Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=-\frac {\left (a+b x^3\right )^{5/3}}{14 a x^{14}}+\frac {9 b \left (a+b x^3\right )^{5/3}}{154 a^2 x^{11}}-\frac {27 b^2 \left (a+b x^3\right )^{5/3}}{616 a^3 x^8}+\frac {81 b^3 \left (a+b x^3\right )^{5/3}}{3080 a^4 x^5} \] Output:
-1/14*(b*x^3+a)^(5/3)/a/x^14+9/154*b*(b*x^3+a)^(5/3)/a^2/x^11-27/616*b^2*( b*x^3+a)^(5/3)/a^3/x^8+81/3080*b^3*(b*x^3+a)^(5/3)/a^4/x^5
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\frac {\left (a+b x^3\right )^{5/3} \left (-220 a^3+180 a^2 b x^3-135 a b^2 x^6+81 b^3 x^9\right )}{3080 a^4 x^{14}} \] Input:
Integrate[(a + b*x^3)^(2/3)/x^15,x]
Output:
((a + b*x^3)^(5/3)*(-220*a^3 + 180*a^2*b*x^3 - 135*a*b^2*x^6 + 81*b^3*x^9) )/(3080*a^4*x^14)
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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \int \frac {\left (b x^3+a\right )^{2/3}}{x^{12}}dx}{14 a}-\frac {\left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \int \frac {\left (b x^3+a\right )^{2/3}}{x^9}dx}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right )}{14 a}-\frac {\left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \left (-\frac {3 b \int \frac {\left (b x^3+a\right )^{2/3}}{x^6}dx}{8 a}-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right )}{14 a}-\frac {\left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {9 b \left (-\frac {6 b \left (\frac {3 b \left (a+b x^3\right )^{5/3}}{40 a^2 x^5}-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right )}{14 a}-\frac {\left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
Input:
Int[(a + b*x^3)^(2/3)/x^15,x]
Output:
-1/14*(a + b*x^3)^(5/3)/(a*x^14) - (9*b*(-1/11*(a + b*x^3)^(5/3)/(a*x^11) - (6*b*(-1/8*(a + b*x^3)^(5/3)/(a*x^8) + (3*b*(a + b*x^3)^(5/3))/(40*a^2*x ^5)))/(11*a)))/(14*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 = 0.59 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-81 b^{3} x^{9}+135 a \,b^{2} x^{6}-180 a^{2} b \,x^{3}+220 a^{3}\right )}{3080 x^{14} a^{4}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-81 b^{3} x^{9}+135 a \,b^{2} x^{6}-180 a^{2} b \,x^{3}+220 a^{3}\right )}{3080 x^{14} a^{4}}\) | \(50\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-81 b^{3} x^{9}+135 a \,b^{2} x^{6}-180 a^{2} b \,x^{3}+220 a^{3}\right )}{3080 x^{14} a^{4}}\) | \(50\) |
trager | \(-\frac {\left (-81 b^{4} x^{12}+54 a \,b^{3} x^{9}-45 a^{2} b^{2} x^{6}+40 a^{3} b \,x^{3}+220 a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 x^{14} a^{4}}\) | \(61\) |
risch | \(-\frac {\left (-81 b^{4} x^{12}+54 a \,b^{3} x^{9}-45 a^{2} b^{2} x^{6}+40 a^{3} b \,x^{3}+220 a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 x^{14} a^{4}}\) | \(61\) |
Input:
int((b*x^3+a)^(2/3)/x^15,x,method=_RETURNVERBOSE)
Output:
-1/3080*(b*x^3+a)^(5/3)*(-81*b^3*x^9+135*a*b^2*x^6-180*a^2*b*x^3+220*a^3)/ x^14/a^4
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\frac {{\left (81 \, b^{4} x^{12} - 54 \, a b^{3} x^{9} + 45 \, a^{2} b^{2} x^{6} - 40 \, a^{3} b x^{3} - 220 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{3080 \, a^{4} x^{14}} \] Input:
integrate((b*x^3+a)^(2/3)/x^15,x, algorithm="fricas")
Output:
1/3080*(81*b^4*x^12 - 54*a*b^3*x^9 + 45*a^2*b^2*x^6 - 40*a^3*b*x^3 - 220*a ^4)*(b*x^3 + a)^(2/3)/(a^4*x^14)
Leaf count of result is larger than twice the leaf count of optimal. 847 vs. \(2 (85) = 170\).
