\(\int \frac {x^8 \sqrt {c+d x^3}}{(a+b x^3)^2} \, dx\) [633]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=-\frac {4 a \sqrt {c+d x^3}}{3 b^3}-\frac {a^2 \sqrt {c+d x^3}}{3 b^3 \left (a+b x^3\right )}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}+\frac {a (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2} \sqrt {b c-a d}} \] Output:

-4/3*a*(d*x^3+c)^(1/2)/b^3-1/3*a^2*(d*x^3+c)^(1/2)/b^3/(b*x^3+a)+2/9*(d*x^ 
3+c)^(3/2)/b^2/d+1/3*a*(-5*a*d+4*b*c)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a* 
d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {c+d x^3} \left (-15 a^2 d+2 a b \left (c-5 d x^3\right )+2 b^2 x^3 \left (c+d x^3\right )\right )}{9 b^3 d \left (a+b x^3\right )}+\frac {a (-4 b c+5 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2} \sqrt {-b c+a d}} \] Input:

Integrate[(x^8*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]
 

Output:

(Sqrt[c + d*x^3]*(-15*a^2*d + 2*a*b*(c - 5*d*x^3) + 2*b^2*x^3*(c + d*x^3)) 
)/(9*b^3*d*(a + b*x^3)) + (a*(-4*b*c + 5*a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x 
^3])/Sqrt[-(b*c) + a*d]])/(3*b^(7/2)*Sqrt[-(b*c) + a*d])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {948, 100, 27, 90, 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.

\(\displaystyle \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt {d x^3+c}}{\left (b x^3+a\right )^2}dx^3\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{3} \left (\frac {\int -\frac {\sqrt {d x^3+c} \left (a (2 b c-3 a d)-2 b (b c-a d) x^3\right )}{2 \left (b x^3+a\right )}dx^3}{b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {\sqrt {d x^3+c} \left (a (2 b c-3 a d)-2 b (b c-a d) x^3\right )}{b x^3+a}dx^3}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-5 a d) \int \frac {\sqrt {d x^3+c}}{b x^3+a}dx^3-\frac {4 \left (c+d x^3\right )^{3/2} (b c-a d)}{3 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-5 a d) \left (\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}+\frac {2 \sqrt {c+d x^3}}{b}\right )-\frac {4 \left (c+d x^3\right )^{3/2} (b c-a d)}{3 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (-\frac {a (4 b c-5 a d) \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}+\frac {2 \sqrt {c+d x^3}}{b}\right )-\frac {4 \left (c+d x^3\right )^{3/2} (b c-a d)}{3 d}}{2 b^2 (b c-a d)}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (-\frac {a^2 \left (c+d x^3\right )^{3/2}}{b^2 \left (a+b x^3\right ) (b c-a d)}-\frac {a (4 b c-5 a d) \left (\frac {2 \sqrt {c+d x^3}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )-\frac {4 \left (c+d x^3\right )^{3/2} (b c-a d)}{3 d}}{2 b^2 (b c-a d)}\right )\)

Input:

Int[(x^8*Sqrt[c + d*x^3])/(a + b*x^3)^2,x]
 

Output:

(-((a^2*(c + d*x^3)^(3/2))/(b^2*(b*c - a*d)*(a + b*x^3))) - ((-4*(b*c - a* 
d)*(c + d*x^3)^(3/2))/(3*d) + a*(4*b*c - 5*a*d)*((2*Sqrt[c + d*x^3])/b - ( 
2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3 
/2)))/(2*b^2*(b*c - a*d)))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 60

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]
 

rule 73

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]
 

rule 90

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]
 

rule 100

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])))
 

rule 221

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]
 

rule 948

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]]
 

