Integrand size = 15, antiderivative size = 92 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac {9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac {27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac {81 b^3 \left (a+b x^3\right )^{2/3}}{440 a^4 x^2} \]
-1/11*(b*x^3+a)^(2/3)/a/x^11+9/88*b*(b*x^3+a)^(2/3)/a^2/x^8-27/220*b^2*(b* x^3+a)^(2/3)/a^3/x^5+81/440*b^3*(b*x^3+a)^(2/3)/a^4/x^2
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-40 a^3+45 a^2 b x^3-54 a b^2 x^6+81 b^3 x^9\right )}{440 a^4 x^{11}} \]
Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {803, 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.
\(\displaystyle \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \int \frac {1}{x^9 \sqrt [3]{b x^3+a}}dx}{11 a}-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {3 b \int \frac {1}{x^6 \sqrt [3]{b x^3+a}}dx}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {3 b \left (-\frac {3 b \int \frac {1}{x^3 \sqrt [3]{b x^3+a}}dx}{5 a}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {9 b \left (-\frac {3 b \left (\frac {3 b \left (a+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
-1/11*(a + b*x^3)^(2/3)/(a*x^11) - (9*b*(-1/8*(a + b*x^3)^(2/3)/(a*x^8) - (3*b*(-1/5*(a + b*x^3)^(2/3)/(a*x^5) + (3*b*(a + b*x^3)^(2/3))/(10*a^2*x^2 )))/(4*a)))/(11*a)
3.6.60.3.1 Defintions of rubi rules used
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 = 4.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
Time = 0.35 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {{\left (81 \, b^{3} x^{9} - 54 \, a b^{2} x^{6} + 45 \, a^{2} b x^{3} - 40 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{440 \, a^{4} x^{11}} \]
Leaf count of result is larger than twice the leaf count of optimal. 692 vs. \(2 (85) = 170\).
Time = 1.06 (sec) , antiderivative size = 692, normalized size of antiderivative = 7.52 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=- \frac {80 a^{6} b^{\frac {29}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} - \frac {150 a^{5} b^{\frac {32}{3}} x^{3} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} - \frac {78 a^{4} b^{\frac {35}{3}} x^{6} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} + \frac {28 a^{3} b^{\frac {38}{3}} x^{9} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} + \frac {252 a^{2} b^{\frac {41}{3}} x^{12} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} + \frac {378 a b^{\frac {44}{3}} x^{15} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} + \frac {162 b^{\frac {47}{3}} x^{18} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {11}{3}\right )}{81 a^{7} b^{9} x^{9} \Gamma \left (\frac {1}{3}\right ) + 243 a^{6} b^{10} x^{12} \Gamma \left (\frac {1}{3}\right ) + 243 a^{5} b^{11} x^{15} \Gamma \left (\frac {1}{3}\right ) + 81 a^{4} b^{12} x^{18} \Gamma \left (\frac {1}{3}\right )} \]
-80*a**6*b**(29/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9 *gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma (1/3) + 81*a**4*b**12*x**18*gamma(1/3)) - 150*a**5*b**(32/3)*x**3*(a/(b*x* *3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**1 0*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18 *gamma(1/3)) - 78*a**4*b**(35/3)*x**6*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3) /(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a** 5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) + 28*a**3*b**(3 8/3)*x**9*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/ 3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 8 1*a**4*b**12*x**18*gamma(1/3)) + 252*a**2*b**(41/3)*x**12*(a/(b*x**3) + 1) **(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12* gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1 /3)) + 378*a*b**(44/3)*x**15*(a/(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7 *b**9*x**9*gamma(1/3) + 243*a**6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x **15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) + 162*b**(47/3)*x**18*(a /(b*x**3) + 1)**(2/3)*gamma(-11/3)/(81*a**7*b**9*x**9*gamma(1/3) + 243*a** 6*b**10*x**12*gamma(1/3) + 243*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12 *x**18*gamma(1/3))
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {\frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{3}}{x^{2}} - \frac {264 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{2}}{x^{5}} + \frac {165 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b}{x^{8}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}}}{x^{11}}}{440 \, a^{4}} \]
1/440*(220*(b*x^3 + a)^(2/3)*b^3/x^2 - 264*(b*x^3 + a)^(5/3)*b^2/x^5 + 165 *(b*x^3 + a)^(8/3)*b/x^8 - 40*(b*x^3 + a)^(11/3)/x^11)/a^4
\[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{12}} \,d x } \]
Time = 5.70 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {9\,b\,{\left (b\,x^3+a\right )}^{2/3}}{88\,a^2\,x^8}-\frac {{\left (b\,x^3+a\right )}^{2/3}}{11\,a\,x^{11}}+\frac {81\,b^3\,{\left (b\,x^3+a\right )}^{2/3}}{440\,a^4\,x^2}-\frac {27\,b^2\,{\left (b\,x^3+a\right )}^{2/3}}{220\,a^3\,x^5} \]