Integrand size = 15, antiderivative size = 92 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac {9 b \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac {27 b^2 \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac {81 b^3 \left (a+b x^3\right )^{2/3}}{440 a^4 x^2} \] Output:
-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 = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-40 a^3+45 a^2 b x^3-54 a b^2 x^6+81 b^3 x^9\right )}{440 a^4 x^{11}} \] Input:
Integrate[1/(x^12*(a + b*x^3)^(1/3)),x]
Output:
((a + b*x^3)^(2/3)*(-40*a^3 + 45*a^2*b*x^3 - 54*a*b^2*x^6 + 81*b^3*x^9))/( 440*a^4*x^11)
Time = 0.36 (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 |
\(\displaystyle -\frac {9 b \left (-\frac {3 b \left (\frac {3 b \left (a+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
Input:
Int[1/(x^12*(a + b*x^3)^(1/3)),x]
Output:
-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)
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 = 0.54 (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 | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 b^{3} x^{9}+54 a \,b^{2} x^{6}-45 a^{2} b \,x^{3}+40 a^{3}\right )}{440 x^{11} a^{4}}\) | \(50\) |
Input:
int(1/x^12/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/440*(b*x^3+a)^(2/3)*(-81*b^3*x^9+54*a*b^2*x^6-45*a^2*b*x^3+40*a^3)/x^11 /a^4
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {{\left (81 \, b^{3} x^{9} - 54 \, a b^{2} x^{6} + 45 \, a^{2} b x^{3} - 40 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{440 \, a^{4} x^{11}} \] Input:
integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
1/440*(81*b^3*x^9 - 54*a*b^2*x^6 + 45*a^2*b*x^3 - 40*a^3)*(b*x^3 + a)^(2/3 )/(a^4*x^11)
Leaf count of result is larger than twice the leaf count of optimal. 692 vs. \(2 (85) = 170\).
Time = 1.15 (sec) , antiderivative size = 692, normalized size of antiderivative = 7.52 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx =\text {Too large to display} \] Input:
integrate(1/x**12/(b*x**3+a)**(1/3),x)
Output:
-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 = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {\frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{3}}{x^{2}} - \frac {264 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{2}}{x^{5}} + \frac {165 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b}{x^{8}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}}}{x^{11}}}{440 \, a^{4}} \] Input:
integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
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
\[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{12}} \,d x } \] Input:
integrate(1/x^12/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(1/3)*x^12), x)
Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {9\,b\,{\left (b\,x^3+a\right )}^{2/3}}{88\,a^2\,x^8}-\frac {{\left (b\,x^3+a\right )}^{2/3}}{11\,a\,x^{11}}+\frac {81\,b^3\,{\left (b\,x^3+a\right )}^{2/3}}{440\,a^4\,x^2}-\frac {27\,b^2\,{\left (b\,x^3+a\right )}^{2/3}}{220\,a^3\,x^5} \] Input:
int(1/(x^12*(a + b*x^3)^(1/3)),x)
Output:
(9*b*(a + b*x^3)^(2/3))/(88*a^2*x^8) - (a + b*x^3)^(2/3)/(11*a*x^11) + (81 *b^3*(a + b*x^3)^(2/3))/(440*a^4*x^2) - (27*b^2*(a + b*x^3)^(2/3))/(220*a^ 3*x^5)
\[ \int \frac {1}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{12}}d x \] Input:
int(1/x^12/(b*x^3+a)^(1/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**12),x)