\(\int \frac {x^5 (a+b x^3)^{2/3}}{c+d x^3} \, dx\) [694]

Optimal result

Integrand size = 24, antiderivative size = 188 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {c \left (a+b x^3\right )^{2/3}}{2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3}}+\frac {c (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^{8/3}}-\frac {c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3}} \] Output:

-1/2*c*(b*x^3+a)^(2/3)/d^2+1/5*(b*x^3+a)^(5/3)/b/d-1/3*c*(-a*d+b*c)^(2/3)* 
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)+1/6*c*(-a*d+b*c)^(2/3)*ln(d*x^3+c)/d^(8/3)-1/2*c*(-a*d+b*c)^(2/3) 
*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(8/3)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.21 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {\frac {3 d^{2/3} \left (a+b x^3\right )^{2/3} \left (-5 b c+2 a d+2 b d x^3\right )}{b}-10 \sqrt {3} c (b c-a d)^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-10 c (b c-a d)^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+5 c (b c-a d)^{2/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{30 d^{8/3}} \] Input:

Integrate[(x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x]
 

Output:

((3*d^(2/3)*(a + b*x^3)^(2/3)*(-5*b*c + 2*a*d + 2*b*d*x^3))/b - 10*Sqrt[3] 
*c*(b*c - a*d)^(2/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d) 
^(1/3))/Sqrt[3]] - 10*c*(b*c - a*d)^(2/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)* 
(a + b*x^3)^(1/3)] + 5*c*(b*c - a*d)^(2/3)*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)])/(30*d^( 
8/3))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {948, 90, 60, 68, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 68

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3}}{5 b d}-\frac {c \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}\right )\)

Input:

Int[(x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x]
 

Output:

((3*(a + b*x^3)^(5/3))/(5*b*d) - (c*((3*(a + b*x^3)^(2/3))/(2*d) - ((b*c - 
 a*d)*(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1 
/3))/Sqrt[3]])/(d^(2/3)*(b*c - a*d)^(1/3))) + Log[c + d*x^3]/(2*d^(2/3)*(b 
*c - a*d)^(1/3)) - (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/ 
(2*d^(2/3)*(b*c - a*d)^(1/3))))/d))/d)/3
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 68

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

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]]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {\frac {6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} d \left (\left (d \,x^{3}-\frac {5 c}{2}\right ) b +a d \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{5}+b c \left (a d -b c \right ) \left (-2 \arctan \left (\frac {2 \sqrt {3}\, \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}}{3}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} b \,d^{3}}\)	\(195\)

Input:

int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (149) = 298\).

Time = 0.19 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.88 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {10 \, \sqrt {3} b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 5 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )} + {\left (b c - a d\right )} \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, b c \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (-d \left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{d^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right ) - 3 \, {\left (2 \, b d x^{3} - 5 \, b c + 2 \, a d\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, b d^{2}} \] Input:

integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

-1/30*(10*sqrt(3)*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)*arctan( 
-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2) 
^(1/3) + sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 5*b*c*(-(b^2*c^2 - 2*a*b*c*d 
+ a^2*d^2)/d^2)^(1/3)*log((b*x^3 + a)^(1/3)*d*(-(b^2*c^2 - 2*a*b*c*d + a^2 
*d^2)/d^2)^(2/3) - (b*x^3 + a)^(2/3)*(b*c - a*d) + (b*c - a*d)*(-(b^2*c^2 
- 2*a*b*c*d + a^2*d^2)/d^2)^(1/3)) - 10*b*c*(-(b^2*c^2 - 2*a*b*c*d + a^2*d 
^2)/d^2)^(1/3)*log(-d*(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/d^2)^(2/3) - (b*x^ 
3 + a)^(1/3)*(b*c - a*d)) - 3*(2*b*d*x^3 - 5*b*c + 2*a*d)*(b*x^3 + a)^(2/3 
))/(b*d^2)
 

Sympy [F]

\[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int \frac {x^{5} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \] Input:

integrate(x**5*(b*x**3+a)**(2/3)/(d*x**3+c),x)
 

Output:

Integral(x**5*(a + b*x**3)**(2/3)/(c + d*x**3), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5*(b*x^3+a)^(2/3)/(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (149) = 298\).

Time = 0.16 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.63 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {{\left (b^{7} c^{2} d^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} - a b^{6} c d^{4} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{7} c d^{5} - a b^{6} d^{6}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{5} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{4} d^{4}}{10 \, b^{5} d^{5}} \] Input:

integrate(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

-1/3*(b^7*c^2*d^3*(-(b*c - a*d)/d)^(1/3) - a*b^6*c*d^4*(-(b*c - a*d)/d)^(1 
/3))*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^( 
1/3)))/(b^7*c*d^5 - a*b^6*d^6) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(2/3)*c*ar 
ctan(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^4 + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*c*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^4 - 
1/10*(5*(b*x^3 + a)^(2/3)*b^5*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^4*d^4)/(b^5*d^ 
5)
 

Mupad [B] (verification not implemented)

Time = 3.66 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.61 \[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b\,d}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {a}{2\,b\,d}+\frac {b^2\,c-a\,b\,d}{2\,b^2\,d^2}\right )-\frac {c\,\ln \left (\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}{d^3}-\frac {c^2\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{9\,d^{16/3}}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}-\frac {c\,\ln \left (\frac {c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}+\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}}+\frac {c\,\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d^3}-\frac {c^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{10/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{2/3}}{3\,d^{8/3}} \] Input:

int((x^5*(a + b*x^3)^(2/3))/(c + d*x^3),x)
 

Output:

(a + b*x^3)^(5/3)/(5*b*d) - (a + b*x^3)^(2/3)*(a/(2*b*d) + (b^2*c - a*b*d) 
/(2*b^2*d^2)) - (c*log(((a + b*x^3)^(1/3)*(b^2*c^4 + a^2*c^2*d^2 - 2*a*b*c 
^3*d))/d^3 - (c^2*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(9*d^(16/3)))*( 
a*d - b*c)^(2/3))/(3*d^(8/3)) - (c*log((c^2*((3^(1/2)*1i)/2 + 1/2)*(a*d - 
b*c)^(7/3))/d^(10/3) + (c^2*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d^3)*((3^(1/2 
)*1i)/2 - 1/2)*(a*d - b*c)^(2/3))/(3*d^(8/3)) + (c*log((c^2*(a + b*x^3)^(1 
/3)*(a*d - b*c)^2)/d^3 - (c^2*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(7/3))/d^ 
(10/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(2/3))/(3*d^(8/3))
 

Reduce [F]

\[ \int \frac {x^5 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {-3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a +2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}+10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b d -10 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c}{10 b d} \] Input:

int(x^5*(b*x^3+a)^(2/3)/(d*x^3+c),x)
 

Output:

( - 3*(a + b*x**3)**(2/3)*a + 2*(a + b*x**3)**(2/3)*b*x**3 + 10*int(((a + 
b*x**3)**(2/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b*d - 10* 
int(((a + b*x**3)**(2/3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b 
**2*c)/(10*b*d)