3.1001 \(\int \frac {1}{x^3 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=52 \[ \frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {\sqrt {b x^2+c x^4}}{3 b x^4} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3, 2016, 2014} \[ \frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {\sqrt {b x^2+c x^4}}{3 b x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3

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

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx &=\int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx\\ &=-\frac {\sqrt {b x^2+c x^4}}{3 b x^4}-\frac {(2 c) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{3 b}\\ &=-\frac {\sqrt {b x^2+c x^4}}{3 b x^4}+\frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.67 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (2 c x^2-b\right )}{3 b^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 31, normalized size = 0.60 \[ \frac {\sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{3 \, b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 57, normalized size = 1.10 \[ \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} + b}{3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 37, normalized size = 0.71 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+b \right )}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^4+b*x^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.09, size = 44, normalized size = 0.85 \[ \frac {2 \, \sqrt {c x^{4} + b x^{2}} c}{3 \, b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}}}{3 \, b x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.47, size = 29, normalized size = 0.56 \[ -\frac {\left (b-2\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2}}{3\,b^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*sqrt(x**2*(b + c*x**2))), x)

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