\(\int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx\) [326]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {1+x^2+x^4}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1604} \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {x^4+x^2+1}}{x} \]

[In]

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

[Out]

Sqrt[1 + x^2 + x^4]/x

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+x^2+x^4}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {1+x^2+x^4}}{x} \]

[In]

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

[Out]

Sqrt[1 + x^2 + x^4]/x

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\sqrt {x^{4}+x^{2}+1}}{x}\)	\(15\)
trager	\(\frac {\sqrt {x^{4}+x^{2}+1}}{x}\)	\(15\)
risch	\(\frac {\sqrt {x^{4}+x^{2}+1}}{x}\)	\(15\)
elliptic	\(\frac {\sqrt {x^{4}+x^{2}+1}}{x}\)	\(15\)
gosper	\(\frac {\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}{\sqrt {x^{4}+x^{2}+1}\, x}\)	\(29\)
pseudoelliptic	\(\frac {\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}{\sqrt {x^{4}+x^{2}+1}\, x}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {x^{4} + x^{2} + 1}}{x} \]

[In]

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

[Out]

sqrt(x^4 + x^2 + 1)/x

Sympy [F]

\[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}{x^{2} \sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}\, dx \]

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)/(x**2*sqrt((x**2 - x + 1)*(x**2 + x + 1))), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {x^{2} + x + 1} \sqrt {x^{2} - x + 1}}{x} \]

[In]

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

[Out]

sqrt(x^2 + x + 1)*sqrt(x^2 - x + 1)/x

Giac [F]

\[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\int { \frac {x^{4} - 1}{\sqrt {x^{4} + x^{2} + 1} x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^4}{x^2 \sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {x^4+x^2+1}}{x} \]

[In]

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

[Out]

(x^2 + x^4 + 1)^(1/2)/x