Integrand size = 15, antiderivative size = 92 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^3}}{10 a x^{10}}+\frac {9 b \sqrt [3]{a+b x^3}}{70 a^2 x^7}-\frac {27 b^2 \sqrt [3]{a+b x^3}}{140 a^3 x^4}+\frac {81 b^3 \sqrt [3]{a+b x^3}}{140 a^4 x} \] Output:
-1/10*(b*x^3+a)^(1/3)/a/x^10+9/70*b*(b*x^3+a)^(1/3)/a^2/x^7-27/140*b^2*(b* x^3+a)^(1/3)/a^3/x^4+81/140*b^3*(b*x^3+a)^(1/3)/a^4/x
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^3} \left (-14 a^3+18 a^2 b x^3-27 a b^2 x^6+81 b^3 x^9\right )}{140 a^4 x^{10}} \] Input:
Integrate[1/(x^11*(a + b*x^3)^(2/3)),x]
Output:
((a + b*x^3)^(1/3)*(-14*a^3 + 18*a^2*b*x^3 - 27*a*b^2*x^6 + 81*b^3*x^9))/( 140*a^4*x^10)
Time = 0.34 (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.
\(\displaystyle \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \int \frac {1}{x^8 \left (b x^3+a\right )^{2/3}}dx}{10 a}-\frac {\sqrt [3]{a+b x^3}}{10 a x^{10}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \int \frac {1}{x^5 \left (b x^3+a\right )^{2/3}}dx}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a}-\frac {\sqrt [3]{a+b x^3}}{10 a x^{10}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \left (-\frac {3 b \int \frac {1}{x^2 \left (b x^3+a\right )^{2/3}}dx}{4 a}-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a}-\frac {\sqrt [3]{a+b x^3}}{10 a x^{10}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \left (\frac {3 b \sqrt [3]{a+b x^3}}{4 a^2 x}-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a}-\frac {\sqrt [3]{a+b x^3}}{10 a x^{10}}\) |
Input:
Int[1/(x^11*(a + b*x^3)^(2/3)),x]
Output:
-1/10*(a + b*x^3)^(1/3)/(a*x^10) - (9*b*(-1/7*(a + b*x^3)^(1/3)/(a*x^7) - (6*b*(-1/4*(a + b*x^3)^(1/3)/(a*x^4) + (3*b*(a + b*x^3)^(1/3))/(4*a^2*x))) /(7*a)))/(10*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 = 0.82 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{9}+27 a \,b^{2} x^{6}-18 a^{2} b \,x^{3}+14 a^{3}\right )}{140 x^{10} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{9}+27 a \,b^{2} x^{6}-18 a^{2} b \,x^{3}+14 a^{3}\right )}{140 x^{10} a^{4}}\) | \(50\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{9}+27 a \,b^{2} x^{6}-18 a^{2} b \,x^{3}+14 a^{3}\right )}{140 x^{10} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{9}+27 a \,b^{2} x^{6}-18 a^{2} b \,x^{3}+14 a^{3}\right )}{140 x^{10} a^{4}}\) | \(50\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-81 b^{3} x^{9}+27 a \,b^{2} x^{6}-18 a^{2} b \,x^{3}+14 a^{3}\right )}{140 x^{10} a^{4}}\) | \(50\) |
Input:
int(1/x^11/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
-1/140*(b*x^3+a)^(1/3)*(-81*b^3*x^9+27*a*b^2*x^6-18*a^2*b*x^3+14*a^3)/x^10 /a^4
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\frac {{\left (81 \, b^{3} x^{9} - 27 \, a b^{2} x^{6} + 18 \, a^{2} b x^{3} - 14 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{140 \, a^{4} x^{10}} \] Input:
integrate(1/x^11/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
1/140*(81*b^3*x^9 - 27*a*b^2*x^6 + 18*a^2*b*x^3 - 14*a^3)*(b*x^3 + a)^(1/3 )/(a^4*x^10)
Leaf count of result is larger than twice the leaf count of optimal. 692 vs. \(2 (83) = 166\).
