Integrand size = 24, antiderivative size = 232 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {a^3}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {(b c+2 a d) \left (a+b x^3\right )^{2/3}}{2 b^3 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^3 d}-\frac {c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3} (b c-a d)^{4/3}}+\frac {c^3 \log \left (c+d x^3\right )}{6 d^{8/3} (b c-a d)^{4/3}}-\frac {c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}} \] Output:
a^3/b^3/(-a*d+b*c)/(b*x^3+a)^(1/3)-1/2*(2*a*d+b*c)*(b*x^3+a)^(2/3)/b^3/d^2 +1/5*(b*x^3+a)^(5/3)/b^3/d-1/3*c^3*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3) /(-a*d+b*c)^(1/3))*3^(1/2))*3^(1/2)/d^(8/3)/(-a*d+b*c)^(4/3)+1/6*c^3*ln(d* x^3+c)/d^(8/3)/(-a*d+b*c)^(4/3)-1/2*c^3*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3 +a)^(1/3))/d^(8/3)/(-a*d+b*c)^(4/3)
Time = 0.92 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.28 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {\frac {3 d^{2/3} \left (18 a^3 d^2+b^3 c x^3 \left (-5 c+2 d x^3\right )+3 a^2 b d \left (-c+2 d x^3\right )-a b^2 \left (5 c^2+c d x^3+2 d^2 x^6\right )\right )}{b^3 (b c-a d) \sqrt [3]{a+b x^3}}-\frac {10 \sqrt {3} c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{(b c-a d)^{4/3}}-\frac {10 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{4/3}}+\frac {5 c^3 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{4/3}}}{30 d^{8/3}} \] Input:
Integrate[x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
Output:
((3*d^(2/3)*(18*a^3*d^2 + b^3*c*x^3*(-5*c + 2*d*x^3) + 3*a^2*b*d*(-c + 2*d *x^3) - a*b^2*(5*c^2 + c*d*x^3 + 2*d^2*x^6)))/(b^3*(b*c - a*d)*(a + b*x^3) ^(1/3)) - (10*Sqrt[3]*c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(b*c - a*d)^(4/3) - (10*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(4/3) + (5*c^3*Log[(b*c - a*d)^(2 /3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2 /3)])/(b*c - a*d)^(4/3))/(30*d^(8/3))
Time = 0.75 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 98, 2009}
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.
| \(\displaystyle \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx^3\) |
| \(\Big \downarrow \) 98 |
| \(\displaystyle \frac {1}{3} \int \left (-\frac {a^3}{b^2 (b c-a d) \left (b x^3+a\right )^{4/3}}+\frac {x^3}{b d \sqrt [3]{b x^3+a}}+\frac {-b c-a d}{b^2 d^2 \sqrt [3]{b x^3+a}}-\frac {c^3}{d^2 (a d-b c) \sqrt [3]{b x^3+a} \left (d x^3+c\right )}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 a^3}{b^3 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\sqrt {3} c^3 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{8/3} (b c-a d)^{4/3}}-\frac {3 \left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^3 d^2}-\frac {3 a \left (a+b x^3\right )^{2/3}}{2 b^3 d}+\frac {3 \left (a+b x^3\right )^{5/3}}{5 b^3 d}+\frac {c^3 \log \left (c+d x^3\right )}{2 d^{8/3} (b c-a d)^{4/3}}-\frac {3 c^3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} (b c-a d)^{4/3}}\right )\) |
Input:
Int[x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
Output:
((3*a^3)/(b^3*(b*c - a*d)*(a + b*x^3)^(1/3)) - (3*a*(a + b*x^3)^(2/3))/(2* b^3*d) - (3*(b*c + a*d)*(a + b*x^3)^(2/3))/(2*b^3*d^2) + (3*(a + b*x^3)^(5 /3))/(5*b^3*d) - (Sqrt[3]*c^3*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b *c - a*d)^(1/3))/Sqrt[3]])/(d^(8/3)*(b*c - a*d)^(4/3)) + (c^3*Log[c + d*x^ 3])/(2*d^(8/3)*(b*c - a*d)^(4/3)) - (3*c^3*Log[(b*c - a*d)^(1/3) + d^(1/3) *(a + b*x^3)^(1/3)])/(2*d^(8/3)*(b*c - a*d)^(4/3)))/3
