3.278 \(\int \frac{1}{x (b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=74 \[ \frac{8 c \sqrt{b x^2+c x^4}}{3 b^3 x^2}-\frac{4 \sqrt{b x^2+c x^4}}{3 b^2 x^4}+\frac{1}{b x^2 \sqrt{b x^2+c x^4}} \]

[Out]

1/(b*x^2*Sqrt[b*x^2 + c*x^4]) - (4*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^4) + (8*c*Sqrt[b*x^2 + c*x^4])/(3*b^3*x^2)

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Rubi [A]  time = 0.129287, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ \frac{8 c \sqrt{b x^2+c x^4}}{3 b^3 x^2}-\frac{4 \sqrt{b x^2+c x^4}}{3 b^2 x^4}+\frac{1}{b x^2 \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^2*Sqrt[b*x^2 + c*x^4]) - (4*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^4) + (8*c*Sqrt[b*x^2 + c*x^4])/(3*b^3*x^2)

Rule 2015

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] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (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])

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 \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac{1}{b x^2 \sqrt{b x^2+c x^4}}+\frac{4 \int \frac{1}{x^3 \sqrt{b x^2+c x^4}} \, dx}{b}\\ &=\frac{1}{b x^2 \sqrt{b x^2+c x^4}}-\frac{4 \sqrt{b x^2+c x^4}}{3 b^2 x^4}-\frac{(8 c) \int \frac{1}{x \sqrt{b x^2+c x^4}} \, dx}{3 b^2}\\ &=\frac{1}{b x^2 \sqrt{b x^2+c x^4}}-\frac{4 \sqrt{b x^2+c x^4}}{3 b^2 x^4}+\frac{8 c \sqrt{b x^2+c x^4}}{3 b^3 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0119983, size = 48, normalized size = 0.65 \[ -\frac{\left (b+c x^2\right ) \left (b^2-4 b c x^2-8 c^2 x^4\right )}{3 b^3 \left (x^2 \left (b+c x^2\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b + c*x^2)*(b^2 - 4*b*c*x^2 - 8*c^2*x^4))/(3*b^3*(x^2*(b + c*x^2))^(3/2))

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Maple [A]  time = 0.046, size = 45, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -8\,{c}^{2}{x}^{4}-4\,bc{x}^{2}+{b}^{2} \right ) }{3\,{b}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(c*x^2+b)*(-8*c^2*x^4-4*b*c*x^2+b^2)/b^3/(c*x^4+b*x^2)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27117, size = 104, normalized size = 1.41 \begin{align*} \frac{{\left (8 \, c^{2} x^{4} + 4 \, b c x^{2} - b^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{3 \,{\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(8*c^2*x^4 + 4*b*c*x^2 - b^2)*sqrt(c*x^4 + b*x^2)/(b^3*c*x^6 + b^4*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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