\(\int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx\) [156]

Optimal result

Integrand size = 20, antiderivative size = 18 \[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\frac {2 \left (x^2+x^6\right )^{3/4}}{3 x^3} \]

[Out]

2/3*(x^6+x^2)^(3/4)/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1604} \[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\frac {2 \left (x^6+x^2\right )^{3/4}}{3 x^3} \]

[In]

Int[(-1 + x^4)/(x^2*(x^2 + x^6)^(1/4)),x]

[Out]

(2*(x^2 + x^6)^(3/4))/(3*x^3)

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 {2 \left (x^2+x^6\right )^{3/4}}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\frac {2 \left (x^2+x^6\right )^{3/4}}{3 x^3} \]

[In]

Integrate[(-1 + x^4)/(x^2*(x^2 + x^6)^(1/4)),x]

[Out]

(2*(x^2 + x^6)^(3/4))/(3*x^3)

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(\frac {2 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)
pseudoelliptic	\(\frac {2 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(17\)
gosper	\(\frac {\frac {2 x^{4}}{3}+\frac {2}{3}}{\left (x^{6}+x^{2}\right )^{\frac {1}{4}} x}\)	\(20\)
risch	\(\frac {\frac {2 x^{4}}{3}+\frac {2}{3}}{x \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\)	\(22\)
meijerg	\(\frac {2 \operatorname {hypergeom}\left (\left [-\frac {3}{8}, \frac {1}{4}\right ], \left [\frac {5}{8}\right ], -x^{4}\right )}{3 x^{\frac {3}{2}}}+\frac {2 x^{\frac {5}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{8}\right ], \left [\frac {13}{8}\right ], -x^{4}\right )}{5}\)	\(34\)

[In]

int((x^4-1)/x^2/(x^6+x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

2/3*(x^6+x^2)^(3/4)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\frac {2 \, {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

integrate((x^4-1)/x^2/(x^6+x^2)^(1/4),x, algorithm="fricas")

[Out]

2/3*(x^6 + x^2)^(3/4)/x^3

Sympy [F]

\[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}{x^{2} \sqrt [4]{x^{2} \left (x^{4} + 1\right )}}\, dx \]

[In]

integrate((x**4-1)/x**2/(x**6+x**2)**(1/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2}} \,d x } \]

[In]

integrate((x^4-1)/x^2/(x^6+x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^4 - 1)/((x^6 + x^2)^(1/4)*x^2), x)

Giac [F]

\[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2}} \,d x } \]

[In]

integrate((x^4-1)/x^2/(x^6+x^2)^(1/4),x, algorithm="giac")

[Out]

integrate((x^4 - 1)/((x^6 + x^2)^(1/4)*x^2), x)

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-1+x^4}{x^2 \sqrt [4]{x^2+x^6}} \, dx=\frac {2\,{\left (x^6+x^2\right )}^{3/4}}{3\,x^3} \]

[In]

int((x^4 - 1)/(x^2*(x^2 + x^6)^(1/4)),x)

[Out]

(2*(x^2 + x^6)^(3/4))/(3*x^3)