\(\int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx\) [1027]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3 x \sqrt {c x^4}} \]

[Out]

-1/3/x/(c*x^4)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1, 15, 30} \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3 x \sqrt {c x^4}} \]

[In]

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

[Out]

-1/3*1/(x*Sqrt[c*x^4])

Rule 1

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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}{x^2 \sqrt {c x^4}} \, dx \\ & = \frac {x^2 \int \frac {1}{x^4} \, dx}{\sqrt {c x^4}} \\ & = -\frac {1}{3 x \sqrt {c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3 x \sqrt {c x^4}} \]

[In]

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

[Out]

-1/3*1/(x*Sqrt[c*x^4])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
gosper	\(-\frac {1}{3 x \sqrt {c \,x^{4}}}\)	\(13\)
default	\(-\frac {1}{3 x \sqrt {c \,x^{4}}}\)	\(13\)
risch	\(-\frac {1}{3 x \sqrt {c \,x^{4}}}\)	\(13\)
trager	\(\frac {\left (x -1\right ) \left (x^{2}+x +1\right ) \sqrt {c \,x^{4}}}{3 c \,x^{5}}\)	\(25\)

[In]

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

[Out]

-1/3/x/(c*x^4)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {\sqrt {c x^{4}}}{3 \, c x^{5}} \]

[In]

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

[Out]

-1/3*sqrt(c*x^4)/(c*x^5)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=- \frac {1}{3 x \sqrt {c x^{4}}} \]

[In]

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

[Out]

-1/(3*x*sqrt(c*x**4))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3 \, \sqrt {c x^{4}} x} \]

[In]

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

[Out]

-1/3/(sqrt(c*x^4)*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3 \, \sqrt {c} x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.50 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+c x^4}} \, dx=-\frac {1}{3\,\sqrt {c}\,x\,\sqrt {x^4}} \]

[In]

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

[Out]

-1/(3*c^(1/2)*x*(x^4)^(1/2))