Integrand size = 22, antiderivative size = 117 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=-\frac {A \left (a+b x^3\right )^{5/3}}{14 a x^{14}}+\frac {(9 A b-14 a B) \left (a+b x^3\right )^{5/3}}{154 a^2 x^{11}}-\frac {3 b (9 A b-14 a B) \left (a+b x^3\right )^{5/3}}{616 a^3 x^8}+\frac {9 b^2 (9 A b-14 a B) \left (a+b x^3\right )^{5/3}}{3080 a^4 x^5} \] Output:
-1/14*A*(b*x^3+a)^(5/3)/a/x^14+1/154*(9*A*b-14*B*a)*(b*x^3+a)^(5/3)/a^2/x^ 11-3/616*b*(9*A*b-14*B*a)*(b*x^3+a)^(5/3)/a^3/x^8+9/3080*b^2*(9*A*b-14*B*a )*(b*x^3+a)^(5/3)/a^4/x^5
Time = 0.50 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=\frac {\left (a+b x^3\right )^{5/3} \left (-220 a^3 A+180 a^2 A b x^3-280 a^3 B x^3-135 a A b^2 x^6+210 a^2 b B x^6+81 A b^3 x^9-126 a b^2 B x^9\right )}{3080 a^4 x^{14}} \] Input:
Integrate[((a + b*x^3)^(2/3)*(A + B*x^3))/x^15,x]
Output:
((a + b*x^3)^(5/3)*(-220*a^3*A + 180*a^2*A*b*x^3 - 280*a^3*B*x^3 - 135*a*A *b^2*x^6 + 210*a^2*b*B*x^6 + 81*A*b^3*x^9 - 126*a*b^2*B*x^9))/(3080*a^4*x^ 14)
Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(9 A b-14 a B) \int \frac {\left (b x^3+a\right )^{2/3}}{x^{12}}dx}{14 a}-\frac {A \left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(9 A b-14 a B) \left (-\frac {6 b \int \frac {\left (b x^3+a\right )^{2/3}}{x^9}dx}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right )}{14 a}-\frac {A \left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(9 A b-14 a B) \left (-\frac {6 b \left (-\frac {3 b \int \frac {\left (b x^3+a\right )^{2/3}}{x^6}dx}{8 a}-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right )}{14 a}-\frac {A \left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (-\frac {6 b \left (\frac {3 b \left (a+b x^3\right )^{5/3}}{40 a^2 x^5}-\frac {\left (a+b x^3\right )^{5/3}}{8 a x^8}\right )}{11 a}-\frac {\left (a+b x^3\right )^{5/3}}{11 a x^{11}}\right ) (9 A b-14 a B)}{14 a}-\frac {A \left (a+b x^3\right )^{5/3}}{14 a x^{14}}\) |
Input:
Int[((a + b*x^3)^(2/3)*(A + B*x^3))/x^15,x]
Output:
-1/14*(A*(a + b*x^3)^(5/3))/(a*x^14) - ((9*A*b - 14*a*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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.94 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (\left (\frac {14 B \,x^{3}}{11}+A \right ) a^{3}-\frac {9 b \,x^{3} \left (\frac {7 B \,x^{3}}{6}+A \right ) a^{2}}{11}+\frac {27 b^{2} \left (\frac {14 B \,x^{3}}{15}+A \right ) x^{6} a}{44}-\frac {81 A \,x^{9} b^{3}}{220}\right )}{14 x^{14} a^{4}}\) | \(74\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-81 A \,x^{9} b^{3}+126 B \,x^{9} a \,b^{2}+135 A \,x^{6} a \,b^{2}-210 B \,x^{6} a^{2} b -180 a^{2} A b \,x^{3}+280 B \,x^{3} a^{3}+220 a^{3} A \right )}{3080 x^{14} a^{4}}\) | \(83\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-81 A \,x^{9} b^{3}+126 B \,x^{9} a \,b^{2}+135 A \,x^{6} a \,b^{2}-210 B \,x^{6} a^{2} b -180 a^{2} A b \,x^{3}+280 B \,x^{3} a^{3}+220 a^{3} A \right )}{3080 x^{14} a^{4}}\) | \(83\) |
trager | \(-\frac {\left (-81 A \,b^{4} x^{12}+126 B a \,b^{3} x^{12}+54 A a \,b^{3} x^{9}-84 B \,a^{2} b^{2} x^{9}-45 A \,a^{2} b^{2} x^{6}+70 B \,a^{3} b \,x^{6}+40 A \,a^{3} b \,x^{3}+280 B \,a^{4} x^{3}+220 A \,a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 x^{14} a^{4}}\) | \(107\) |
risch | \(-\frac {\left (-81 A \,b^{4} x^{12}+126 B a \,b^{3} x^{12}+54 A a \,b^{3} x^{9}-84 B \,a^{2} b^{2} x^{9}-45 A \,a^{2} b^{2} x^{6}+70 B \,a^{3} b \,x^{6}+40 A \,a^{3} b \,x^{3}+280 B \,a^{4} x^{3}+220 A \,a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{3080 x^{14} a^{4}}\) | \(107\) |
Input:
int((b*x^3+a)^(2/3)*(B*x^3+A)/x^15,x,method=_RETURNVERBOSE)
Output:
-1/14*(b*x^3+a)^(5/3)*((14/11*B*x^3+A)*a^3-9/11*b*x^3*(7/6*B*x^3+A)*a^2+27 /44*b^2*(14/15*B*x^3+A)*x^6*a-81/220*A*x^9*b^3)/x^14/a^4
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=-\frac {{\left (9 \, {\left (14 \, B a b^{3} - 9 \, A b^{4}\right )} x^{12} - 6 \, {\left (14 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{9} + 5 \, {\left (14 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{6} + 220 \, A a^{4} + 40 \, {\left (7 \, B a^{4} + A a^{3} b\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{3080 \, a^{4} x^{14}} \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^15,x, algorithm="fricas")
Output:
-1/3080*(9*(14*B*a*b^3 - 9*A*b^4)*x^12 - 6*(14*B*a^2*b^2 - 9*A*a*b^3)*x^9 + 5*(14*B*a^3*b - 9*A*a^2*b^2)*x^6 + 220*A*a^4 + 40*(7*B*a^4 + A*a^3*b)*x^ 3)*(b*x^3 + a)^(2/3)/(a^4*x^14)
Leaf count of result is larger than twice the leaf count of optimal. 1392 vs. \(2 (112) = 224\).
