\(\int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx\) [1010]

Optimal result

Integrand size = 29, antiderivative size = 21 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a+c x^4}}{2 a x^2} \]

[Out]

-1/2*(c*x^4+a)^(1/2)/a/x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4, 270} \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a+c x^4}}{2 a x^2} \]

[In]

Int[1/(x^3*Sqrt[a + (2 + 2*b - 2*(1 + b))*x^2 + c*x^4]),x]

[Out]

-1/2*Sqrt[a + c*x^4]/(a*x^2)

Rule 4

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + c*x^(2*n))^p, x] /; Fre
eQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[b, 0]

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^3 \sqrt {a+c x^4}} \, dx \\ & = -\frac {\sqrt {a+c x^4}}{2 a x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a+c x^4}}{2 a x^2} \]

[In]

Integrate[1/(x^3*Sqrt[a + (2 + 2*b - 2*(1 + b))*x^2 + c*x^4]),x]

[Out]

-1/2*Sqrt[a + c*x^4]/(a*x^2)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
gosper	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)
default	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)
trager	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)
risch	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)
elliptic	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)
pseudoelliptic	\(-\frac {\sqrt {c \,x^{4}+a}}{2 a \,x^{2}}\)	\(18\)

[In]

int(1/x^3/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(c*x^4+a)^(1/2)/a/x^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {c x^{4} + a}}{2 \, a x^{2}} \]

[In]

integrate(1/x^3/(c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(c*x^4 + a)/(a*x^2)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=- \frac {\sqrt {c} \sqrt {\frac {a}{c x^{4}} + 1}}{2 a} \]

[In]

integrate(1/x**3/(c*x**4+a)**(1/2),x)

[Out]

-sqrt(c)*sqrt(a/(c*x**4) + 1)/(2*a)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {c x^{4} + a}}{2 \, a x^{2}} \]

[In]

integrate(1/x^3/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(c*x^4 + a)/(a*x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\sqrt {c}}{{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2} - a} \]

[In]

integrate(1/x^3/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

sqrt(c)/((sqrt(c)*x^2 - sqrt(c*x^4 + a))^2 - a)

Mupad [B] (verification not implemented)

Time = 12.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {c\,x^4+a}}{2\,a\,x^2} \]

[In]

int(1/(x^3*(a + c*x^4)^(1/2)),x)

[Out]

-(a + c*x^4)^(1/2)/(2*a*x^2)