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

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

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Rubi [A]  time = 0.13, antiderivative size = 74, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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])

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*}

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Mathematica [A]  time = 0.01, size = 48, normalized size = 0.65 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 55, normalized size = 0.74 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-b^2+4 b c x^2+8 c^2 x^4\right )}{3 b^3 x^4 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 3.97, size = 54, normalized size = 0.73 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.01, size = 45, normalized size = 0.61 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right )}{3 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{3}} \end {gather*}

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 [A]  time = 1.49, size = 65, normalized size = 0.88 \begin {gather*} \frac {8 \, c^{2} x^{2}}{3 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{3 \, \sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{3 \, \sqrt {c x^{4} + b x^{2}} b x^{2}} \end {gather*}

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]

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

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mupad [B]  time = 4.24, size = 51, normalized size = 0.69 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (-b^2+4\,b\,c\,x^2+8\,c^2\,x^4\right )}{3\,b^3\,x^4\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

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

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

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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