\(\int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) [783]

Optimal result

Integrand size = 28, antiderivative size = 172 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {a \sqrt [3]{a+b x^3}}{b^2 d}-\frac {\left (a+b x^3\right )^{4/3}}{4 b^2 d}+\frac {\sqrt [3]{2} a^{4/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^2 d}+\frac {a^{4/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^2 d}-\frac {a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.09 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {15 a \sqrt [3]{a+b x^3}+3 b x^3 \sqrt [3]{a+b x^3}-4 \sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{2} a^{4/3} \log \left (-2 \sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}\right )-2 \sqrt [3]{2} a^{4/3} \log \left (2 a^{2/3}+2^{2/3} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{12 b^2 d} \] Input:

Integrate[(x^5*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
 

Output:

-1/12*(15*a*(a + b*x^3)^(1/3) + 3*b*x^3*(a + b*x^3)^(1/3) - 4*2^(1/3)*Sqrt 
[3]*a^(4/3)*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 4* 
2^(1/3)*a^(4/3)*Log[-2*a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3)] - 2*2^(1/3)*a^ 
(4/3)*Log[2*a^(2/3) + 2^(2/3)*a^(1/3)*(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x 
^3)^(2/3)])/(b^2*d)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {948, 27, 90, 60, 69, 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.

\(\displaystyle \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {a \left (2 a \left (\frac {\sqrt {3} \arctan \left (\frac {\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{2^{2/3} a^{2/3} b}+\frac {\log \left (a-b x^3\right )}{2\ 2^{2/3} a^{2/3} b}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2\ 2^{2/3} a^{2/3} b}\right )-\frac {3 \sqrt [3]{a+b x^3}}{b}\right )}{b}-\frac {3 \left (a+b x^3\right )^{4/3}}{4 b^2}}{3 d}\)

Input:

Int[(x^5*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
 

Output:

((-3*(a + b*x^3)^(4/3))/(4*b^2) + (a*((-3*(a + b*x^3)^(1/3))/b + 2*a*((Sqr 
t[3]*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/(2^(2/3)*a 
^(2/3)*b) + Log[a - b*x^3]/(2*2^(2/3)*a^(2/3)*b) - (3*Log[2^(1/3)*a^(1/3) 
- (a + b*x^3)^(1/3)])/(2*2^(2/3)*a^(2/3)*b))))/b)/(3*d)
 

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 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 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(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 = 1.70 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {\left (-3 b \,x^{3}-15 a \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}+2 a^{\frac {4}{3}} \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )\right ) 2^{\frac {1}{3}}}{12 b^{2} d}\)	\(125\)

Input:

int(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
 

Output:

1/12*((-3*b*x^3-15*a)*(b*x^3+a)^(1/3)+2*a^(4/3)*(2*arctan(1/3*(a^(1/3)+2^( 
2/3)*(b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)+ln((b*x^3+a)^(2/3)+2^(1/3)* 
a^(1/3)*(b*x^3+a)^(1/3)+2^(2/3)*a^(2/3))-2*ln((b*x^3+a)^(1/3)-2^(1/3)*a^(1 
/3)))*2^(1/3))/b^2/d
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 2 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \log \left (2^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right ) - 4 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a \log \left (2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (b x^{3} + 5 \, a\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, b^{2} d} \] Input:

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
 

Output:

-1/12*(4*sqrt(3)*2^(1/3)*(-a)^(1/3)*a*arctan(1/3*(sqrt(3)*2^(2/3)*(b*x^3 + 
 a)^(1/3)*(-a)^(2/3) + sqrt(3)*a)/a) + 2*2^(1/3)*(-a)^(1/3)*a*log(2^(2/3)* 
(-a)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(2/3)) - 4 
*2^(1/3)*(-a)^(1/3)*a*log(2^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(1/3)) + 3*(b*x 
^3 + 5*a)*(b*x^3 + a)^(1/3))/(b^2*d)
 

Sympy [F]

\[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{5} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \] Input:

integrate(x**5*(b*x**3+a)**(1/3)/(-b*d*x**3+a*d),x)
 

Output:

-Integral(x**5*(a + b*x**3)**(1/3)/(-a + b*x**3), x)/d
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{d} + \frac {2 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{d} - \frac {4 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}{d} - \frac {3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {4}{3}} + 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a\right )}}{d}}{12 \, b^{2}} \] Input:

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
 

Output:

1/12*(4*sqrt(3)*2^(1/3)*a^(4/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3 
) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/d + 2*2^(1/3)*a^(4/3)*log(2^(2/3)*a^(2/3 
) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/d - 4*2^(1/3)*a 
^(4/3)*log(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3))/d - 3*((b*x^3 + a)^(4/3) 
+ 4*(b*x^3 + a)^(1/3)*a)/d)/b^2
 

Giac [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.01 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {\sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{3 \, b^{2} d} + \frac {2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{6 \, b^{2} d} - \frac {2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left ({\left | -2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \right |}\right )}{3 \, b^{2} d} - \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{6} d^{3} + 4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{6} d^{3}}{4 \, b^{8} d^{4}} \] Input:

integrate(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="giac")
 

Output:

1/3*sqrt(3)*2^(1/3)*a^(4/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 
2*(b*x^3 + a)^(1/3))/a^(1/3))/(b^2*d) + 1/6*2^(1/3)*a^(4/3)*log(2^(2/3)*a^ 
(2/3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/(b^2*d) - 1 
/3*2^(1/3)*a^(4/3)*log(abs(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3)))/(b^2*d) 
- 1/4*((b*x^3 + a)^(4/3)*b^6*d^3 + 4*(b*x^3 + a)^(1/3)*a*b^6*d^3)/(b^8*d^4 
)
 

Mupad [B] (verification not implemented)

Time = 3.49 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.16 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b^2\,d}-\frac {a\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{3\,b^2\,d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {6\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}-\frac {6\,2^{1/3}\,a^{7/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^2\,d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {6\,a^2\,{\left (b\,x^3+a\right )}^{1/3}}{b^2\,d}+\frac {18\,2^{1/3}\,a^{7/3}\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^2\,d}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^2\,d} \] Input:

int((x^5*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x)
 

Output:

(2^(1/3)*a^(4/3)*log((6*a^2*(a + b*x^3)^(1/3))/(b^2*d) + (18*2^(1/3)*a^(7/ 
3)*((3^(1/2)*1i)/6 + 1/6))/(b^2*d))*((3^(1/2)*1i)/6 + 1/6))/(b^2*d) - (a*( 
a + b*x^3)^(1/3))/(b^2*d) - (2^(1/3)*a^(4/3)*log((a + b*x^3)^(1/3) - 2^(1/ 
3)*a^(1/3)))/(3*b^2*d) - (2^(1/3)*a^(4/3)*log((6*a^2*(a + b*x^3)^(1/3))/(b 
^2*d) - (6*2^(1/3)*a^(7/3)*((3^(1/2)*1i)/2 - 1/2))/(b^2*d))*((3^(1/2)*1i)/ 
2 - 1/2))/(3*b^2*d) - (a + b*x^3)^(4/3)/(4*b^2*d)
 

Reduce [F]

\[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a -\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}+8 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{-b^{2} x^{6}+a^{2}}d x \right ) a \,b^{2}}{4 b^{2} d} \] Input:

int(x^5*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)
 

Output:

(3*(a + b*x**3)**(1/3)*a - (a + b*x**3)**(1/3)*b*x**3 + 8*int(((a + b*x**3 
)**(1/3)*x**5)/(a**2 - b**2*x**6),x)*a*b**2)/(4*b**2*d)