\(\int \frac {-1+x^4}{x^6 (1+x^4)^{3/4}} \, dx\) [217]

Optimal result

Integrand size = 18, antiderivative size = 23 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\left (1-9 x^4\right ) \sqrt [4]{1+x^4}}{5 x^5} \]

[Out]

1/5*(-9*x^4+1)*(x^4+1)^(1/4)/x^5

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {464, 270} \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{x^4+1}}{5 x^5}-\frac {9 \sqrt [4]{x^4+1}}{5 x} \]

[In]

Int[(-1 + x^4)/(x^6*(1 + x^4)^(3/4)),x]

[Out]

(1 + x^4)^(1/4)/(5*x^5) - (9*(1 + x^4)^(1/4))/(5*x)

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 {\sqrt [4]{1+x^4}}{5 x^5}+\frac {9}{5} \int \frac {1}{x^2 \left (1+x^4\right )^{3/4}} \, dx \\ & = \frac {\sqrt [4]{1+x^4}}{5 x^5}-\frac {9 \sqrt [4]{1+x^4}}{5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\left (1-9 x^4\right ) \sqrt [4]{1+x^4}}{5 x^5} \]

[In]

Integrate[(-1 + x^4)/(x^6*(1 + x^4)^(3/4)),x]

[Out]

((1 - 9*x^4)*(1 + x^4)^(1/4))/(5*x^5)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
gosper	\(-\frac {\left (9 x^{4}-1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{5 x^{5}}\)	\(20\)
trager	\(-\frac {\left (9 x^{4}-1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{5 x^{5}}\)	\(20\)
pseudoelliptic	\(-\frac {\left (9 x^{4}-1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{5 x^{5}}\)	\(20\)
risch	\(-\frac {9 x^{8}+8 x^{4}-1}{5 \left (x^{4}+1\right )^{\frac {3}{4}} x^{5}}\)	\(25\)
meijerg	\(-\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}+\frac {\left (-4 x^{4}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{5 x^{5}}\)	\(33\)

[In]

int((x^4-1)/x^6/(x^4+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-1/5*(9*x^4-1)*(x^4+1)^(1/4)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=-\frac {{\left (9 \, x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \]

[In]

integrate((x^4-1)/x^6/(x^4+1)^(3/4),x, algorithm="fricas")

[Out]

-1/5*(9*x^4 - 1)*(x^4 + 1)^(1/4)/x^5

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (20) = 40\).

Time = 0.88 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{1 + \frac {1}{x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt [4]{x^{4} + 1} \Gamma \left (- \frac {5}{4}\right )}{4 x \Gamma \left (\frac {3}{4}\right )} + \frac {\sqrt [4]{x^{4} + 1} \Gamma \left (- \frac {5}{4}\right )}{16 x^{5} \Gamma \left (\frac {3}{4}\right )} \]

[In]

integrate((x**4-1)/x**6/(x**4+1)**(3/4),x)

[Out]

(1 + x**(-4))**(1/4)*gamma(-1/4)/(4*gamma(3/4)) - (x**4 + 1)**(1/4)*gamma(-5/4)/(4*x*gamma(3/4)) + (x**4 + 1)*
*(1/4)*gamma(-5/4)/(16*x**5*gamma(3/4))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=-\frac {2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} + \frac {{\left (x^{4} + 1\right )}^{\frac {5}{4}}}{5 \, x^{5}} \]

[In]

integrate((x^4-1)/x^6/(x^4+1)^(3/4),x, algorithm="maxima")

[Out]

-2*(x^4 + 1)^(1/4)/x + 1/5*(x^4 + 1)^(5/4)/x^5

Giac [F]

\[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\int { \frac {x^{4} - 1}{{\left (x^{4} + 1\right )}^{\frac {3}{4}} x^{6}} \,d x } \]

[In]

integrate((x^4-1)/x^6/(x^4+1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^4 - 1)/((x^4 + 1)^(3/4)*x^6), x)

Mupad [B] (verification not implemented)

Time = 5.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1+x^4}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {{\left (x^4+1\right )}^{1/4}-9\,x^4\,{\left (x^4+1\right )}^{1/4}}{5\,x^5} \]

[In]

int((x^4 - 1)/(x^6*(x^4 + 1)^(3/4)),x)

[Out]

((x^4 + 1)^(1/4) - 9*x^4*(x^4 + 1)^(1/4))/(5*x^5)