Integrand size = 24, antiderivative size = 154 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 a^2 (b c-a d) \sqrt {c+d x^3}}{3 b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}-\frac {2 a^2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}} \] Output:
2/3*a^2*(-a*d+b*c)*(d*x^3+c)^(1/2)/b^4+2/9*a^2*(d*x^3+c)^(3/2)/b^3-2/15*(a *d+b*c)*(d*x^3+c)^(5/2)/b^2/d^2+2/21*(d*x^3+c)^(7/2)/b/d^2-2/3*a^2*(-a*d+b *c)^(3/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9/2)
Time = 0.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.93 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \sqrt {c+d x^3} \left (-105 a^3 d^3-21 a b^2 d \left (c+d x^3\right )^2-3 b^3 \left (2 c-5 d x^3\right ) \left (c+d x^3\right )^2+35 a^2 b d^2 \left (4 c+d x^3\right )\right )}{315 b^4 d^2}+\frac {2 a^2 (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{9/2}} \] Input:
Integrate[(x^8*(c + d*x^3)^(3/2))/(a + b*x^3),x]
Output:
(2*Sqrt[c + d*x^3]*(-105*a^3*d^3 - 21*a*b^2*d*(c + d*x^3)^2 - 3*b^3*(2*c - 5*d*x^3)*(c + d*x^3)^2 + 35*a^2*b*d^2*(4*c + d*x^3)))/(315*b^4*d^2) + (2* a^2*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a* d]])/(3*b^(9/2))
Time = 0.53 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, 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.
| \(\displaystyle \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \left (d x^3+c\right )^{3/2}}{b x^3+a}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {\left (d x^3+c\right )^{5/2}}{b d}+\frac {(-b c-a d) \left (d x^3+c\right )^{3/2}}{b^2 d}+\frac {a^2 \left (d x^3+c\right )^{3/2}}{b^2 \left (b x^3+a\right )}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {2 a^2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{9/2}}+\frac {2 a^2 \sqrt {c+d x^3} (b c-a d)}{b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{3 b^3}-\frac {2 \left (c+d x^3\right )^{5/2} (a d+b c)}{5 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{7 b d^2}\right )\) |
Input:
Int[(x^8*(c + d*x^3)^(3/2))/(a + b*x^3),x]
Output:
((2*a^2*(b*c - a*d)*Sqrt[c + d*x^3])/b^4 + (2*a^2*(c + d*x^3)^(3/2))/(3*b^ 3) - (2*(b*c + a*d)*(c + d*x^3)^(5/2))/(5*b^2*d^2) + (2*(c + d*x^3)^(7/2)) /(7*b*d^2) - (2*a^2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sq rt[b*c - a*d]])/b^(9/2))/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 = 1.77 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.97
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (\sqrt {\left (a d -b c \right ) b}\, \left (\frac {2 \left (d \,x^{3}+c \right )^{2} \left (-\frac {5 d \,x^{3}}{2}+c \right ) b^{3}}{35}+\frac {a d \left (d \,x^{3}+c \right )^{2} b^{2}}{5}-\frac {4 a^{2} d^{2} \left (\frac {d \,x^{3}}{4}+c \right ) b}{3}+a^{3} d^{3}\right ) \sqrt {d \,x^{3}+c}-a^{2} d^{2} \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{3 \sqrt {\left (a d -b c \right ) b}\, d^{2} b^{4}}\) | \(149\) |
| default | \(\frac {\frac {2 d \,x^{9} \sqrt {d \,x^{3}+c}}{21}+\frac {16 c \,x^{6} \sqrt {d \,x^{3}+c}}{105}+\frac {2 c^{2} x^{3} \sqrt {d \,x^{3}+c}}{105 d}-\frac {4 c^{3} \sqrt {d \,x^{3}+c}}{105 d^{2}}}{b}-\frac {2 a^{2} \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 b^{4} \sqrt {\left (a d -b c \right ) b}}-\frac {2 a \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 b^{2} d}\) | \(188\) |
| risch | \(-\frac {2 \left (-15 b^{3} d^{3} x^{9}+21 a \,b^{2} d^{3} x^{6}-24 b^{3} c \,d^{2} x^{6}-35 a^{2} b \,d^{3} x^{3}+42 a \,b^{2} c \,d^{2} x^{3}-3 b^{3} c^{2} d \,x^{3}+105 a^{3} d^{3}-140 a^{2} b c \,d^{2}+21 a \,b^{2} c^{2} d +6 b^{3} c^{3}\right ) \sqrt {d \,x^{3}+c}}{315 d^{2} b^{4}}+\frac {2 a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{4} \sqrt {\left (a d -b c \right ) b}}\) | \(193\) |
