\(\int \frac {x^5}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\) [765]

Optimal result

Integrand size = 24, antiderivative size = 174 \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {a}{b (b c-a d) \sqrt [3]{a+b x^3}}-\frac {c \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{2/3} (b c-a d)^{4/3}}+\frac {c \log \left (c+d x^3\right )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} (b c-a d)^{4/3}} \] Output:

a/b/(-a*d+b*c)/(b*x^3+a)^(1/3)-1/3*c*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^(2/3)/(-a*d+b*c)^(4/3)+1/6*c*ln(d* 
x^3+c)/d^(2/3)/(-a*d+b*c)^(4/3)-1/2*c*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a 
)^(1/3))/d^(2/3)/(-a*d+b*c)^(4/3)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.30 \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {1}{6} \left (\frac {6 a}{\left (b^2 c-a b d\right ) \sqrt [3]{a+b x^3}}-\frac {2 \sqrt {3} c \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{2/3} (b c-a d)^{4/3}}-\frac {2 c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{d^{2/3} (b c-a d)^{4/3}}+\frac {c \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{d^{2/3} (b c-a d)^{4/3}}\right ) \] Input:

Integrate[x^5/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((6*a)/((b^2*c - a*b*d)*(a + b*x^3)^(1/3)) - (2*Sqrt[3]*c*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 
)^(4/3)) - (2*c*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(d^(2/ 
3)*(b*c - a*d)^(4/3)) + (c*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)])/(d^(2/3)*(b*c - a*d)^(4 
/3)))/6
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 87, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 68

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[x^5/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((3*a)/(b*(b*c - a*d)*(a + b*x^3)^(1/3)) + (c*(-((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))))/(b 
*c - a*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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 = 1.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(-\frac {\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) b c \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3}+\frac {\ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right ) b c \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{3}-\frac {\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right ) b c \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{6}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} a d}{\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (a d -b c \right ) d b}\)	\(224\)

Input:

int(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (141) = 282\).

Time = 0.13 (sec) , antiderivative size = 872, normalized size of antiderivative = 5.01 \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

integrate(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{5}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**5/(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(x**5/((a + b*x**3)**(4/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5/(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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (141) = 282\).

Time = 0.15 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.73 \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {\frac {6 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d^{2} - 2 \, \sqrt {3} a b c d^{3} + \sqrt {3} a^{2} d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}} + \frac {2 \, b c \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {6 \, a}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}}{6 \, b} \] Input:

integrate(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.52 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.37 \[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {a}{b\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {c^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {{\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {{\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \] Input:

int(x^5/((a + b*x^3)^(4/3)*(c + d*x^3)),x)
 

Output:

(log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b*c^3*d) - ((c - 3^(1/2)*c*1i)^2*(9*a^ 
4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c 
*d^5))/(36*d^(4/3)*(a*d - b*c)^(8/3)))*(c - 3^(1/2)*c*1i))/(6*d^(2/3)*(a*d 
 - b*c)^(4/3)) - (c*log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b*c^3*d) - (c^2*(9* 
a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b 
*c*d^5))/(9*d^(4/3)*(a*d - b*c)^(8/3))))/(3*d^(2/3)*(a*d - b*c)^(4/3)) - a 
/(b*(a + b*x^3)^(1/3)*(a*d - b*c)) + (log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b 
*c^3*d) - ((c + 3^(1/2)*c*1i)^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3* 
d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(36*d^(4/3)*(a*d - b*c)^(8/3)) 
)*(c + 3^(1/2)*c*1i))/(6*d^(2/3)*(a*d - b*c)^(4/3))
 

Reduce [F]

\[ \int \frac {x^5}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{5}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{6}}d x \] Input:

int(x^5/(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(x**5/((a + b*x**3)**(1/3)*a*c + (a + b*x**3)**(1/3)*a*d*x**3 + (a + b* 
x**3)**(1/3)*b*c*x**3 + (a + b*x**3)**(1/3)*b*d*x**6),x)