\(\int \frac {1}{x^6 (1+x^4)^{3/4}} \, dx\) [216]

Optimal result

Integrand size = 13, antiderivative size = 23 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{1+x^4} \left (-1+4 x^4\right )}{5 x^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {277, 270} \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {4 \sqrt [4]{x^4+1}}{5 x}-\frac {\sqrt [4]{x^4+1}}{5 x^5} \]

[In]

Int[1/(x^6*(1 + x^4)^(3/4)),x]

[Out]

-1/5*(1 + x^4)^(1/4)/x^5 + (4*(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 277

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{1+x^4} \left (-1+4 x^4\right )}{5 x^5} \]

[In]

Integrate[1/(x^6*(1 + x^4)^(3/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {{\left (4 \, x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \]

[In]

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

[Out]

1/5*(4*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. 48 vs. \(2 (19) = 38\).

Time = 0.46 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{x^{4} + 1} \Gamma \left (- \frac {5}{4}\right )}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt [4]{x^{4} + 1} \Gamma \left (- \frac {5}{4}\right )}{16 x^{5} \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

(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 = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x} - \frac {{\left (x^{4} + 1\right )}^{\frac {5}{4}}}{5 \, x^{5}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(1/((x^4 + 1)^(3/4)*x^6), x)

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^6 \left (1+x^4\right )^{3/4}} \, dx=-\frac {{\left (x^4+1\right )}^{1/4}-4\,x^4\,{\left (x^4+1\right )}^{1/4}}{5\,x^5} \]

[In]

int(1/(x^6*(x^4 + 1)^(3/4)),x)

[Out]

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