2*x^2 + 3*x^4))/(1 + x^2 + x^4)^(3/2)

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IntegrateAlgebraic [A]  time = 0.37, size = 29, normalized size = 1.00 \begin {gather*} -\frac {x \left (3+2 x^2+3 x^4\right )}{3 \left (1+x^2+x^4\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)*(1 + x^4))/(1 + x^2 + x^4)^(5/2),x]

[Out]

-1/3*(x*(3 + 2*x^2 + 3*x^4))/(1 + x^2 + x^4)^(3/2)

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fricas [A]  time = 0.46, size = 48, normalized size = 1.66 \begin {gather*} -\frac {{\left (3 \, x^{5} + 2 \, x^{3} + 3 \, x\right )} \sqrt {x^{4} + x^{2} + 1}}{3 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)/(x^4+x^2+1)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(3*x^5 + 2*x^3 + 3*x)*sqrt(x^4 + x^2 + 1)/(x^8 + 2*x^6 + 3*x^4 + 2*x^2 + 1)

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giac [A]  time = 0.42, size = 26, normalized size = 0.90 \begin {gather*} -\frac {{\left ({\left (3 \, x^{2} + 2\right )} x^{2} + 3\right )} x}{3 \, {\left (x^{4} + x^{2} + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)/(x^4+x^2+1)^(5/2),x, algorithm="giac")

[Out]

-1/3*((3*x^2 + 2)*x^2 + 3)*x/(x^4 + x^2 + 1)^(3/2)

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maple [A]  time = 0.09, size = 26, normalized size = 0.90

method	result	size
trager	\(-\frac {x \left (3 x^{4}+2 x^{2}+3\right )}{3 \left (x^{4}+x^{2}+1\right )^{\frac {3}{2}}}\)	\(26\)
risch	\(-\frac {x \left (3 x^{4}+2 x^{2}+3\right )}{3 \left (x^{4}+x^{2}+1\right )^{\frac {3}{2}}}\)	\(26\)
gosper	\(-\frac {\left (3 x^{4}+2 x^{2}+3\right ) x \left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}{3 \left (x^{4}+x^{2}+1\right )^{\frac {5}{2}}}\)	\(40\)
elliptic	\(\frac {\left (\frac {\sqrt {2}\, x^{3}}{3 \left (x^{4}+x^{2}+1\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, x}{\sqrt {x^{4}+x^{2}+1}}\right ) \sqrt {2}}{2}\)	\(41\)
default	\(\frac {\frac {2}{9} x^{3}+\frac {1}{9} x}{\left (x^{4}+x^{2}+1\right )^{\frac {3}{2}}}-\frac {2 \left (\frac {7}{27} x^{3}+\frac {11}{27} x \right )}{\sqrt {x^{4}+x^{2}+1}}-\frac {\frac {1}{9} x -\frac {1}{9} x^{3}}{\left (x^{4}+x^{2}+1\right )^{\frac {3}{2}}}+\frac {\frac {14}{27} x^{3}-\frac {5}{27} x}{\sqrt {x^{4}+x^{2}+1}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)*(x^4+1)/(x^4+x^2+1)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*x*(3*x^4+2*x^2+3)/(x^4+x^2+1)^(3/2)

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maxima [B]  time = 0.51, size = 56, normalized size = 1.93 \begin {gather*} -\frac {{\left (3 \, x^{5} + 2 \, x^{3} + 3 \, x\right )} \sqrt {x^{2} + x + 1} \sqrt {x^{2} - x + 1}}{3 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)*(x^4+1)/(x^4+x^2+1)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*x^5 + 2*x^3 + 3*x)*sqrt(x^2 + x + 1)*sqrt(x^2 - x + 1)/(x^8 + 2*x^6 + 3*x^4 + 2*x^2 + 1)

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mupad [B]  time = 0.13, size = 25, normalized size = 0.86 \begin {gather*} -\frac {x\,\left (3\,x^4+2\,x^2+3\right )}{3\,{\left (x^4+x^2+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)*(x^4 + 1))/(x^2 + x^4 + 1)^(5/2),x)

[Out]

-(x*(2*x^2 + 3*x^4 + 3))/(3*(x^2 + x^4 + 1)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}{\left (\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)*(x**4+1)/(x**4+x**2+1)**(5/2),x)

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)/((x**2 - x + 1)*(x**2 + x + 1))**(5/2), x)

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