Integrand size = 24, antiderivative size = 251 \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=-\frac {c^2 (b c-a d) \sqrt [3]{a+b x^3}}{d^4}+\frac {c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{7/3}}{7 b^2 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{13/3}}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}} \] Output:
-c^2*(-a*d+b*c)*(b*x^3+a)^(1/3)/d^4+1/4*c^2*(b*x^3+a)^(4/3)/d^3-1/7*(a*d+b *c)*(b*x^3+a)^(7/3)/b^2/d^2+1/10*(b*x^3+a)^(10/3)/b^2/d-1/3*c^2*(-a*d+b*c) ^(4/3)*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^(13/3)-1/6*c^2*(-a*d+b*c)^(4/3)*ln(d*x^3+c)/d^(13/3)+1/2*c^2*(-a *d+b*c)^(4/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(13/3)
Time = 0.61 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.23 \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (-6 a^3 d^3+2 a^2 b d^2 \left (-10 c+d x^3\right )+a b^2 d \left (175 c^2-40 c d x^3+22 d^2 x^6\right )+b^3 \left (-140 c^3+35 c^2 d x^3-20 c d^2 x^6+14 d^3 x^9\right )\right )}{b^2}-140 \sqrt {3} c^2 (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+140 c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-70 c^2 (b c-a d)^{4/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{420 d^{13/3}} \] Input:
Integrate[(x^8*(a + b*x^3)^(4/3))/(c + d*x^3),x]
Output:
((3*d^(1/3)*(a + b*x^3)^(1/3)*(-6*a^3*d^3 + 2*a^2*b*d^2*(-10*c + d*x^3) + a*b^2*d*(175*c^2 - 40*c*d*x^3 + 22*d^2*x^6) + b^3*(-140*c^3 + 35*c^2*d*x^3 - 20*c*d^2*x^6 + 14*d^3*x^9)))/b^2 - 140*Sqrt[3]*c^2*(b*c - a*d)^(4/3)*Ar cTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] + 140* c^2*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] - 70*c^2*(b*c - a*d)^(4/3)*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)])/(420*d^(13/3))
Time = 0.78 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 99, 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.
| \(\displaystyle \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \left (b x^3+a\right )^{4/3}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {\left (b x^3+a\right )^{7/3}}{b d}+\frac {(-b c-a d) \left (b x^3+a\right )^{4/3}}{b d^2}+\frac {c^2 \left (b x^3+a\right )^{4/3}}{d^2 \left (d x^3+c\right )}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {\sqrt {3} c^2 (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{13/3}}-\frac {3 \left (a+b x^3\right )^{7/3} (a d+b c)}{7 b^2 d^2}+\frac {3 \left (a+b x^3\right )^{10/3}}{10 b^2 d}-\frac {c^2 (b c-a d)^{4/3} \log \left (c+d x^3\right )}{2 d^{13/3}}+\frac {3 c^2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac {3 c^2 \sqrt [3]{a+b x^3} (b c-a d)}{d^4}+\frac {3 c^2 \left (a+b x^3\right )^{4/3}}{4 d^3}\right )\) |
Input:
Int[(x^8*(a + b*x^3)^(4/3))/(c + d*x^3),x]
Output:
((-3*c^2*(b*c - a*d)*(a + b*x^3)^(1/3))/d^4 + (3*c^2*(a + b*x^3)^(4/3))/(4 *d^3) - (3*(b*c + a*d)*(a + b*x^3)^(7/3))/(7*b^2*d^2) + (3*(a + b*x^3)^(10 /3))/(10*b^2*d) - (Sqrt[3]*c^2*(b*c - a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/d^(13/3) - (c^2*(b*c - a*d)^ (4/3)*Log[c + d*x^3])/(2*d^(13/3)) + (3*c^2*(b*c - a*d)^(4/3)*Log[(b*c - a *d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(13/3)))/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 1.92 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.04
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (d \left (\left (b \,x^{3}+a \right )^{2} \left (-\frac {7 b \,x^{3}}{3}+a \right ) d^{3}+\frac {10 b c \left (b \,x^{3}+a \right )^{2} d^{2}}{3}-\frac {175 c^{2} \left (\frac {b \,x^{3}}{5}+a \right ) b^{2} d}{6}+\frac {70 b^{3} c^{3}}{3}\right ) \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\frac {35 b^{2} c^{2} \left (a d -b c \right )^{2} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{9}\right )}{70 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{5} b^{2}}\) | \(260\) |
Input:
int(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
