Integrand size = 24, antiderivative size = 189 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=-\frac {a (4 b c-7 a d) \sqrt {c+d x^3}}{3 b^4}-\frac {a (4 b c-7 a d) \left (c+d x^3\right )^{3/2}}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {a (4 b c-7 a d) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}} \]
-1/9*a*(-7*a*d+4*b*c)*(d*x^3+c)^(3/2)/b^3/(-a*d+b*c)+2/15*(d*x^3+c)^(5/2)/ b^2/d-1/3*a^2*(d*x^3+c)^(5/2)/b^2/(-a*d+b*c)/(b*x^3+a)+1/3*a*(-7*a*d+4*b*c )*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/b^(9/ 2)-1/3*a*(-7*a*d+4*b*c)*(d*x^3+c)^(1/2)/b^4
Time = 0.56 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.86 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {c+d x^3} \left (105 a^3 d^2+6 b^3 x^3 \left (c+d x^3\right )^2+5 a^2 b d \left (-19 c+14 d x^3\right )+2 a b^2 \left (3 c^2-34 c d x^3-7 d^2 x^6\right )\right )}{45 b^4 d \left (a+b x^3\right )}+\frac {a (4 b c-7 a d) \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{9/2}} \]
(Sqrt[c + d*x^3]*(105*a^3*d^2 + 6*b^3*x^3*(c + d*x^3)^2 + 5*a^2*b*d*(-19*c + 14*d*x^3) + 2*a*b^2*(3*c^2 - 34*c*d*x^3 - 7*d^2*x^6)))/(45*b^4*d*(a + b *x^3)) + (a*(4*b*c - 7*a*d)*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*Sqrt[c + d* x^3])/Sqrt[-(b*c) + a*d]])/(3*b^(9/2))
Time = 0.32 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 100, 27, 90, 60, 60, 73, 221}
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.
| \(\displaystyle \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \left (d x^3+c\right )^{3/2}}{\left (b x^3+a\right )^2}dx^3\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {1}{3} \left (\frac {\int -\frac {\left (d x^3+c\right )^{3/2} \left (a (2 b c-5 a d)-2 b (b c-a d) x^3\right )}{2 \left (b x^3+a\right )}dx^3}{b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {\left (d x^3+c\right )^{3/2} \left (a (2 b c-5 a d)-2 b (b c-a d) x^3\right )}{b x^3+a}dx^3}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-7 a d) \int \frac {\left (d x^3+c\right )^{3/2}}{b x^3+a}dx^3-\frac {4 \left (c+d x^3\right )^{5/2} (b c-a d)}{5 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-7 a d) \left (\frac {(b c-a d) \int \frac {\sqrt {d x^3+c}}{b x^3+a}dx^3}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )-\frac {4 \left (c+d x^3\right )^{5/2} (b c-a d)}{5 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )-\frac {4 \left (c+d x^3\right )^{5/2} (b c-a d)}{5 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )-\frac {4 \left (c+d x^3\right )^{5/2} (b c-a d)}{5 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (-\frac {a^2 \left (c+d x^3\right )^{5/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}-\frac {a (4 b c-7 a d) \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x^3}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )-\frac {4 \left (c+d x^3\right )^{5/2} (b c-a d)}{5 d}}{2 b^2 (b c-a d)}\right )\) |
(-((a^2*(c + d*x^3)^(5/2))/(b^2*(b*c - a*d)*(a + b*x^3))) - ((-4*(b*c - a* d)*(c + d*x^3)^(5/2))/(5*d) + a*(4*b*c - 7*a*d)*((2*(c + d*x^3)^(3/2))/(3* b) + ((b*c - a*d)*((2*Sqrt[c + d*x^3])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqr t[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3/2)))/b))/(2*b^2*(b*c - a*d))) /3
3.5.70.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 = 4.65 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.89
| method | result | size |
