\(\int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx\) [229]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=\frac {\left (-1+x^4\right )^{3/4} \left (-3+10 x^4\right )}{21 x^7} \]

[Out]

1/21*(x^4-1)^(3/4)*(10*x^4-3)/x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {464, 270} \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=\frac {10 \left (x^4-1\right )^{3/4}}{21 x^3}-\frac {\left (x^4-1\right )^{3/4}}{7 x^7} \]

[In]

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

[Out]

-1/7*(-1 + x^4)^(3/4)/x^7 + (10*(-1 + x^4)^(3/4))/(21*x^3)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {10}{7} \int \frac {1}{x^4 \sqrt [4]{-1+x^4}} \, dx \\ & = -\frac {\left (-1+x^4\right )^{3/4}}{7 x^7}+\frac {10 \left (-1+x^4\right )^{3/4}}{21 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=\frac {\left (-1+x^4\right )^{3/4} \left (-3+10 x^4\right )}{21 x^7} \]

[In]

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

[Out]

((-1 + x^4)^(3/4)*(-3 + 10*x^4))/(21*x^7)

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
trager	\(\frac {\left (x^{4}-1\right )^{\frac {3}{4}} \left (10 x^{4}-3\right )}{21 x^{7}}\)	\(20\)
pseudoelliptic	\(\frac {\left (x^{4}-1\right )^{\frac {3}{4}} \left (10 x^{4}-3\right )}{21 x^{7}}\)	\(20\)
risch	\(\frac {10 x^{8}-13 x^{4}+3}{21 x^{7} \left (x^{4}-1\right )^{\frac {1}{4}}}\)	\(25\)
gosper	\(\frac {\left (x^{2}+1\right ) \left (x -1\right ) \left (1+x \right ) \left (10 x^{4}-3\right )}{21 x^{7} \left (x^{4}-1\right )^{\frac {1}{4}}}\)	\(31\)
meijerg	\(\frac {{\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \left (1+\frac {4 x^{4}}{3}\right ) \left (-x^{4}+1\right )^{\frac {3}{4}}}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{7}}-\frac {2 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \left (-x^{4}+1\right )^{\frac {3}{4}}}{3 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{3}}\)	\(73\)

[In]

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

[Out]

1/21*(x^4-1)^(3/4)*(10*x^4-3)/x^7

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=\frac {{\left (10 \, x^{4} - 3\right )} {\left (x^{4} - 1\right )}^{\frac {3}{4}}}{21 \, x^{7}} \]

[In]

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

[Out]

1/21*(10*x^4 - 3)*(x^4 - 1)^(3/4)/x^7

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 190, normalized size of antiderivative = 8.26 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=2 \left (\begin {cases} - \frac {\left (-1 + \frac {1}{x^{4}}\right )^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} \Gamma \left (- \frac {3}{4}\right )}{4 \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > 1 \\- \frac {\left (1 - \frac {1}{x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases}\right ) - \begin {cases} - \frac {\left (-1 + \frac {1}{x^{4}}\right )^{\frac {3}{4}} e^{- \frac {i \pi }{4}} \Gamma \left (- \frac {7}{4}\right )}{4 \Gamma \left (\frac {1}{4}\right )} - \frac {3 \left (-1 + \frac {1}{x^{4}}\right )^{\frac {3}{4}} e^{- \frac {i \pi }{4}} \Gamma \left (- \frac {7}{4}\right )}{16 x^{4} \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > 1 \\\frac {\left (1 - \frac {1}{x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{4 \Gamma \left (\frac {1}{4}\right )} + \frac {3 \left (1 - \frac {1}{x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 x^{4} \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

2*Piecewise((-(-1 + x**(-4))**(3/4)*exp(3*I*pi/4)*gamma(-3/4)/(4*gamma(1/4)), 1/Abs(x**4) > 1), (-(1 - 1/x**4)
**(3/4)*gamma(-3/4)/(4*gamma(1/4)), True)) - Piecewise((-(-1 + x**(-4))**(3/4)*exp(-I*pi/4)*gamma(-7/4)/(4*gam
ma(1/4)) - 3*(-1 + x**(-4))**(3/4)*exp(-I*pi/4)*gamma(-7/4)/(16*x**4*gamma(1/4)), 1/Abs(x**4) > 1), ((1 - 1/x*
*4)**(3/4)*gamma(-7/4)/(4*gamma(1/4)) + 3*(1 - 1/x**4)**(3/4)*gamma(-7/4)/(16*x**4*gamma(1/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=\frac {{\left (x^{4} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} + \frac {{\left (x^{4} - 1\right )}^{\frac {7}{4}}}{7 \, x^{7}} \]

[In]

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

[Out]

1/3*(x^4 - 1)^(3/4)/x^3 + 1/7*(x^4 - 1)^(7/4)/x^7

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.57 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-1+2 x^4}{x^8 \sqrt [4]{-1+x^4}} \, dx=-\frac {3\,{\left (x^4-1\right )}^{3/4}-10\,x^4\,{\left (x^4-1\right )}^{3/4}}{21\,x^7} \]

[In]

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

[Out]

-(3*(x^4 - 1)^(3/4) - 10*x^4*(x^4 - 1)^(3/4))/(21*x^7)