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]

-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)

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Rubi [A]  time = 0.0837913, antiderivative size = 52, normalized size of antiderivative = 1., 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 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])

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])

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*}

Mathematica [A]  time = 0.0146715, 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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Maple [A]  time = 0.043, size = 37, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -2\,c{x}^{2}+b \right ) }{3\,{b}^{2}{x}^{2}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.50202, size = 66, normalized size = 1.27 \begin{align*} \frac{\sqrt{c x^{4} + b x^{2}}{\left (2 \, c x^{2} - b\right )}}{3 \, b^{2} x^{4}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{x^{2} \left (b + c x^{2}\right )}}\, dx \end{align*}

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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Giac [A]  time = 1.19894, size = 36, normalized size = 0.69 \begin{align*} -\frac{{\left (c + \frac{b}{x^{2}}\right )}^{\frac{3}{2}} - 3 \, \sqrt{c + \frac{b}{x^{2}}} c}{3 \, b^{2}} \end{align*}

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