| pseudoelliptic | \(-\frac {7 \left (\left (a d -\frac {4 b c}{7}\right ) d \left (b \,x^{3}+a \right ) a \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\sqrt {d \,x^{3}+c}\, \left (\frac {2 x^{3} \left (d \,x^{3}+c \right )^{2} b^{3}}{35}+\frac {2 \left (-\frac {7}{3} d^{2} x^{6}-\frac {34}{3} c d \,x^{3}+c^{2}\right ) a \,b^{2}}{35}-\frac {19 d \,a^{2} \left (-\frac {14 d \,x^{3}}{19}+c \right ) b}{21}+a^{3} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, b^{4} d \left (b \,x^{3}+a \right )}\) | \(169\) |
| default | \(\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 b^{2} d}+\frac {a^{2} \left (-d \left (b \,x^{3}+a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (2 d \,x^{3}-c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{b^{4} \sqrt {\left (a d -b c \right ) b}\, \left (b \,x^{3}+a \right )}+\frac {4 a \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {d \,x^{3}+c}\, \left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {\left (a d -b c \right ) b}\right )}{3 b^{4} \sqrt {\left (a d -b c \right ) b}}\) | \(223\) |
| risch | \(\frac {2 \left (3 b^{2} d^{2} x^{6}-10 x^{3} a b \,d^{2}+6 x^{3} b^{2} c d +45 a^{2} d^{2}-40 a b c d +3 b^{2} c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d \,b^{4}}-\frac {a \left (\frac {2 \left (4 a^{2} d^{2}-6 a b c d +2 b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \sqrt {\left (a d -b c \right ) b}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (b \,x^{3}+a \right )+\sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right ) \left (b \,x^{3}+a \right )}\right )}{b^{4}}\) | \(252\) |
| elliptic | \(\frac {a^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{4} \left (b \,x^{3}+a \right )}+\frac {2 d \,x^{6} \sqrt {d \,x^{3}+c}}{15 b^{2}}+\frac {2 \left (-\frac {2 \left (a d -b c \right ) d}{b^{3}}-\frac {4 c d}{5 b^{2}}\right ) x^{3} \sqrt {d \,x^{3}+c}}{9 d}+\frac {2 \left (\frac {3 a^{2} d^{2}-4 a b c d +b^{2} c^{2}}{b^{4}}-\frac {2 \left (-\frac {2 \left (a d -b c \right ) d}{b^{3}}-\frac {4 c d}{5 b^{2}}\right ) c}{3 d}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i a \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (7 a^{2} d^{2}-11 a b c d +4 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 ) \Pi \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}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +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 )}{6 b^{4} d^{2}}\) | \(614\) |
-7/3*((a*d-4/7*b*c)*d*(b*x^3+a)*a*(a*d-b*c)*arctan(b*(d*x^3+c)^(1/2)/((a*d -b*c)*b)^(1/2))-(d*x^3+c)^(1/2)*(2/35*x^3*(d*x^3+c)^2*b^3+2/35*(-7/3*d^2*x ^6-34/3*c*d*x^3+c^2)*a*b^2-19/21*d*a^2*(-14/19*d*x^3+c)*b+a^3*d^2)*((a*d-b *c)*b)^(1/2))/((a*d-b*c)*b)^(1/2)/b^4/d/(b*x^3+a)
Time = 0.38 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.34 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\left [-\frac {15 \, {\left (4 \, a^{2} b c d - 7 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 7 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (6 \, b^{3} d^{2} x^{9} + 2 \, {\left (6 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{6} + 6 \, a b^{2} c^{2} - 95 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 34 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{90 \, {\left (b^{5} d x^{3} + a b^{4} d\right )}}, \frac {15 \, {\left (4 \, a^{2} b c d - 7 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 7 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (6 \, b^{3} d^{2} x^{9} + 2 \, {\left (6 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{6} + 6 \, a b^{2} c^{2} - 95 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 34 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{45 \, {\left (b^{5} d x^{3} + a b^{4} d\right )}}\right ] \]