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(-\frac {5 \left (-a d \left (b \,x^{3}+a \right ) \left (a d -\frac {4 b c}{5}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {d \,x^{3}+c}\, \left (-\frac {2 x^{3} \left (d \,x^{3}+c \right ) b^{2}}{15}-\frac {2 a \left (-5 d \,x^{3}+c \right ) b}{15}+d \,a^{2}\right ) \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, d \,b^{3} \left (b \,x^{3}+a \right )}\)	\(129\)
default	\(\frac {2 \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9 b^{2} d}+\frac {a^{2} \left (-\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{3}}-\frac {4 a \left (\sqrt {d \,x^{3}+c}-\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{3}}\)	\(148\)
risch	\(-\frac {2 \left (-b d \,x^{3}+6 a d -b c \right ) \sqrt {d \,x^{3}+c}}{9 d \,b^{3}}+\frac {5 a^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) d}{3 b^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {4 a \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) c}{3 b^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {a^{2} \sqrt {d \,x^{3}+c}}{3 b^{3} \left (b \,x^{3}+a \right )}\)	\(149\)
elliptic	\(-\frac {a^{2} \sqrt {d \,x^{3}+c}}{3 b^{3} \left (b \,x^{3}+a \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 b^{2}}+\frac {2 \left (-\frac {2 a d -b c}{b^{3}}-\frac {2 c}{3 b^{2}}\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 (5 a d -4 b c \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 )}{6 b^{3} d^{2}}\)	\(518\)

Input:

int(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-5/3/((a*d-b*c)*b)^(1/2)*(-a*d*(b*x^3+a)*(a*d-4/5*b*c)*arctan(b*(d*x^3+c)^ 
(1/2)/((a*d-b*c)*b)^(1/2))+(d*x^3+c)^(1/2)*(-2/15*x^3*(d*x^3+c)*b^2-2/15*a 
*(-5*d*x^3+c)*b+d*a^2)*((a*d-b*c)*b)^(1/2))/d/b^3/(b*x^3+a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (108) = 216\).

Time = 0.10 (sec) , antiderivative size = 469, normalized size of antiderivative = 3.55 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\left [-\frac {3 \, {\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 5 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \, {\left (2 \, {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \, {\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{18 \, {\left (a b^{5} c d - a^{2} b^{4} d^{2} + {\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}, -\frac {3 \, {\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 5 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \, {\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{9 \, {\left (a b^{5} c d - a^{2} b^{4} d^{2} + {\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}\right ] \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="fricas")
 

Output:

[-1/18*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*a^2*b*d^2)*x^3)*sqrt 
(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*sqrt(b^2*c 
- a*b*d))/(b*x^3 + a)) - 2*(2*(b^4*c*d - a*b^3*d^2)*x^6 + 2*a*b^3*c^2 - 17 
*a^2*b^2*c*d + 15*a^3*b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^ 
3)*sqrt(d*x^3 + c))/(a*b^5*c*d - a^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*x^3), 
 -1/9*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*a^2*b*d^2)*x^3)*sqrt( 
-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c 
)) - (2*(b^4*c*d - a*b^3*d^2)*x^6 + 2*a*b^3*c^2 - 17*a^2*b^2*c*d + 15*a^3* 
b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(a 
*b^5*c*d - a^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*x^3)]
 

Sympy [F]

\[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\int \frac {x^{8} \sqrt {c + d x^{3}}}{\left (a + b x^{3}\right )^{2}}\, dx \] Input:

integrate(x**8*(d*x**3+c)**(1/2)/(b*x**3+a)**2,x)
 

Output:

Integral(x**8*sqrt(c + d*x**3)/(a + b*x**3)**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,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
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.03 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=-\frac {\sqrt {d x^{3} + c} a^{2} d}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{3}} - \frac {{\left (4 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{4} d^{2} - 6 \, \sqrt {d x^{3} + c} a b^{3} d^{3}\right )}}{9 \, b^{6} d^{3}} \] Input:

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="giac")
 

Output:

