Integrand size = 22, antiderivative size = 117 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=-\frac {A \left (a+b x^3\right )^{2/3}}{11 a x^{11}}+\frac {(9 A b-11 a B) \left (a+b x^3\right )^{2/3}}{88 a^2 x^8}-\frac {3 b (9 A b-11 a B) \left (a+b x^3\right )^{2/3}}{220 a^3 x^5}+\frac {9 b^2 (9 A b-11 a B) \left (a+b x^3\right )^{2/3}}{440 a^4 x^2} \] Output:
-1/11*A*(b*x^3+a)^(2/3)/a/x^11+1/88*(9*A*b-11*B*a)*(b*x^3+a)^(2/3)/a^2/x^8 -3/220*b*(9*A*b-11*B*a)*(b*x^3+a)^(2/3)/a^3/x^5+9/440*b^2*(9*A*b-11*B*a)*( b*x^3+a)^(2/3)/a^4/x^2
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-40 a^3 A+45 a^2 A b x^3-55 a^3 B x^3-54 a A b^2 x^6+66 a^2 b B x^6+81 A b^3 x^9-99 a b^2 B x^9\right )}{440 a^4 x^{11}} \] Input:
Integrate[(A + B*x^3)/(x^12*(a + b*x^3)^(1/3)),x]
Output:
((a + b*x^3)^(2/3)*(-40*a^3*A + 45*a^2*A*b*x^3 - 55*a^3*B*x^3 - 54*a*A*b^2 *x^6 + 66*a^2*b*B*x^6 + 81*A*b^3*x^9 - 99*a*b^2*B*x^9))/(440*a^4*x^11)
Time = 0.39 (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.
\(\displaystyle \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(9 A b-11 a B) \int \frac {1}{x^9 \sqrt [3]{b x^3+a}}dx}{11 a}-\frac {A \left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(9 A b-11 a B) \left (-\frac {3 b \int \frac {1}{x^6 \sqrt [3]{b x^3+a}}dx}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {A \left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(9 A b-11 a B) \left (-\frac {3 b \left (-\frac {3 b \int \frac {1}{x^3 \sqrt [3]{b x^3+a}}dx}{5 a}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right )}{11 a}-\frac {A \left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (-\frac {3 b \left (\frac {3 b \left (a+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\right ) (9 A b-11 a B)}{11 a}-\frac {A \left (a+b x^3\right )^{2/3}}{11 a x^{11}}\) |
Input:
Int[(A + B*x^3)/(x^12*(a + b*x^3)^(1/3)),x]
Output:
-1/11*(A*(a + b*x^3)^(2/3))/(a*x^11) - ((9*A*b - 11*a*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]
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 = 1.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(-\frac {\left (\left (\frac {11 B \,x^{3}}{8}+A \right ) a^{3}-\frac {9 b \left (\frac {22 B \,x^{3}}{15}+A \right ) x^{3} a^{2}}{8}+\frac {27 b^{2} x^{6} \left (\frac {11 B \,x^{3}}{6}+A \right ) a}{20}-\frac {81 A \,x^{9} b^{3}}{40}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{11 x^{11} a^{4}}\) | \(74\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 A \,x^{9} b^{3}+99 B \,x^{9} a \,b^{2}+54 A \,x^{6} a \,b^{2}-66 B \,x^{6} a^{2} b -45 a^{2} A b \,x^{3}+55 B \,x^{3} a^{3}+40 a^{3} A \right )}{440 x^{11} a^{4}}\) | \(83\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 A \,x^{9} b^{3}+99 B \,x^{9} a \,b^{2}+54 A \,x^{6} a \,b^{2}-66 B \,x^{6} a^{2} b -45 a^{2} A b \,x^{3}+55 B \,x^{3} a^{3}+40 a^{3} A \right )}{440 x^{11} a^{4}}\) | \(83\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 A \,x^{9} b^{3}+99 B \,x^{9} a \,b^{2}+54 A \,x^{6} a \,b^{2}-66 B \,x^{6} a^{2} b -45 a^{2} A b \,x^{3}+55 B \,x^{3} a^{3}+40 a^{3} A \right )}{440 x^{11} a^{4}}\) | \(83\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-81 A \,x^{9} b^{3}+99 B \,x^{9} a \,b^{2}+54 A \,x^{6} a \,b^{2}-66 B \,x^{6} a^{2} b -45 a^{2} A b \,x^{3}+55 B \,x^{3} a^{3}+40 a^{3} A \right )}{440 x^{11} a^{4}}\) | \(83\) |
Input:
int((B*x^3+A)/x^12/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/11*((11/8*B*x^3+A)*a^3-9/8*b*(22/15*B*x^3+A)*x^3*a^2+27/20*b^2*x^6*(11/ 6*B*x^3+A)*a-81/40*A*x^9*b^3)*(b*x^3+a)^(2/3)/x^11/a^4
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=-\frac {{\left (9 \, {\left (11 \, B a b^{2} - 9 \, A b^{3}\right )} x^{9} - 6 \, {\left (11 \, B a^{2} b - 9 \, A a b^{2}\right )} x^{6} + 40 \, A a^{3} + 5 \, {\left (11 \, B a^{3} - 9 \, A a^{2} b\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{440 \, a^{4} x^{11}} \] Input:
integrate((B*x^3+A)/x^12/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
-1/440*(9*(11*B*a*b^2 - 9*A*b^3)*x^9 - 6*(11*B*a^2*b - 9*A*a*b^2)*x^6 + 40 *A*a^3 + 5*(11*B*a^3 - 9*A*a^2*b)*x^3)*(b*x^3 + a)^(2/3)/(a^4*x^11)
Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (112) = 224\).