| elliptic | \(\frac {2 d \,x^{9} \sqrt {d \,x^{3}+c}}{21 b}+\frac {2 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {6 c d}{7 b}\right ) x^{6} \sqrt {d \,x^{3}+c}}{15 d}+\frac {2 \left (\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3}}-\frac {4 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {6 c d}{7 b}\right ) c}{5 d}\right ) x^{3} \sqrt {d \,x^{3}+c}}{9 d}+\frac {2 \left (-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{4}}-\frac {2 \left (\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3}}-\frac {4 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {6 c d}{7 b}\right ) c}{5 d}\right ) c}{3 d}\right ) \sqrt {d \,x^{3}+c}}{3 d}-\frac {i a^{2} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{3 b^{4} d^{2}}\) | \(684\) |
Input:
int(x^8*(d*x^3+c)^(3/2)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
-2/3/((a*d-b*c)*b)^(1/2)*(((a*d-b*c)*b)^(1/2)*(2/35*(d*x^3+c)^2*(-5/2*d*x^ 3+c)*b^3+1/5*a*d*(d*x^3+c)^2*b^2-4/3*a^2*d^2*(1/4*d*x^3+c)*b+a^3*d^3)*(d*x ^3+c)^(1/2)-a^2*d^2*(a*d-b*c)^2*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/ 2)))/d^2/b^4
Time = 0.12 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.66 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\left [-\frac {105 \, {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (15 \, b^{3} d^{3} x^{9} + 3 \, {\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + {\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{315 \, b^{4} d^{2}}, -\frac {2 \, {\left (105 \, {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (15 \, b^{3} d^{3} x^{9} + 3 \, {\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + {\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{315 \, b^{4} d^{2}}\right ] \] Input:
integrate(x^8*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="fricas")
Output:
[-1/315*(105*(a^2*b*c*d^2 - a^3*d^3)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2* b*c - a*d + 2*sqrt(d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b*x^3 + a)) - 2*(15* b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c^3 - 21*a*b^2*c^2 *d + 140*a^2*b*c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a*b^2*c*d^2 + 35*a^ 2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^4*d^2), -2/315*(105*(a^2*b*c*d^2 - a^3*d ^3)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b *c - a*d)) - (15*b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c ^3 - 21*a*b^2*c^2*d + 140*a^2*b*c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a* b^2*c*d^2 + 35*a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^4*d^2)]
Time = 41.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.31 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b^{3}} + \frac {a^{2} d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{5} \sqrt {\frac {a d - b c}{b}}} + \frac {\left (c + d x^{3}\right )^{\frac {7}{2}}}{21 b d} + \frac {\left (c + d x^{3}\right )^{\frac {5}{2}} \left (- a d - b c\right )}{15 b^{2} d} + \frac {\sqrt {c + d x^{3}} \left (- a^{3} d^{2} + a^{2} b c d\right )}{3 b^{4}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\frac {a^{2} \left (\begin {cases} \frac {x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{3} \right )}}{b} & \text {otherwise} \end {cases}\right )}{3 b^{2}} - \frac {a x^{3}}{3 b^{2}} + \frac {x^{6}}{6 b}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(d*x**3+c)**(3/2)/(b*x**3+a),x)
Output:
Piecewise((2*(a**2*d*(c + d*x**3)**(3/2)/(9*b**3) + a**2*d*(a*d - b*c)**2* atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**5*sqrt((a*d - b*c)/b)) + (c + d*x**3)**(7/2)/(21*b*d) + (c + d*x**3)**(5/2)*(-a*d - b*c)/(15*b**2*d ) + sqrt(c + d*x**3)*(-a**3*d**2 + a**2*b*c*d)/(3*b**4))/d, Ne(d, 0)), (c* *(3/2)*(a**2*Piecewise((x**3/a, Eq(b, 0)), (log(a + b*x**3)/b, True))/(3*b **2) - a*x**3/(3*b**2) + x**6/(6*b)), True))
