3.6.40 \(\int \frac {(a+b x^3)^{2/3}}{x^6} \, dx\) [540]

Optimal. Leaf size=21 \[ -\frac {\left (a+b x^3\right )^{5/3}}{5 a x^5} \]

[Out]

-1/5*(b*x^3+a)^(5/3)/a/x^5

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Rubi [A]
time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \begin {gather*} -\frac {\left (a+b x^3\right )^{5/3}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/x^6,x]

[Out]

-1/5*(a + b*x^3)^(5/3)/(a*x^5)

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^{2/3}}{x^6} \, dx &=-\frac {\left (a+b x^3\right )^{5/3}}{5 a x^5}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 21, normalized size = 1.00 \begin {gather*} -\frac {\left (a+b x^3\right )^{5/3}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/x^6,x]

[Out]

-1/5*(a + b*x^3)^(5/3)/(a*x^5)

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Maple [A]
time = 0.14, size = 18, normalized size = 0.86

method result size
gosper \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 a \,x^{5}}\) \(18\)
trager \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 a \,x^{5}}\) \(18\)
risch \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{5 a \,x^{5}}\) \(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(2/3)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5*(b*x^3+a)^(5/3)/a/x^5

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Maxima [A]
time = 0.29, size = 17, normalized size = 0.81 \begin {gather*} -\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^6,x, algorithm="maxima")

[Out]

-1/5*(b*x^3 + a)^(5/3)/(a*x^5)

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Fricas [A]
time = 0.36, size = 17, normalized size = 0.81 \begin {gather*} -\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^6,x, algorithm="fricas")

[Out]

-1/5*(b*x^3 + a)^(5/3)/(a*x^5)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (17) = 34\).
time = 0.43, size = 68, normalized size = 3.24 \begin {gather*} \frac {b^{\frac {2}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {b^{\frac {5}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 a \Gamma \left (- \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(2/3)/x**6,x)

[Out]

b**(2/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)) + b**(5/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/
3)/(3*a*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((b*x^3+a)^(2/3)/x^6,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(2/3)/x^6, x)

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Mupad [B]
time = 1.12, size = 17, normalized size = 0.81 \begin {gather*} -\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,a\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(2/3)/x^6,x)

[Out]

-(a + b*x^3)^(5/3)/(5*a*x^5)

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