\(\int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx\) [1033]

Optimal result

Integrand size = 22, antiderivative size = 12 \[ \int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=\frac {x^2}{2 \sqrt {a}} \]

[Out]

1/2*x^2/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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2, 12, 30} \[ \int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=\frac {x^2}{2 \sqrt {a}} \]

[In]

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

[Out]

x^2/(2*Sqrt[a])

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 {x}{\sqrt {a}} \, dx \\ & = \frac {\int x \, dx}{\sqrt {a}} \\ & = \frac {x^2}{2 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

x^2/(2*Sqrt[a])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {x^{2}}{2 \sqrt {a}}\)	\(9\)
default	\(\frac {x^{2}}{2 \sqrt {a}}\)	\(9\)
norman	\(\frac {x^{2}}{2 \sqrt {a}}\)	\(9\)
parallelrisch	\(\frac {x^{2}}{2 \sqrt {a}}\)	\(9\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=\frac {x^{2}}{2 \, \sqrt {a}} \]

[In]

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

[Out]

1/2*x^2/sqrt(a)

Sympy [A] (verification not implemented)

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

[In]

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

[Out]

x**2/(2*sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=\frac {x^{2}}{2 \, \sqrt {a}} \]

[In]

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

[Out]

1/2*x^2/sqrt(a)

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

1/2*x^2/sqrt(a)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{\sqrt {a+(2+2 c-2 (1+c)) x^4}} \, dx=\frac {x^2}{2\,\sqrt {a}} \]

[In]

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

[Out]

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