\(\int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx\) [520]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7}+\frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4} \]

[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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {277, 270} \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=\frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4}-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7} \]

[In]

Int[(a + b*x^3)^(1/3)/x^8,x]

[Out]

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

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=\frac {\sqrt [3]{a+b x^3} \left (-4 a^2-a b x^3+3 b^2 x^6\right )}{28 a^2 x^7} \]

[In]

Integrate[(a + b*x^3)^(1/3)/x^8,x]

[Out]

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

Maple [A] (verified)

Time = 3.92 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (-3 b \,x^{3}+4 a \right )}{28 a^{2} x^{7}}\)	\(28\)
pseudoelliptic	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (-3 b \,x^{3}+4 a \right )}{28 a^{2} x^{7}}\)	\(28\)
trager	\(-\frac {\left (-3 b^{2} x^{6}+a b \,x^{3}+4 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{28 a^{2} x^{7}}\)	\(38\)
risch	\(-\frac {\left (-3 b^{2} x^{6}+a b \,x^{3}+4 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{28 a^{2} x^{7}}\)	\(38\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=\frac {{\left (3 \, b^{2} x^{6} - a b x^{3} - 4 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{28 \, a^{2} x^{7}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (37) = 74\).

Time = 0.59 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=- \frac {4 \sqrt [3]{b} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{9 x^{6} \Gamma \left (- \frac {1}{3}\right )} - \frac {b^{\frac {4}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{9 a x^{3} \Gamma \left (- \frac {1}{3}\right )} + \frac {b^{\frac {7}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{3 a^{2} \Gamma \left (- \frac {1}{3}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=\frac {\frac {7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b}{x^{4}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}}}{x^{7}}}{28 \, a^{2}} \]

[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

Giac [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{8}} \,d x } \]

[In]

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

[Out]

integrate((b*x^3 + a)^(1/3)/x^8, x)

Mupad [B] (verification not implemented)

Time = 6.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^8} \, dx=-\frac {7\,a\,{\left (b\,x^3+a\right )}^{4/3}-3\,{\left (b\,x^3+a\right )}^{7/3}}{28\,a^2\,x^7} \]

[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)