Time = 1.49 (sec) , antiderivative size = 847, normalized size of antiderivative = 9.21 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx =\text {Too large to display} \] Input:
integrate((b*x**3+a)**(2/3)/x**15,x)
Output:
-440*a**7*b**(29/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**7*b**9*x** 12*gamma(-2/3) + 243*a**6*b**10*x**15*gamma(-2/3) + 243*a**5*b**11*x**18*g amma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) - 1400*a**6*b**(32/3)*x**3*( a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**7*b**9*x**12*gamma(-2/3) + 243* a**6*b**10*x**15*gamma(-2/3) + 243*a**5*b**11*x**18*gamma(-2/3) + 81*a**4* b**12*x**21*gamma(-2/3)) - 1470*a**5*b**(35/3)*x**6*(a/(b*x**3) + 1)**(2/3 )*gamma(-14/3)/(81*a**7*b**9*x**12*gamma(-2/3) + 243*a**6*b**10*x**15*gamm a(-2/3) + 243*a**5*b**11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/ 3)) - 518*a**4*b**(38/3)*x**9*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a** 7*b**9*x**12*gamma(-2/3) + 243*a**6*b**10*x**15*gamma(-2/3) + 243*a**5*b** 11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) + 28*a**3*b**(41/3 )*x**12*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**7*b**9*x**12*gamma(-2/ 3) + 243*a**6*b**10*x**15*gamma(-2/3) + 243*a**5*b**11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) + 252*a**2*b**(44/3)*x**15*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**7*b**9*x**12*gamma(-2/3) + 243*a**6*b**10*x **15*gamma(-2/3) + 243*a**5*b**11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21* gamma(-2/3)) + 378*a*b**(47/3)*x**18*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/ (81*a**7*b**9*x**12*gamma(-2/3) + 243*a**6*b**10*x**15*gamma(-2/3) + 243*a **5*b**11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) + 162*b**(5 0/3)*x**21*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**7*b**9*x**12*gam...
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\frac {\frac {616 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{3}}{x^{5}} - \frac {1155 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{2}}{x^{8}} + \frac {840 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b}{x^{11}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {14}{3}}}{x^{14}}}{3080 \, a^{4}} \] Input:
integrate((b*x^3+a)^(2/3)/x^15,x, algorithm="maxima")
Output:
1/3080*(616*(b*x^3 + a)^(5/3)*b^3/x^5 - 1155*(b*x^3 + a)^(8/3)*b^2/x^8 + 8 40*(b*x^3 + a)^(11/3)*b/x^11 - 220*(b*x^3 + a)^(14/3)/x^14)/a^4
\[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{15}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)/x^15,x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(2/3)/x^15, x)
Time = 0.73 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\frac {81\,b^4\,{\left (b\,x^3+a\right )}^{2/3}}{3080\,a^4\,x^2}-\frac {b\,{\left (b\,x^3+a\right )}^{2/3}}{77\,a\,x^{11}}-\frac {{\left (b\,x^3+a\right )}^{2/3}}{14\,x^{14}}-\frac {27\,b^3\,{\left (b\,x^3+a\right )}^{2/3}}{1540\,a^3\,x^5}+\frac {9\,b^2\,{\left (b\,x^3+a\right )}^{2/3}}{616\,a^2\,x^8} \] Input:
int((a + b*x^3)^(2/3)/x^15,x)
Output:
(81*b^4*(a + b*x^3)^(2/3))/(3080*a^4*x^2) - (b*(a + b*x^3)^(2/3))/(77*a*x^ 11) - (a + b*x^3)^(2/3)/(14*x^14) - (27*b^3*(a + b*x^3)^(2/3))/(1540*a^3*x ^5) + (9*b^2*(a + b*x^3)^(2/3))/(616*a^2*x^8)
Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65 \[ \int \frac {\left (a+b x^3\right )^{2/3}}{x^{15}} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (81 b^{4} x^{12}-54 a \,b^{3} x^{9}+45 a^{2} b^{2} x^{6}-40 a^{3} b \,x^{3}-220 a^{4}\right )}{3080 a^{4} x^{14}} \] Input:
int((b*x^3+a)^(2/3)/x^15,x)
Output:
((a + b*x**3)**(2/3)*( - 220*a**4 - 40*a**3*b*x**3 + 45*a**2*b**2*x**6 - 5 4*a*b**3*x**9 + 81*b**4*x**12))/(3080*a**4*x**14)