3.2273 \(\int \frac{1}{x^3 (a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=50 \[ \frac{3 b \sqrt [3]{a+b x^{3/2}}}{2 a^2 \sqrt{x}}-\frac{\sqrt [3]{a+b x^{3/2}}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.012822, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {271, 264} \[ \frac{3 b \sqrt [3]{a+b x^{3/2}}}{2 a^2 \sqrt{x}}-\frac{\sqrt [3]{a+b x^{3/2}}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^(3/2))^(2/3)),x]

[Out]

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

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0124195, size = 33, normalized size = 0.66 \[ -\frac{\left (a-3 b x^{3/2}\right ) \sqrt [3]{a+b x^{3/2}}}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( a+b{x}^{{\frac{3}{2}}} \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.949018, size = 47, normalized size = 0.94 \begin{align*} \frac{\frac{4 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} b}{\sqrt{x}} - \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}}}{x^{2}}}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 4.13987, size = 74, normalized size = 1.48 \begin{align*} \frac{{\left (3 \, b x^{\frac{3}{2}} - a\right )}{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}}}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.42057, size = 76, normalized size = 1.52 \begin{align*} - \frac{2 \sqrt [3]{b} \sqrt [3]{\frac{a}{b x^{\frac{3}{2}}} + 1} \Gamma \left (- \frac{4}{3}\right )}{9 a x^{\frac{3}{2}} \Gamma \left (\frac{2}{3}\right )} + \frac{2 b^{\frac{4}{3}} \sqrt [3]{\frac{a}{b x^{\frac{3}{2}}} + 1} \Gamma \left (- \frac{4}{3}\right )}{3 a^{2} \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b**(1/3)*(a/(b*x**(3/2)) + 1)**(1/3)*gamma(-4/3)/(9*a*x**(3/2)*gamma(2/3)) + 2*b**(4/3)*(a/(b*x**(3/2)) + 1
)**(1/3)*gamma(-4/3)/(3*a**2*gamma(2/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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