Exception generated. \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^8*(d*x^3+c)^(3/2)/(b*x^3+a),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 = 0.12 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.25 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{4}} + \frac {2 \, {\left (15 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} b^{6} d^{12} - 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{6} c d^{12} - 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} a b^{5} d^{13} + 35 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a^{2} b^{4} d^{14} + 105 \, \sqrt {d x^{3} + c} a^{2} b^{4} c d^{14} - 105 \, \sqrt {d x^{3} + c} a^{3} b^{3} d^{15}\right )}}{315 \, b^{7} d^{14}} \] Input:
integrate(x^8*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="giac")
Output:
2/3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b ^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^4) + 2/315*(15*(d*x^3 + c)^(7/2)*b^ 6*d^12 - 21*(d*x^3 + c)^(5/2)*b^6*c*d^12 - 21*(d*x^3 + c)^(5/2)*a*b^5*d^13 + 35*(d*x^3 + c)^(3/2)*a^2*b^4*d^14 + 105*sqrt(d*x^3 + c)*a^2*b^4*c*d^14 - 105*sqrt(d*x^3 + c)*a^3*b^3*d^15)/(b^7*d^14)
Time = 4.61 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.14 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2\,d\,x^9\,\sqrt {d\,x^3+c}}{21\,b}-\frac {\left (\frac {2\,a\,\left (\frac {c^2}{b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}\right )}{b}+\frac {2\,c\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {4\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{5\,d}\right )}{3\,d}\right )\,\sqrt {d\,x^3+c}}{3\,d}+\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {4\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{5\,d}\right )}{9\,d}-\frac {x^6\,\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{15\,d}+\frac {a^2\,\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d-\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{9/2}} \] Input:
int((x^8*(c + d*x^3)^(3/2))/(a + b*x^3),x)
Output:
(2*d*x^9*(c + d*x^3)^(1/2))/(21*b) - (((2*a*(c^2/b + (a*((a*d^2)/b^2 - (2* c*d)/b))/b))/b + (2*c*((2*c^2)/b + (2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (4* c*((2*a*d^2)/b^2 - (16*c*d)/(7*b)))/(5*d)))/(3*d))*(c + d*x^3)^(1/2))/(3*d ) + (x^3*(c + d*x^3)^(1/2)*((2*c^2)/b + (2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (4*c*((2*a*d^2)/b^2 - (16*c*d)/(7*b)))/(5*d)))/(9*d) - (x^6*(c + d*x^3)^ (1/2)*((2*a*d^2)/b^2 - (16*c*d)/(7*b)))/(15*d) + (a^2*log((a^2*d^2 + 2*b^2 *c^2 - b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/2)*2i - a*b*d^2*x^3 + b^2* c*d*x^3 - 3*a*b*c*d)/(a + b*x^3))*(a*d - b*c)^(3/2)*1i)/(3*b^(9/2))
\[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {-140 \sqrt {d \,x^{3}+c}\, a^{2} c \,d^{2}+70 \sqrt {d \,x^{3}+c}\, a^{2} d^{3} x^{3}+168 \sqrt {d \,x^{3}+c}\, a b \,c^{2} d -84 \sqrt {d \,x^{3}+c}\, a b c \,d^{2} x^{3}-42 \sqrt {d \,x^{3}+c}\, a b \,d^{3} x^{6}-12 \sqrt {d \,x^{3}+c}\, b^{2} c^{3}+6 \sqrt {d \,x^{3}+c}\, b^{2} c^{2} d \,x^{3}+48 \sqrt {d \,x^{3}+c}\, b^{2} c \,d^{2} x^{6}+30 \sqrt {d \,x^{3}+c}\, b^{2} d^{3} x^{9}-315 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{3} d^{4}+630 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} b c \,d^{3}-315 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,b^{2} c^{2} d^{2}}{315 b^{3} d^{2}} \] Input:
int(x^8*(d*x^3+c)^(3/2)/(b*x^3+a),x)
Output:
( - 140*sqrt(c + d*x**3)*a**2*c*d**2 + 70*sqrt(c + d*x**3)*a**2*d**3*x**3 + 168*sqrt(c + d*x**3)*a*b*c**2*d - 84*sqrt(c + d*x**3)*a*b*c*d**2*x**3 - 42*sqrt(c + d*x**3)*a*b*d**3*x**6 - 12*sqrt(c + d*x**3)*b**2*c**3 + 6*sqrt (c + d*x**3)*b**2*c**2*d*x**3 + 48*sqrt(c + d*x**3)*b**2*c*d**2*x**6 + 30* sqrt(c + d*x**3)*b**2*d**3*x**9 - 315*int((sqrt(c + d*x**3)*x**5)/(a*c + a *d*x**3 + b*c*x**3 + b*d*x**6),x)*a**3*d**4 + 630*int((sqrt(c + d*x**3)*x* *5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a**2*b*c*d**3 - 315*int((sqr t(c + d*x**3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b**2*c**2* d**2)/(315*b**3*d**2)