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 1.51 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.25
| method | result | size |
| pseudoelliptic | \(-\frac {9 \left (-\frac {5 \ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right ) b^{3} c^{3} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{54}+\frac {5 \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) b^{3} c^{3} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{27}+\frac {5 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right ) b^{3} c^{3} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{27}+\left (-\frac {5 c \left (-\frac {2 d \,x^{3}}{5}+c \right ) x^{3} b^{3}}{18}-\frac {5 a \left (\frac {2}{5} d^{2} x^{6}+\frac {1}{5} c d \,x^{3}+c^{2}\right ) b^{2}}{18}-\frac {a^{2} d \left (-2 d \,x^{3}+c \right ) b}{6}+a^{3} d^{2}\right ) d \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{5 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} d^{3} \left (a d -b c \right ) b^{3}}\) | \(289\) |
Input:
int(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
-9/5/((a*d-b*c)/d)^(1/3)/(b*x^3+a)^(1/3)*(-5/54*ln((b*x^3+a)^(2/3)+((a*d-b *c)/d)^(1/3)*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(2/3))*b^3*c^3*(b*x^3+a)^(1/3)+ 5/27*3^(1/2)*arctan(2/3*3^(1/2)/((a*d-b*c)/d)^(1/3)*(b*x^3+a)^(1/3)+1/3*3^ (1/2))*b^3*c^3*(b*x^3+a)^(1/3)+5/27*ln((b*x^3+a)^(1/3)-((a*d-b*c)/d)^(1/3) )*b^3*c^3*(b*x^3+a)^(1/3)+(-5/18*c*(-2/5*d*x^3+c)*x^3*b^3-5/18*a*(2/5*d^2* x^6+1/5*c*d*x^3+c^2)*b^2-1/6*a^2*d*(-2*d*x^3+c)*b+a^3*d^2)*d*((a*d-b*c)/d) ^(1/3))/d^3/(a*d-b*c)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (191) = 382\).
Time = 0.16 (sec) , antiderivative size = 1141, normalized size of antiderivative = 4.92 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
Output:
[-1/30*(15*sqrt(1/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4*d - a*b^4*c ^3*d^2)*x^3)*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 - 3*sqrt(1/3)*(2*(b*c*d^2 - a*d^3)^(2/3)*(b*x^3 + a)^(2/3 ) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) - (b*c*d^2 - a*d^3)^(1/3)*(b*c - a*d ))*sqrt(-(b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d)) - 3*(b*c*d^2 - a*d^3)^(2/3)* (b*x^3 + a)^(1/3))/(d*x^3 + c)) - 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a *d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a )^(1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) + 10*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d ^2 - a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) + 3*( 5*a*b^3*c^3*d^2 - 2*a^2*b^2*c^2*d^3 - 21*a^3*b*c*d^4 + 18*a^4*d^5 - 2*(b^4 *c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^6 + (5*b^4*c^3*d^2 - 4*a*b^3*c^2 *d^3 - 7*a^2*b^2*c*d^4 + 6*a^3*b*d^5)*x^3)*(b*x^3 + a)^(2/3))/(a*b^5*c^2*d ^4 - 2*a^2*b^4*c*d^5 + a^3*b^3*d^6 + (b^6*c^2*d^4 - 2*a*b^5*c*d^5 + a^2*b^ 4*d^6)*x^3), 1/30*(30*sqrt(1/3)*(a*b^4*c^4*d - a^2*b^3*c^3*d^2 + (b^5*c^4* d - a*b^4*c^3*d^2)*x^3)*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))*arctan(s qrt(1/3)*(2*(b*x^3 + a)^(1/3)*d - (b*c*d^2 - a*d^3)^(1/3))*sqrt((b*c*d^2 - a*d^3)^(1/3)/(b*c - a*d))/d) + 5*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d ^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 - (b*c*d^2 - a*d^3)^(1/3)*(b*x^3 + a)^ (1/3)*d + (b*c*d^2 - a*d^3)^(2/3)) - 10*(b^4*c^3*x^3 + a*b^3*c^3)*(b*c*d^2 - a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d + (b*c*d^2 - a*d^3)^(1/3)) - 3*...
