\(\int \frac {(1+x^8) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx\) [202]

Optimal result

Integrand size = 23, antiderivative size = 21 \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {\left (-1-2 x^4+x^8\right )^{3/2}}{6 x^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1604} \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {\left (x^8-2 x^4-1\right )^{3/2}}{6 x^6} \]

[In]

Int[((1 + x^8)*Sqrt[-1 - 2*x^4 + x^8])/x^7,x]

[Out]

(-1 - 2*x^4 + x^8)^(3/2)/(6*x^6)

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 {\left (-1-2 x^4+x^8\right )^{3/2}}{6 x^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {\left (-1-2 x^4+x^8\right )^{3/2}}{6 x^6} \]

[In]

Integrate[((1 + x^8)*Sqrt[-1 - 2*x^4 + x^8])/x^7,x]

[Out]

(-1 - 2*x^4 + x^8)^(3/2)/(6*x^6)

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
gosper	\(\frac {\left (x^{8}-2 x^{4}-1\right )^{\frac {3}{2}}}{6 x^{6}}\)	\(18\)
trager	\(\frac {\left (x^{8}-2 x^{4}-1\right )^{\frac {3}{2}}}{6 x^{6}}\)	\(18\)
pseudoelliptic	\(\frac {\left (x^{8}-2 x^{4}-1\right ) \sqrt {\frac {x^{8}-2 x^{4}-1}{x^{2}}}}{6 x^{5}}\)	\(32\)
risch	\(\frac {x^{16}-4 x^{12}+2 x^{8}+4 x^{4}+1}{6 x^{6} \sqrt {x^{8}-2 x^{4}-1}}\)	\(38\)

[In]

int((x^8+1)*(x^8-2*x^4-1)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {{\left (x^{8} - 2 \, x^{4} - 1\right )}^{\frac {3}{2}}}{6 \, x^{6}} \]

[In]

integrate((x^8+1)*(x^8-2*x^4-1)^(1/2)/x^7,x, algorithm="fricas")

[Out]

1/6*(x^8 - 2*x^4 - 1)^(3/2)/x^6

Sympy [F]

\[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\int \frac {\left (x^{8} + 1\right ) \sqrt {x^{8} - 2 x^{4} - 1}}{x^{7}}\, dx \]

[In]

integrate((x**8+1)*(x**8-2*x**4-1)**(1/2)/x**7,x)

[Out]

Integral((x**8 + 1)*sqrt(x**8 - 2*x**4 - 1)/x**7, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {{\left (x^{8} - 2 \, x^{4} - 1\right )}^{\frac {3}{2}}}{6 \, x^{6}} \]

[In]

integrate((x^8+1)*(x^8-2*x^4-1)^(1/2)/x^7,x, algorithm="maxima")

[Out]

1/6*(x^8 - 2*x^4 - 1)^(3/2)/x^6

Giac [F]

\[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\int { \frac {\sqrt {x^{8} - 2 \, x^{4} - 1} {\left (x^{8} + 1\right )}}{x^{7}} \,d x } \]

[In]

integrate((x^8+1)*(x^8-2*x^4-1)^(1/2)/x^7,x, algorithm="giac")

[Out]

integrate(sqrt(x^8 - 2*x^4 - 1)*(x^8 + 1)/x^7, x)

Mupad [B] (verification not implemented)

Time = 5.71 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {\left (1+x^8\right ) \sqrt {-1-2 x^4+x^8}}{x^7} \, dx=\frac {{\left (x^8-2\,x^4-1\right )}^{3/2}}{6\,x^6} \]

[In]

int(((x^8 + 1)*(x^8 - 2*x^4 - 1)^(1/2))/x^7,x)

[Out]

(x^8 - 2*x^4 - 1)^(3/2)/(6*x^6)