\(\int \frac {(1+x^4) \sqrt [4]{-x^2+x^6}}{x^4} \, dx\) [186]

Optimal result

Integrand size = 22, antiderivative size = 20 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (-x^2+x^6\right )^{5/4}}{5 x^5} \]

[Out]

2/5*(x^6-x^2)^(5/4)/x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1604} \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (x^6-x^2\right )^{5/4}}{5 x^5} \]

[In]

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

[Out]

(2*(-x^2 + x^6)^(5/4))/(5*x^5)

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 )^{5/4}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \left (x^2 \left (-1+x^4\right )\right )^{5/4}}{5 x^5} \]

[In]

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

[Out]

(2*(x^2*(-1 + x^4))^(5/4))/(5*x^5)

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
trager	\(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(22\)
pseudoelliptic	\(\frac {2 \left (x^{4}-1\right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(22\)
gosper	\(\frac {2 \left (x^{2}+1\right ) \left (x -1\right ) \left (1+x \right ) \left (x^{6}-x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(28\)
risch	\(\frac {2 \left (x^{2} \left (x^{4}-1\right )\right )^{\frac {1}{4}} \left (x^{8}-2 x^{4}+1\right )}{5 x^{3} \left (x^{4}-1\right )}\)	\(34\)
meijerg	\(-\frac {2 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{8}, -\frac {1}{4}\right ], \left [\frac {3}{8}\right ], x^{4}\right )}{5 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {5}{2}}}+\frac {2 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{8}\right ], \left [\frac {11}{8}\right ], x^{4}\right )}{3 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2 \, {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}}{5 \, x^{3}} \]

[In]

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

[Out]

2/5*(x^6 - x^2)^(1/4)*(x^4 - 1)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^6}}{x^4} \, dx=\frac {2\,\left (x^4-1\right )\,{\left (x^6-x^2\right )}^{1/4}}{5\,x^3} \]

[In]

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

[Out]

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