-1/3*sqrt(d*x^3 + c)*a^2*d/(((d*x^3 + c)*b - b*c + a*d)*b^3) - 1/3*(4*a*b* 
c - 5*a^2*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + 
 a*b*d)*b^3) + 2/9*((d*x^3 + c)^(3/2)*b^4*d^2 - 6*sqrt(d*x^3 + c)*a*b^3*d^ 
3)/(b^6*d^3)
 

Mupad [B] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.53 \[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,b^2}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {4\,c}{3\,b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}+\frac {2\,a\,d}{b^3}\right )}{3\,d}+\frac {a^2\,\left (\frac {2\,a\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {2\,b\,c}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )\,\sqrt {d\,x^3+c}}{b^2\,\left (b\,x^3+a\right )}+\frac {a\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (5\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{7/2}\,\sqrt {a\,d-b\,c}} \] Input:

int((x^8*(c + d*x^3)^(1/2))/(a + b*x^3)^2,x)
 

Output:

(2*x^3*(c + d*x^3)^(1/2))/(9*b^2) - ((c + d*x^3)^(1/2)*((4*c)/(3*b^2) - (2 
*b^2*c - 2*a*b*d)/b^4 + (2*a*d)/b^3))/(3*d) + (a*log((2*b*c - a*d + b^(1/2 
)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i + b*d*x^3)/(a + b*x^3))*(5*a*d - 
4*b*c)*1i)/(6*b^(7/2)*(a*d - b*c)^(1/2)) + (a^2*((2*a*d)/(3*(2*b^2*c - 2*a 
*b*d)) - (2*b*c)/(3*(2*b^2*c - 2*a*b*d)))*(c + d*x^3)^(1/2))/(b^2*(a + b*x 
^3))
 

Reduce [F]

\[ \int \frac {x^8 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x)
 

Output:

(20*sqrt(c + d*x**3)*a**2*c*d - 10*sqrt(c + d*x**3)*a**2*d**2*x**3 - 4*sqr 
t(c + d*x**3)*a*b*c**2 + 22*sqrt(c + d*x**3)*a*b*c*d*x**3 + 2*sqrt(c + d*x 
**3)*a*b*d**2*x**6 - 4*sqrt(c + d*x**3)*b**2*c**2*x**3 - 4*sqrt(c + d*x**3 
)*b**2*c*d*x**6 + 45*int((sqrt(c + d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x** 
3 - 2*a**2*b*c**2 + 2*a**2*b*d**2*x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d 
*x**6 + a*b**2*d**2*x**9 - 2*b**3*c**2*x**6 - 2*b**3*c*d*x**9),x)*a**5*d** 
4 - 171*int((sqrt(c + d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x**3 - 2*a**2*b* 
c**2 + 2*a**2*b*d**2*x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + a*b** 
2*d**2*x**9 - 2*b**3*c**2*x**6 - 2*b**3*c*d*x**9),x)*a**4*b*c*d**3 + 45*in 
t((sqrt(c + d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x**3 - 2*a**2*b*c**2 + 2*a 
**2*b*d**2*x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + a*b**2*d**2*x** 
9 - 2*b**3*c**2*x**6 - 2*b**3*c*d*x**9),x)*a**4*b*d**4*x**3 + 198*int((sqr 
t(c + d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x**3 - 2*a**2*b*c**2 + 2*a**2*b* 
d**2*x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 - 2* 
b**3*c**2*x**6 - 2*b**3*c*d*x**9),x)*a**3*b**2*c**2*d**2 - 171*int((sqrt(c 
 + d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x**3 - 2*a**2*b*c**2 + 2*a**2*b*d** 
2*x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 - 2*b** 
3*c**2*x**6 - 2*b**3*c*d*x**9),x)*a**3*b**2*c*d**3*x**3 - 72*int((sqrt(c + 
 d*x**3)*x**5)/(a**3*c*d + a**3*d**2*x**3 - 2*a**2*b*c**2 + 2*a**2*b*d**2* 
x**6 - 4*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 - 2*b*...