-3/70/((a*d-b*c)/d)^(2/3)*(d*((b*x^3+a)^2*(-7/3*b*x^3+a)*d^3+10/3*b*c*(b*x ^3+a)^2*d^2-175/6*c^2*(1/5*b*x^3+a)*b^2*d+70/3*b^3*c^3)*((a*d-b*c)/d)^(2/3 )*(b*x^3+a)^(1/3)+35/9*b^2*c^2*(a*d-b*c)^2*(2*arctan(1/3*3^(1/2)*(2*(b*x^3 +a)^(1/3)+((a*d-b*c)/d)^(1/3))/((a*d-b*c)/d)^(1/3))*3^(1/2)+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))-2*ln((b*x^3+ a)^(1/3)-((a*d-b*c)/d)^(1/3))))/d^5/b^2
Time = 0.21 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.47 \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {140 \, \sqrt {3} {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 70 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 140 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (14 \, b^{3} d^{3} x^{9} - 2 \, {\left (10 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{6} - 140 \, b^{3} c^{3} + 175 \, a b^{2} c^{2} d - 20 \, a^{2} b c d^{2} - 6 \, a^{3} d^{3} + {\left (35 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{420 \, b^{2} d^{4}} \] Input:
integrate(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
Output:
1/420*(140*sqrt(3)*(b^3*c^3 - a*b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*arctan(- 1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*(-(b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 70*(b^3*c^3 - a*b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*lo g((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)) - 140*(b^3*c^3 - a*b^2*c^2*d)*(-(b*c - a*d)/d)^(1/3)*log((b *x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) + 3*(14*b^3*d^3*x^9 - 2*(10*b^3* c*d^2 - 11*a*b^2*d^3)*x^6 - 140*b^3*c^3 + 175*a*b^2*c^2*d - 20*a^2*b*c*d^2 - 6*a^3*d^3 + (35*b^3*c^2*d - 40*a*b^2*c*d^2 + 2*a^2*b*d^3)*x^3)*(b*x^3 + a)^(1/3))/(b^2*d^4)
\[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\int \frac {x^{8} \left (a + b x^{3}\right )^{\frac {4}{3}}}{c + d x^{3}}\, dx \] Input:
integrate(x**8*(b*x**3+a)**(4/3)/(d*x**3+c),x)
Output:
Integral(x**8*(a + b*x**3)**(4/3)/(c + d*x**3), x)
Exception generated. \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^8*(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 = 0.15 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.57 \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=-\frac {{\left (b^{24} c^{4} d^{6} - 2 \, a b^{23} c^{3} d^{7} + a^{2} b^{22} c^{2} d^{8}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{23} c d^{10} - a b^{22} d^{11}\right )}} + \frac {\sqrt {3} {\left (b c^{3} - a c^{2} d\right )} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{5}} + \frac {{\left (b c^{3} - a c^{2} d\right )} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{5}} - \frac {140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{21} c^{3} d^{6} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{20} c^{2} d^{7} - 140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{20} c^{2} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{19} c d^{8} - 14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} b^{18} d^{9} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a b^{18} d^{9}}{140 \, b^{20} d^{10}} \] Input:
integrate(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
Output:
-1/3*(b^24*c^4*d^6 - 2*a*b^23*c^3*d^7 + a^2*b^22*c^2*d^8)*(-(b*c - a*d)/d) ^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^23*c*d^10 - a*b^22*d^11) + 1/3*sqrt(3)*(b*c^3 - a*c^2*d)*(-b*c*d^2 + a*d^3)^(1/3)*arc tan(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))/d^5 + 1/6*(b*c^3 - a*c^2*d)*(-b*c*d^2 + a*d^3)^(1/3)*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))/d^5 - 1/140*(140*(b*x^3 + a)^(1/3)*b^21*c^3*d^6 - 35*(b*x^3 + a)^ (4/3)*b^20*c^2*d^7 - 140*(b*x^3 + a)^(1/3)*a*b^20*c^2*d^7 + 20*(b*x^3 + a) ^(7/3)*b^19*c*d^8 - 14*(b*x^3 + a)^(10/3)*b^18*d^9 + 20*(b*x^3 + a)^(7/3)* a*b^18*d^9)/(b^20*d^10)
