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

Optimal. Leaf size=19 \[ -\frac {\sqrt {a+b x^2}}{a x} \]

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Rubi [A]  time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {5, 264} \begin {gather*} -\frac {\sqrt {a+b x^2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x^2]/(a*x))

Rule 5

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

Rule 264

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*} \int \frac {1}{x^2 \sqrt {a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx &=\int \frac {1}{x^2 \sqrt {a+b x^2}} \, dx\\ &=-\frac {\sqrt {a+b x^2}}{a x}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+b x^2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x^2]/(a*x))

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IntegrateAlgebraic [A]  time = 0.05, size = 19, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+b x^2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x^2]/(a*x))

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fricas [A]  time = 0.97, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {b x^{2} + a}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b*x^2 + a)/(a*x)

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giac [A]  time = 0.18, size = 30, normalized size = 1.58 \begin {gather*} \frac {2 \, \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)

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maple [A]  time = 0.00, size = 18, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {b \,x^{2}+a}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^(1/2),x)

[Out]

-(b*x^2+a)^(1/2)/a/x

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maxima [A]  time = 1.01, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {b x^{2} + a}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b*x^2 + a)/(a*x)

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mupad [B]  time = 0.04, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {b\,x^2+a}}{a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^(1/2)),x)

[Out]

-(a + b*x^2)^(1/2)/(a*x)

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sympy [A]  time = 0.75, size = 19, normalized size = 1.00 \begin {gather*} - \frac {\sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b)*sqrt(a/(b*x**2) + 1)/a

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