\(\int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx\) [1038]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 12 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \sqrt {a} x^3} \]

[Out]

-1/3/x^3/a^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2, 12, 30} \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \sqrt {a} x^3} \]

[In]

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

[Out]

-1/3*1/(Sqrt[a]*x^3)

Rule 2

Int[(u_.)*((a_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*a^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[b, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {a} x^4} \, dx \\ & = \frac {\int \frac {1}{x^4} \, dx}{\sqrt {a}} \\ & = -\frac {1}{3 \sqrt {a} x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \sqrt {a} x^3} \]

[In]

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

[Out]

-1/3*1/(Sqrt[a]*x^3)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method result size
gosper \(-\frac {1}{3 x^{3} \sqrt {a}}\) \(9\)
default \(-\frac {1}{3 x^{3} \sqrt {a}}\) \(9\)
norman \(-\frac {1}{3 x^{3} \sqrt {a}}\) \(9\)
parallelrisch \(-\frac {1}{3 x^{3} \sqrt {a}}\) \(9\)

[In]

int(1/x^4/a^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/x^3/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \, \sqrt {a} x^{3}} \]

[In]

integrate(1/x^4/a^(1/2),x, algorithm="fricas")

[Out]

-1/3/(sqrt(a)*x^3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=- \frac {1}{3 \sqrt {a} x^{3}} \]

[In]

integrate(1/x**4/a**(1/2),x)

[Out]

-1/(3*sqrt(a)*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \, \sqrt {a} x^{3}} \]

[In]

integrate(1/x^4/a^(1/2),x, algorithm="maxima")

[Out]

-1/3/(sqrt(a)*x^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3 \, \sqrt {a} x^{3}} \]

[In]

integrate(1/x^4/a^(1/2),x, algorithm="giac")

[Out]

-1/3/(sqrt(a)*x^3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=-\frac {1}{3\,\sqrt {a}\,x^3} \]

[In]

int(1/(a^(1/2)*x^4),x)

[Out]

-1/(3*a^(1/2)*x^3)