Integrand size = 15, antiderivative size = 68 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=-\frac {\left (a+b x^3\right )^{4/3}}{10 a x^{10}}+\frac {3 b \left (a+b x^3\right )^{4/3}}{35 a^2 x^7}-\frac {9 b^2 \left (a+b x^3\right )^{4/3}}{140 a^3 x^4} \] Output:
-1/10*(b*x^3+a)^(4/3)/a/x^10+3/35*b*(b*x^3+a)^(4/3)/a^2/x^7-9/140*b^2*(b*x ^3+a)^(4/3)/a^3/x^4
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=\frac {\sqrt [3]{a+b x^3} \left (-14 a^3-2 a^2 b x^3+3 a b^2 x^6-9 b^3 x^9\right )}{140 a^3 x^{10}} \] Input:
Integrate[(a + b*x^3)^(1/3)/x^11,x]
Output:
((a + b*x^3)^(1/3)*(-14*a^3 - 2*a^2*b*x^3 + 3*a*b^2*x^6 - 9*b^3*x^9))/(140 *a^3*x^10)
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {803, 803, 796}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {3 b \int \frac {\sqrt [3]{b x^3+a}}{x^8}dx}{5 a}-\frac {\left (a+b x^3\right )^{4/3}}{10 a x^{10}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {3 b \left (-\frac {3 b \int \frac {\sqrt [3]{b x^3+a}}{x^5}dx}{7 a}-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7}\right )}{5 a}-\frac {\left (a+b x^3\right )^{4/3}}{10 a x^{10}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {3 b \left (\frac {3 b \left (a+b x^3\right )^{4/3}}{28 a^2 x^4}-\frac {\left (a+b x^3\right )^{4/3}}{7 a x^7}\right )}{5 a}-\frac {\left (a+b x^3\right )^{4/3}}{10 a x^{10}}\) |
Input:
Int[(a + b*x^3)^(1/3)/x^11,x]
Output:
-1/10*(a + b*x^3)^(4/3)/(a*x^10) - (3*b*(-1/7*(a + b*x^3)^(4/3)/(a*x^7) + (3*b*(a + b*x^3)^(4/3))/(28*a^2*x^4)))/(5*a)
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]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[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] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Time = 0.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (9 b^{2} x^{6}-12 a b \,x^{3}+14 a^{2}\right )}{140 a^{3} x^{10}}\) | \(39\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (9 b^{2} x^{6}-12 a b \,x^{3}+14 a^{2}\right )}{140 a^{3} x^{10}}\) | \(39\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (9 b^{2} x^{6}-12 a b \,x^{3}+14 a^{2}\right )}{140 a^{3} x^{10}}\) | \(39\) |
trager | \(-\frac {\left (9 b^{3} x^{9}-3 a \,b^{2} x^{6}+2 a^{2} b \,x^{3}+14 a^{3}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{140 a^{3} x^{10}}\) | \(50\) |
risch | \(-\frac {\left (9 b^{3} x^{9}-3 a \,b^{2} x^{6}+2 a^{2} b \,x^{3}+14 a^{3}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{140 a^{3} x^{10}}\) | \(50\) |
Input:
int((b*x^3+a)^(1/3)/x^11,x,method=_RETURNVERBOSE)
Output:
-1/140*(b*x^3+a)^(4/3)*(9*b^2*x^6-12*a*b*x^3+14*a^2)/a^3/x^10
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=-\frac {{\left (9 \, b^{3} x^{9} - 3 \, a b^{2} x^{6} + 2 \, a^{2} b x^{3} + 14 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{140 \, a^{3} x^{10}} \] Input:
integrate((b*x^3+a)^(1/3)/x^11,x, algorithm="fricas")
Output:
-1/140*(9*b^3*x^9 - 3*a*b^2*x^6 + 2*a^2*b*x^3 + 14*a^3)*(b*x^3 + a)^(1/3)/ (a^3*x^10)
Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (61) = 122\).