Time = 4.20 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.90 \[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\left (\frac {a^2}{4\,b^2\,d}+\frac {\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,\left (b^3\,c-a\,b^2\,d\right )}{4\,b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{4/3}-\left (\frac {2\,a}{7\,b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{7\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{7/3}+\frac {{\left (b\,x^3+a\right )}^{10/3}}{10\,b^2\,d}+\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}{d^2}-\frac {c^2\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{13/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{13/3}}-\frac {\left (\frac {a^2}{b^2\,d}+\frac {\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,\left (b^3\,c-a\,b^2\,d\right )}{b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b^3\,c-a\,b^2\,d\right )}{b^2\,d}-\frac {c^2\,\ln \left (\frac {3\,c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{7/3}}+\frac {3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{13/3}}+\frac {c^2\,\ln \left (\frac {3\,c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^2}-\frac {9\,c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{7/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{13/3}} \] Input:
int((x^8*(a + b*x^3)^(4/3))/(c + d*x^3),x)
Output:
(a^2/(4*b^2*d) + (((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^2))*(b^3*c - a *b^2*d))/(4*b^2*d))*(a + b*x^3)^(4/3) - ((2*a)/(7*b^2*d) + (b^3*c - a*b^2* d)/(7*b^4*d^2))*(a + b*x^3)^(7/3) + (a + b*x^3)^(10/3)/(10*b^2*d) + (c^2*l og((3*(a + b*x^3)^(1/3)*(b^2*c^4 + a^2*c^2*d^2 - 2*a*b*c^3*d))/d^2 - (c^2* (a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(13/3)))*(a*d - b*c)^(4/3))/ (3*d^(13/3)) - ((a^2/(b^2*d) + (((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^ 2))*(b^3*c - a*b^2*d))/(b^2*d))*(a + b*x^3)^(1/3)*(b^3*c - a*b^2*d))/(b^2* d) - (c^2*log((3*c^2*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(7/3) + ( 3*c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^2)*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(4/3))/(3*d^(13/3)) + (c^2*log((3*c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2 )/d^2 - (9*c^2*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(7/3))/d^(7/3))*((3^(1/2 )*1i)/6 - 1/6)*(a*d - b*c)^(4/3))/d^(13/3)
\[ \int \frac {x^8 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {-6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} d^{2}+120 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b c d +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b \,d^{2} x^{3}-105 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} c^{2}-40 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} c d \,x^{3}+22 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} d^{2} x^{6}+35 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} c^{2} x^{3}-20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} c d \,x^{6}+14 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} d^{2} x^{9}-140 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b^{2} c \,d^{2}+280 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{3} c^{2} d -140 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{4} c^{3}}{140 b^{2} d^{3}} \] Input:
int(x^8*(b*x^3+a)^(4/3)/(d*x^3+c),x)
Output:
( - 6*(a + b*x**3)**(1/3)*a**3*d**2 + 120*(a + b*x**3)**(1/3)*a**2*b*c*d + 2*(a + b*x**3)**(1/3)*a**2*b*d**2*x**3 - 105*(a + b*x**3)**(1/3)*a*b**2*c **2 - 40*(a + b*x**3)**(1/3)*a*b**2*c*d*x**3 + 22*(a + b*x**3)**(1/3)*a*b* *2*d**2*x**6 + 35*(a + b*x**3)**(1/3)*b**3*c**2*x**3 - 20*(a + b*x**3)**(1 /3)*b**3*c*d*x**6 + 14*(a + b*x**3)**(1/3)*b**3*d**2*x**9 - 140*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a**2*b**2*c *d**2 + 280*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b* d*x**6),x)*a*b**3*c**2*d - 140*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d*x **3 + b*c*x**3 + b*d*x**6),x)*b**4*c**3)/(140*b**2*d**3)