\[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:
integrate(x**11/(b*x**3+a)**(4/3)/(d*x**3+c),x)
Output:
Integral(x**11/((a + b*x**3)**(4/3)*(c + d*x**3)), x)
Exception generated. \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.16 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.60 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d^{4} - 2 \, \sqrt {3} a b c d^{5} + \sqrt {3} a^{2} d^{6}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{2} c^{2} d^{4} - 2 \, a b c d^{5} + a^{2} d^{6}\right )}} - \frac {c^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}} + \frac {a^{3}}{{\left (b^{4} c - a b^{3} d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{13} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{12} d^{4} + 10 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{12} d^{4}}{10 \, b^{15} d^{5}} \] Input:
integrate(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
Output:
-(-b*c*d^2 + a*d^3)^(2/3)*c^3*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (- (b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/(sqrt(3)*b^2*c^2*d^4 - 2*sqr t(3)*a*b*c*d^5 + sqrt(3)*a^2*d^6) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c^3*log(( b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d )/d)^(2/3))/(b^2*c^2*d^4 - 2*a*b*c*d^5 + a^2*d^6) - 1/3*c^3*(-(b*c - a*d)/ d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4) + a^3/((b^4*c - a*b^3*d)*(b*x^3 + a)^(1/3)) - 1/ 10*(5*(b*x^3 + a)^(2/3)*b^13*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^12*d^4 + 10*(b* x^3 + a)^(2/3)*a*b^12*d^4)/(b^15*d^5)
Time = 4.24 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.12 \[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b^3\,d}-\left (\frac {3\,a}{2\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{2\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\frac {a^3}{b^3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}-\frac {c^3\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {c^6\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{9\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {{\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{36\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c^3+\sqrt {3}\,c^3\,1{}\mathrm {i}\right )}{6\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {c^3\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^6\,d^4-b\,c^7\,d^3\right )-\frac {c^6\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (9\,a^4\,d^{12}-36\,a^3\,b\,c\,d^{11}+54\,a^2\,b^2\,c^2\,d^{10}-36\,a\,b^3\,c^3\,d^9+9\,b^4\,c^4\,d^8\right )}{9\,d^{16/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \] Input:
int(x^11/((a + b*x^3)^(4/3)*(c + d*x^3)),x)
Output:
(a + b*x^3)^(5/3)/(5*b^3*d) - ((3*a)/(2*b^3*d) + (b^4*c - a*b^3*d)/(2*b^6* d^2))*(a + b*x^3)^(2/3) - a^3/(b^3*(a + b*x^3)^(1/3)*(a*d - b*c)) - (c^3*l og((a + b*x^3)^(1/3)*(a*c^6*d^4 - b*c^7*d^3) - (c^6*(9*a^4*d^12 + 9*b^4*c^ 4*d^8 - 36*a*b^3*c^3*d^9 + 54*a^2*b^2*c^2*d^10 - 36*a^3*b*c*d^11))/(9*d^(1 6/3)*(a*d - b*c)^(8/3))))/(3*d^(8/3)*(a*d - b*c)^(4/3)) + (log((a + b*x^3) ^(1/3)*(a*c^6*d^4 - b*c^7*d^3) - ((3^(1/2)*c^3*1i + c^3)^2*(9*a^4*d^12 + 9 *b^4*c^4*d^8 - 36*a*b^3*c^3*d^9 + 54*a^2*b^2*c^2*d^10 - 36*a^3*b*c*d^11))/ (36*d^(16/3)*(a*d - b*c)^(8/3)))*(3^(1/2)*c^3*1i + c^3))/(6*d^(8/3)*(a*d - b*c)^(4/3)) - (c^3*log((a + b*x^3)^(1/3)*(a*c^6*d^4 - b*c^7*d^3) - (c^6*( (3^(1/2)*1i)/2 - 1/2)^2*(9*a^4*d^12 + 9*b^4*c^4*d^8 - 36*a*b^3*c^3*d^9 + 5 4*a^2*b^2*c^2*d^10 - 36*a^3*b*c*d^11))/(9*d^(16/3)*(a*d - b*c)^(8/3)))*((3 ^(1/2)*1i)/2 - 1/2))/(3*d^(8/3)*(a*d - b*c)^(4/3))
\[ \int \frac {x^{11}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{6}}d x \] Input:
int(x^11/(b*x^3+a)^(4/3)/(d*x^3+c),x)
Output:
int(x**11/((a + b*x**3)**(1/3)*a*c + (a + b*x**3)**(1/3)*a*d*x**3 + (a + b *x**3)**(1/3)*b*c*x**3 + (a + b*x**3)**(1/3)*b*d*x**6),x)