[-1/90*(15*(4*a^2*b*c*d - 7*a^3*d^2 + (4*a*b^2*c*d - 7*a^2*b*d^2)*x^3)*sqr t((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*(6*b^3*d^2*x^9 + 2*(6*b^3*c*d - 7*a*b^2*d^2) *x^6 + 6*a*b^2*c^2 - 95*a^2*b*c*d + 105*a^3*d^2 + 2*(3*b^3*c^2 - 34*a*b^2* c*d + 35*a^2*b*d^2)*x^3)*sqrt(d*x^3 + c))/(b^5*d*x^3 + a*b^4*d), 1/45*(15* (4*a^2*b*c*d - 7*a^3*d^2 + (4*a*b^2*c*d - 7*a^2*b*d^2)*x^3)*sqrt(-(b*c - a *d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (6*b^ 3*d^2*x^9 + 2*(6*b^3*c*d - 7*a*b^2*d^2)*x^6 + 6*a*b^2*c^2 - 95*a^2*b*c*d + 105*a^3*d^2 + 2*(3*b^3*c^2 - 34*a*b^2*c*d + 35*a^2*b*d^2)*x^3)*sqrt(d*x^3 + c))/(b^5*d*x^3 + a*b^4*d)]
Timed out. \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.12 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=-\frac {{\left (4 \, a b^{2} c^{2} - 11 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{4}} - \frac {\sqrt {d x^{3} + c} a^{2} b c d - \sqrt {d x^{3} + c} a^{3} d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{8} d^{4} - 10 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b^{7} d^{5} - 30 \, \sqrt {d x^{3} + c} a b^{7} c d^{5} + 45 \, \sqrt {d x^{3} + c} a^{2} b^{6} d^{6}\right )}}{45 \, b^{10} d^{5}} \]
-1/3*(4*a*b^2*c^2 - 11*a^2*b*c*d + 7*a^3*d^2)*arctan(sqrt(d*x^3 + c)*b/sqr t(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^4) - 1/3*(sqrt(d*x^3 + c)*a^2*b *c*d - sqrt(d*x^3 + c)*a^3*d^2)/(((d*x^3 + c)*b - b*c + a*d)*b^4) + 2/45*( 3*(d*x^3 + c)^(5/2)*b^8*d^4 - 10*(d*x^3 + c)^(3/2)*a*b^7*d^5 - 30*sqrt(d*x ^3 + c)*a*b^7*c*d^5 + 45*sqrt(d*x^3 + c)*a^2*b^6*d^6)/(b^10*d^5)
Time = 12.04 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.75 \[ \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {2\,c\,\left (\frac {2\,d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {2\,a\,d^2}{b^3}+\frac {8\,c\,d}{5\,b^2}\right )}{3\,d}+\frac {2\,a\,\left (\frac {d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {a\,d^2}{b^3}\right )}{b}\right )}{3\,d}+\frac {2\,d\,x^6\,\sqrt {d\,x^3+c}}{15\,b^2}-\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {2\,a\,d^2}{b^3}+\frac {8\,c\,d}{5\,b^2}\right )}{9\,d}-\frac {a^2\,\left (\frac {2\,b\,c^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {a\,\left (\frac {2\,a\,d^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {4\,b\,c\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )}{b}\right )\,\sqrt {d\,x^3+c}}{b^2\,\left (b\,x^3+a\right )}+\frac {a\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,\left (7\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{9/2}} \]
((c + d*x^3)^(1/2)*((2*(a*d - b*c)^2)/b^4 + (2*c*((2*d*(a*d - 2*b*c))/b^3 + (2*a*d^2)/b^3 + (8*c*d)/(5*b^2)))/(3*d) + (2*a*((d*(a*d - 2*b*c))/b^3 + (a*d^2)/b^3))/b))/(3*d) + (2*d*x^6*(c + d*x^3)^(1/2))/(15*b^2) - (x^3*(c + d*x^3)^(1/2)*((2*d*(a*d - 2*b*c))/b^3 + (2*a*d^2)/b^3 + (8*c*d)/(5*b^2))) /(9*d) + (a*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2) *2i - b*d*x^3)/(a + b*x^3))*(a*d - b*c)^(1/2)*(7*a*d - 4*b*c)*1i)/(6*b^(9/ 2)) - (a^2*((2*b*c^2)/(3*(2*b^2*c - 2*a*b*d)) + (a*((2*a*d^2)/(3*(2*b^2*c - 2*a*b*d)) - (4*b*c*d)/(3*(2*b^2*c - 2*a*b*d))))/b)*(c + d*x^3)^(1/2))/(b ^2*(a + b*x^3))