Time = 1.15 (sec) , antiderivative size = 692, normalized size of antiderivative = 7.52 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx =\text {Too large to display} \] Input:
integrate(1/x**11/(b*x**3+a)**(2/3),x)
Output:
-28*a**6*b**(28/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/(81*a**7*b**9*x**9 *gamma(2/3) + 243*a**6*b**10*x**12*gamma(2/3) + 243*a**5*b**11*x**15*gamma (2/3) + 81*a**4*b**12*x**18*gamma(2/3)) - 48*a**5*b**(31/3)*x**3*(a/(b*x** 3) + 1)**(1/3)*gamma(-10/3)/(81*a**7*b**9*x**9*gamma(2/3) + 243*a**6*b**10 *x**12*gamma(2/3) + 243*a**5*b**11*x**15*gamma(2/3) + 81*a**4*b**12*x**18* gamma(2/3)) - 30*a**4*b**(34/3)*x**6*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/ (81*a**7*b**9*x**9*gamma(2/3) + 243*a**6*b**10*x**12*gamma(2/3) + 243*a**5 *b**11*x**15*gamma(2/3) + 81*a**4*b**12*x**18*gamma(2/3)) + 80*a**3*b**(37 /3)*x**9*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/(81*a**7*b**9*x**9*gamma(2/3 ) + 243*a**6*b**10*x**12*gamma(2/3) + 243*a**5*b**11*x**15*gamma(2/3) + 81 *a**4*b**12*x**18*gamma(2/3)) + 360*a**2*b**(40/3)*x**12*(a/(b*x**3) + 1)* *(1/3)*gamma(-10/3)/(81*a**7*b**9*x**9*gamma(2/3) + 243*a**6*b**10*x**12*g amma(2/3) + 243*a**5*b**11*x**15*gamma(2/3) + 81*a**4*b**12*x**18*gamma(2/ 3)) + 432*a*b**(43/3)*x**15*(a/(b*x**3) + 1)**(1/3)*gamma(-10/3)/(81*a**7* b**9*x**9*gamma(2/3) + 243*a**6*b**10*x**12*gamma(2/3) + 243*a**5*b**11*x* *15*gamma(2/3) + 81*a**4*b**12*x**18*gamma(2/3)) + 162*b**(46/3)*x**18*(a/ (b*x**3) + 1)**(1/3)*gamma(-10/3)/(81*a**7*b**9*x**9*gamma(2/3) + 243*a**6 *b**10*x**12*gamma(2/3) + 243*a**5*b**11*x**15*gamma(2/3) + 81*a**4*b**12* x**18*gamma(2/3))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\frac {\frac {140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{3}}{x} - \frac {105 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2}}{x^{4}} + \frac {60 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b}{x^{7}} - \frac {14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}}}{x^{10}}}{140 \, a^{4}} \] Input:
integrate(1/x^11/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
1/140*(140*(b*x^3 + a)^(1/3)*b^3/x - 105*(b*x^3 + a)^(4/3)*b^2/x^4 + 60*(b *x^3 + a)^(7/3)*b/x^7 - 14*(b*x^3 + a)^(10/3)/x^10)/a^4
\[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{11}} \,d x } \] Input:
integrate(1/x^11/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(2/3)*x^11), x)
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\frac {9\,b\,{\left (b\,x^3+a\right )}^{1/3}}{70\,a^2\,x^7}-\frac {{\left (b\,x^3+a\right )}^{1/3}}{10\,a\,x^{10}}+\frac {81\,b^3\,{\left (b\,x^3+a\right )}^{1/3}}{140\,a^4\,x}-\frac {27\,b^2\,{\left (b\,x^3+a\right )}^{1/3}}{140\,a^3\,x^4} \] Input:
int(1/(x^11*(a + b*x^3)^(2/3)),x)
Output:
(9*b*(a + b*x^3)^(1/3))/(70*a^2*x^7) - (a + b*x^3)^(1/3)/(10*a*x^10) + (81 *b^3*(a + b*x^3)^(1/3))/(140*a^4*x) - (27*b^2*(a + b*x^3)^(1/3))/(140*a^3* x^4)
\[ \int \frac {1}{x^{11} \left (a+b x^3\right )^{2/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{11}}d x \] Input:
int(1/x^11/(b*x^3+a)^(2/3),x)
Output:
int(1/((a + b*x**3)**(2/3)*x**11),x)