Time = 2.72 (sec) , antiderivative size = 1120, normalized size of antiderivative = 9.57 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\text {Too large to display} \] Input:
integrate((B*x**3+A)/x**12/(b*x**3+a)**(1/3),x)
Output:
-80*A*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*gam ma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) - 150*A*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**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)) - 78*A*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) + 2 43*a**5*b**11*x**15*gamma(1/3) + 81*a**4*b**12*x**18*gamma(1/3)) + 28*A*a* *3*b**(38/3)*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) + 81*a**4*b**12*x**18*gamma(1/3)) + 252*A*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*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*A*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)) + 10*B*a**4*b**(14/3)*(a/(b*x**3) + 1)**( 2/3)*gamma(-8/3)/(27*a**5*b**4*x**6*gamma(1/3) + 54*a**4*b**5*x**9*gamm...
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=-\frac {B {\left (\frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{2}}{x^{2}} - \frac {16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b}{x^{5}} + \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{x^{8}}\right )}}{40 \, a^{3}} + \frac {{\left (\frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{3}}{x^{2}} - \frac {264 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{2}}{x^{5}} + \frac {165 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b}{x^{8}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}}}{x^{11}}\right )} A}{440 \, a^{4}} \] Input:
integrate((B*x^3+A)/x^12/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
-1/40*B*(20*(b*x^3 + a)^(2/3)*b^2/x^2 - 16*(b*x^3 + a)^(5/3)*b/x^5 + 5*(b* x^3 + a)^(8/3)/x^8)/a^3 + 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/a^4
\[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{12}} \,d x } \] Input:
integrate((B*x^3+A)/x^12/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(1/3)*x^12), x)
Time = 1.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\frac {{\left (b\,x^3+a\right )}^{2/3}\,\left (9\,A\,b-11\,B\,a\right )}{88\,a^2\,x^8}+\frac {{\left (b\,x^3+a\right )}^{2/3}\,\left (81\,A\,b^3-99\,B\,a\,b^2\right )}{440\,a^4\,x^2}-\frac {\left (27\,A\,b^2-33\,B\,a\,b\right )\,{\left (b\,x^3+a\right )}^{2/3}}{220\,a^3\,x^5}-\frac {A\,{\left (b\,x^3+a\right )}^{2/3}}{11\,a\,x^{11}} \] Input:
int((A + B*x^3)/(x^12*(a + b*x^3)^(1/3)),x)
Output:
((a + b*x^3)^(2/3)*(9*A*b - 11*B*a))/(88*a^2*x^8) + ((a + b*x^3)^(2/3)*(81 *A*b^3 - 99*B*a*b^2))/(440*a^4*x^2) - ((27*A*b^2 - 33*B*a*b)*(a + b*x^3)^( 2/3))/(220*a^3*x^5) - (A*(a + b*x^3)^(2/3))/(11*a*x^11)
\[ \int \frac {A+B x^3}{x^{12} \sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{12}}d x \right ) a +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{9}}d x \right ) b \] Input:
int((B*x^3+A)/x^12/(b*x^3+a)^(1/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**12),x)*a + int(1/((a + b*x**3)**(1/3)*x**9), x)*b