3.520 \(\int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx\)

Optimal. Leaf size=44 \[ \frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4}-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7} \]

[Out]

-1/7*(b*x^3+a)^(4/3)/a/x^7+3/28*b*(b*x^3+a)^(4/3)/a^2/x^4

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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 = {271, 264} \[ \frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4}-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/x^8,x]

[Out]

-(a + b*x^3)^(4/3)/(7*a*x^7) + (3*b*(a + b*x^3)^(4/3))/(28*a^2*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 {\sqrt [3]{a+b x^3}}{x^8} \, dx &=-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7}-\frac {(3 b) \int \frac {\sqrt [3]{a+b x^3}}{x^5} \, dx}{7 a}\\ &=-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7}+\frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.70 \[ \frac {\left (a+b x^3\right )^{4/3} \left (3 b x^3-4 a\right )}{28 a^2 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/x^8,x]

[Out]

((a + b*x^3)^(4/3)*(-4*a + 3*b*x^3))/(28*a^2*x^7)

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fricas [A]  time = 1.09, size = 38, normalized size = 0.86 \[ \frac {{\left (3 \, b^{2} x^{6} - a b x^{3} - 4 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{28 \, a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^8,x, algorithm="fricas")

[Out]

1/28*(3*b^2*x^6 - a*b*x^3 - 4*a^2)*(b*x^3 + a)^(1/3)/(a^2*x^7)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^8,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/x^8, x)

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maple [A]  time = 0.01, size = 28, normalized size = 0.64 \[ -\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (-3 b \,x^{3}+4 a \right )}{28 a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^8,x)

[Out]

-1/28*(b*x^3+a)^(4/3)*(-3*b*x^3+4*a)/a^2/x^7

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maxima [A]  time = 1.34, size = 35, normalized size = 0.80 \[ \frac {\frac {7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b}{x^{4}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}}}{x^{7}}}{28 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^8,x, algorithm="maxima")

[Out]

1/28*(7*(b*x^3 + a)^(4/3)*b/x^4 - 4*(b*x^3 + a)^(7/3)/x^7)/a^2

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mupad [B]  time = 1.23, size = 32, normalized size = 0.73 \[ -\frac {7\,a\,{\left (b\,x^3+a\right )}^{4/3}-3\,{\left (b\,x^3+a\right )}^{7/3}}{28\,a^2\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/x^8,x)

[Out]

-(7*a*(a + b*x^3)^(4/3) - 3*(a + b*x^3)^(7/3))/(28*a^2*x^7)

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sympy [B]  time = 1.23, size = 109, normalized size = 2.48 \[ - \frac {4 \sqrt [3]{b} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right )} - \frac {b^{\frac {4}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{9 a x^{3} \Gamma \left (- \frac {1}{3}\right )} + \frac {b^{\frac {7}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{3 a^{2} \Gamma \left (- \frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**8,x)

[Out]

-4*b**(1/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(9*x**6*gamma(-1/3)) - b**(4/3)*(a/(b*x**3) + 1)**(1/3)*gamma(
-7/3)/(9*a*x**3*gamma(-1/3)) + b**(7/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(3*a**2*gamma(-1/3))

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