\(\int \frac {(a+b x^3)^{2/3}}{x^9} \, dx\) [541]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}+\frac {3 b \left (a+b x^3\right )^{5/3}}{40 a^2 x^5} \]

[Out]

-1/8*(b*x^3+a)^(5/3)/a/x^8+3/40*b*(b*x^3+a)^(5/3)/a^2/x^5

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} \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=\frac {3 b \left (a+b x^3\right )^{5/3}}{40 a^2 x^5}-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8} \]

[In]

Int[(a + b*x^3)^(2/3)/x^9,x]

[Out]

-1/8*(a + b*x^3)^(5/3)/(a*x^8) + (3*b*(a + b*x^3)^(5/3))/(40*a^2*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]

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 )^{5/3}}{8 a x^8}-\frac {(3 b) \int \frac {\left (a+b x^3\right )^{2/3}}{x^6} \, dx}{8 a} \\ & = -\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}+\frac {3 b \left (a+b x^3\right )^{5/3}}{40 a^2 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-5 a^2-2 a b x^3+3 b^2 x^6\right )}{40 a^2 x^8} \]

[In]

Integrate[(a + b*x^3)^(2/3)/x^9,x]

[Out]

((a + b*x^3)^(2/3)*(-5*a^2 - 2*a*b*x^3 + 3*b^2*x^6))/(40*a^2*x^8)

Maple [A] (verified)

Time = 4.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-3 b \,x^{3}+5 a \right )}{40 x^{8} a^{2}}\)	\(28\)
pseudoelliptic	\(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-3 b \,x^{3}+5 a \right )}{40 x^{8} a^{2}}\)	\(28\)
trager	\(-\frac {\left (-3 b^{2} x^{6}+2 a b \,x^{3}+5 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{40 x^{8} a^{2}}\)	\(39\)
risch	\(-\frac {\left (-3 b^{2} x^{6}+2 a b \,x^{3}+5 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{40 x^{8} a^{2}}\)	\(39\)

[In]

int((b*x^3+a)^(2/3)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/40*(b*x^3+a)^(5/3)*(-3*b*x^3+5*a)/x^8/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=\frac {{\left (3 \, b^{2} x^{6} - 2 \, a b x^{3} - 5 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, a^{2} x^{8}} \]

[In]

integrate((b*x^3+a)^(2/3)/x^9,x, algorithm="fricas")

[Out]

1/40*(3*b^2*x^6 - 2*a*b*x^3 - 5*a^2)*(b*x^3 + a)^(2/3)/(a^2*x^8)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (37) = 74\).

Time = 0.64 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.50 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=- \frac {5 b^{\frac {2}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{6} \Gamma \left (- \frac {2}{3}\right )} - \frac {2 b^{\frac {5}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 a x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {b^{\frac {8}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{3 a^{2} \Gamma \left (- \frac {2}{3}\right )} \]

[In]

integrate((b*x**3+a)**(2/3)/x**9,x)

[Out]

-5*b**(2/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(9*x**6*gamma(-2/3)) - 2*b**(5/3)*(a/(b*x**3) + 1)**(2/3)*gamm
a(-8/3)/(9*a*x**3*gamma(-2/3)) + b**(8/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(3*a**2*gamma(-2/3))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=\frac {\frac {8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b}{x^{5}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{x^{8}}}{40 \, a^{2}} \]

[In]

integrate((b*x^3+a)^(2/3)/x^9,x, algorithm="maxima")

[Out]

1/40*(8*(b*x^3 + a)^(5/3)*b/x^5 - 5*(b*x^3 + a)^(8/3)/x^8)/a^2

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \]

[In]

integrate((b*x^3+a)^(2/3)/x^9,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(2/3)/x^9, x)

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^9} \, dx=-\frac {{\left (b\,x^3+a\right )}^{2/3}\,\left (5\,a^2+2\,a\,b\,x^3-3\,b^2\,x^6\right )}{40\,a^2\,x^8} \]

[In]

int((a + b*x^3)^(2/3)/x^9,x)

[Out]

-((a + b*x^3)^(2/3)*(5*a^2 - 3*b^2*x^6 + 2*a*b*x^3))/(40*a^2*x^8)