3.6.71 \(\int \frac {1}{x^5 (a+b x^3)^{2/3}} \, dx\) [571]

Optimal. Leaf size=44 \[ -\frac {\sqrt [3]{a+b x^3}}{4 a x^4}+\frac {3 b \sqrt [3]{a+b x^3}}{4 a^2 x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 44, 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 = {277, 270} \begin {gather*} \frac {3 b \sqrt [3]{a+b x^3}}{4 a^2 x}-\frac {\sqrt [3]{a+b x^3}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^3)^(2/3)),x]

[Out]

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

Rule 270

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 277

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^5 \left (a+b x^3\right )^{2/3}} \, dx &=-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}-\frac {(3 b) \int \frac {1}{x^2 \left (a+b x^3\right )^{2/3}} \, dx}{4 a}\\ &=-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}+\frac {3 b \sqrt [3]{a+b x^3}}{4 a^2 x}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 31, normalized size = 0.70 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-a+3 b x^3\right )}{4 a^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^3)^(2/3)),x]

[Out]

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

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Maple [A]
time = 0.15, size = 26, normalized size = 0.59

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-3 b \,x^{3}+a \right )}{4 a^{2} x^{4}}\)	\(26\)
trager	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-3 b \,x^{3}+a \right )}{4 a^{2} x^{4}}\)	\(26\)
risch	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-3 b \,x^{3}+a \right )}{4 a^{2} x^{4}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 35, normalized size = 0.80 \begin {gather*} \frac {\frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x} - \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{x^{4}}}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 27, normalized size = 0.61 \begin {gather*} \frac {{\left (3 \, b x^{3} - a\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{4 \, a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.46, size = 68, normalized size = 1.55 \begin {gather*} - \frac {\sqrt [3]{b} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {4}{3}\right )}{9 a x^{3} \Gamma \left (\frac {2}{3}\right )} + \frac {b^{\frac {4}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {4}{3}\right )}{3 a^{2} \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**3+a)**(2/3),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*x^5), x)

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Mupad [B]
time = 1.08, size = 25, normalized size = 0.57 \begin {gather*} -\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a-3\,b\,x^3\right )}{4\,a^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^3)^(2/3)),x)

[Out]

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

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