Time = 0.88 (sec) , antiderivative size = 520, normalized size of antiderivative = 7.65 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=\frac {28 a^{5} b^{\frac {13}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} + \frac {60 a^{4} b^{\frac {16}{3}} x^{3} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} + \frac {30 a^{3} b^{\frac {19}{3}} x^{6} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} + \frac {10 a^{2} b^{\frac {22}{3}} x^{9} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} + \frac {30 a b^{\frac {25}{3}} x^{12} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} + \frac {18 b^{\frac {28}{3}} x^{15} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {10}{3}\right )}{27 a^{5} b^{4} x^{9} \Gamma \left (- \frac {1}{3}\right ) + 54 a^{4} b^{5} x^{12} \Gamma \left (- \frac {1}{3}\right ) + 27 a^{3} b^{6} x^{15} \Gamma \left (- \frac {1}{3}\right )} \] Input:
integrate((b*x**3+a)**(1/3)/x**11,x)
Output:
28*a**5*b**(13/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/(27*a**5*b**4*x**9* gamma(-1/3) + 54*a**4*b**5*x**12*gamma(-1/3) + 27*a**3*b**6*x**15*gamma(-1 /3)) + 60*a**4*b**(16/3)*x**3*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/(27*a** 5*b**4*x**9*gamma(-1/3) + 54*a**4*b**5*x**12*gamma(-1/3) + 27*a**3*b**6*x* *15*gamma(-1/3)) + 30*a**3*b**(19/3)*x**6*(a/(b*x**3) + 1)**(1/3)*gamma(-1 0/3)/(27*a**5*b**4*x**9*gamma(-1/3) + 54*a**4*b**5*x**12*gamma(-1/3) + 27* a**3*b**6*x**15*gamma(-1/3)) + 10*a**2*b**(22/3)*x**9*(a/(b*x**3) + 1)**(1 /3)*gamma(-10/3)/(27*a**5*b**4*x**9*gamma(-1/3) + 54*a**4*b**5*x**12*gamma (-1/3) + 27*a**3*b**6*x**15*gamma(-1/3)) + 30*a*b**(25/3)*x**12*(a/(b*x**3 ) + 1)**(1/3)*gamma(-10/3)/(27*a**5*b**4*x**9*gamma(-1/3) + 54*a**4*b**5*x **12*gamma(-1/3) + 27*a**3*b**6*x**15*gamma(-1/3)) + 18*b**(28/3)*x**15*(a /(b*x**3) + 1)**(1/3)*gamma(-10/3)/(27*a**5*b**4*x**9*gamma(-1/3) + 54*a** 4*b**5*x**12*gamma(-1/3) + 27*a**3*b**6*x**15*gamma(-1/3))
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=-\frac {\frac {35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2}}{x^{4}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b}{x^{7}} + \frac {14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}}}{x^{10}}}{140 \, a^{3}} \] Input:
integrate((b*x^3+a)^(1/3)/x^11,x, algorithm="maxima")
Output:
-1/140*(35*(b*x^3 + a)^(4/3)*b^2/x^4 - 40*(b*x^3 + a)^(7/3)*b/x^7 + 14*(b* x^3 + a)^(10/3)/x^10)/a^3
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{11}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)/x^11,x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(1/3)/x^11, x)
Time = 0.52 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=\frac {3\,b^2\,{\left (b\,x^3+a\right )}^{1/3}}{140\,a^2\,x^4}-\frac {b\,{\left (b\,x^3+a\right )}^{1/3}}{70\,a\,x^7}-\frac {9\,b^3\,{\left (b\,x^3+a\right )}^{1/3}}{140\,a^3\,x}-\frac {{\left (b\,x^3+a\right )}^{1/3}}{10\,x^{10}} \] Input:
int((a + b*x^3)^(1/3)/x^11,x)
Output:
(3*b^2*(a + b*x^3)^(1/3))/(140*a^2*x^4) - (b*(a + b*x^3)^(1/3))/(70*a*x^7) - (9*b^3*(a + b*x^3)^(1/3))/(140*a^3*x) - (a + b*x^3)^(1/3)/(10*x^10)
Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11}} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-9 b^{3} x^{9}+3 a \,b^{2} x^{6}-2 a^{2} b \,x^{3}-14 a^{3}\right )}{140 a^{3} x^{10}} \] Input:
int((b*x^3+a)^(1/3)/x^11,x)
Output:
((a + b*x**3)**(1/3)*( - 14*a**3 - 2*a**2*b*x**3 + 3*a*b**2*x**6 - 9*b**3* x**9))/(140*a**3*x**10)