Time = 3.13 (sec) , antiderivative size = 1392, normalized size of antiderivative = 11.90 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=\text {Too large to display} \] Input:
integrate((b*x**3+a)**(2/3)*(B*x**3+A)/x**15,x)
Output:
-440*A*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 *gamma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) - 1400*A*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*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**1 5*gamma(-2/3) + 243*a**5*b**11*x**18*gamma(-2/3) + 81*a**4*b**12*x**21*gam ma(-2/3)) - 518*A*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*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*gam ma(-2/3) + 81*a**4*b**12*x**21*gamma(-2/3)) + 252*A*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*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*A*b**(50/3)*x**21*(a/(b*x**3) + 1)**(2/3)*gamma(-14/3)/(81*a**...
Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=-\frac {B {\left (\frac {44 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{2}}{x^{5}} - \frac {55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b}{x^{8}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}}}{x^{11}}\right )}}{220 \, a^{3}} + \frac {A {\left (\frac {616 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{3}}{x^{5}} - \frac {1155 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{2}}{x^{8}} + \frac {840 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b}{x^{11}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {14}{3}}}{x^{14}}\right )}}{3080 \, a^{4}} \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^15,x, algorithm="maxima")
Output:
-1/220*B*(44*(b*x^3 + a)^(5/3)*b^2/x^5 - 55*(b*x^3 + a)^(8/3)*b/x^8 + 20*( b*x^3 + a)^(11/3)/x^11)/a^3 + 1/3080*A*(616*(b*x^3 + a)^(5/3)*b^3/x^5 - 11 55*(b*x^3 + a)^(8/3)*b^2/x^8 + 840*(b*x^3 + a)^(11/3)*b/x^11 - 220*(b*x^3 + a)^(14/3)/x^14)/a^4
\[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{15}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^15,x, algorithm="giac")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(2/3)/x^15, x)
Time = 2.04 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=\frac {81\,A\,b^4\,{\left (b\,x^3+a\right )}^{2/3}}{3080\,a^4\,x^2}-\frac {B\,{\left (b\,x^3+a\right )}^{2/3}}{11\,x^{11}}-\frac {A\,b\,{\left (b\,x^3+a\right )}^{2/3}}{77\,a\,x^{11}}-\frac {B\,b\,{\left (b\,x^3+a\right )}^{2/3}}{44\,a\,x^8}-\frac {A\,{\left (b\,x^3+a\right )}^{2/3}}{14\,x^{14}}-\frac {27\,A\,b^3\,{\left (b\,x^3+a\right )}^{2/3}}{1540\,a^3\,x^5}+\frac {9\,A\,b^2\,{\left (b\,x^3+a\right )}^{2/3}}{616\,a^2\,x^8}-\frac {9\,B\,b^3\,{\left (b\,x^3+a\right )}^{2/3}}{220\,a^3\,x^2}+\frac {3\,B\,b^2\,{\left (b\,x^3+a\right )}^{2/3}}{110\,a^2\,x^5} \] Input:
int(((A + B*x^3)*(a + b*x^3)^(2/3))/x^15,x)
Output:
(81*A*b^4*(a + b*x^3)^(2/3))/(3080*a^4*x^2) - (B*(a + b*x^3)^(2/3))/(11*x^ 11) - (A*b*(a + b*x^3)^(2/3))/(77*a*x^11) - (B*b*(a + b*x^3)^(2/3))/(44*a* x^8) - (A*(a + b*x^3)^(2/3))/(14*x^14) - (27*A*b^3*(a + b*x^3)^(2/3))/(154 0*a^3*x^5) + (9*A*b^2*(a + b*x^3)^(2/3))/(616*a^2*x^8) - (9*B*b^3*(a + b*x ^3)^(2/3))/(220*a^3*x^2) + (3*B*b^2*(a + b*x^3)^(2/3))/(110*a^2*x^5)
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^{15}} \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-9 b^{4} x^{12}+6 a \,b^{3} x^{9}-5 a^{2} b^{2} x^{6}-64 a^{3} b \,x^{3}-44 a^{4}\right )}{616 a^{3} x^{14}} \] Input:
int((b*x^3+a)^(2/3)*(B*x^3+A)/x^15,x)
Output:
((a + b*x**3)**(2/3)*( - 44*a**4 - 64*a**3*b*x**3 - 5*a**2*b**2*x**6 + 6*a *b**3*x**9 - 9*b**4*x**12))/(616*a**3*x**14)