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

Optimal. Leaf size=42 \[ -\frac {b x^2}{a^2 \sqrt {a+b x^4}}-\frac {1}{2 a x^2 \sqrt {a+b x^4}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {271, 264} \[ -\frac {b x^2}{a^2 \sqrt {a+b x^4}}-\frac {1}{2 a x^2 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)^(3/2)),x]

[Out]

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

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]

Rule 271

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.69 \[ -\frac {a+2 b x^4}{2 a^2 x^2 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 37, normalized size = 0.88 \[ -\frac {{\left (2 \, b x^{4} + a\right )} \sqrt {b x^{4} + a}}{2 \, {\left (a^{2} b x^{6} + a^{3} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 53, normalized size = 1.26 \[ -\frac {b x^{2}}{2 \, \sqrt {b x^{4} + a} a^{2}} + \frac {\sqrt {b}}{{\left ({\left (\sqrt {b} x^{2} - \sqrt {b x^{4} + a}\right )}^{2} - a\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 26, normalized size = 0.62 \[ -\frac {2 b \,x^{4}+a}{2 \sqrt {b \,x^{4}+a}\, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 36, normalized size = 0.86 \[ -\frac {b x^{2}}{2 \, \sqrt {b x^{4} + a} a^{2}} - \frac {\sqrt {b x^{4} + a}}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 25, normalized size = 0.60 \[ -\frac {2\,b\,x^4+a}{2\,a^2\,x^2\,\sqrt {b\,x^4+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^4)^(3/2)),x)

[Out]

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

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sympy [A]  time = 1.34, size = 46, normalized size = 1.10 \[ - \frac {1}{2 a \sqrt {b} x^{4} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {\sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{4}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a*sqrt(b)*x**4*sqrt(a/(b*x**4) + 1)) - sqrt(b)/(a**2*sqrt(a/(b*x**4) + 1))

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