\(\int \frac {A+B x^3}{x^4 (a+b x^3)^{2/3}} \, dx\) [347]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {A \sqrt [3]{a+b x^3}}{3 a x^3}+\frac {(2 A b-3 a B) \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3}}+\frac {(2 A b-3 a B) \log (x)}{6 a^{5/3}}-\frac {(2 A b-3 a B) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{5/3}} \] Output:

-1/3*A*(b*x^3+a)^(1/3)/a/x^3+1/9*(2*A*b-3*B*a)*arctan(1/3*(a^(1/3)+2*(b*x^ 
3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/a^(5/3)+1/6*(2*A*b-3*B*a)*ln(x)/a^(5/ 
3)-1/6*(2*A*b-3*B*a)*ln(a^(1/3)-(b*x^3+a)^(1/3))/a^(5/3)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {-\frac {6 a^{2/3} A \sqrt [3]{a+b x^3}}{x^3}+2 \sqrt {3} (2 A b-3 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 (-2 A b+3 a B) \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )+(2 A b-3 a B) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 a^{5/3}} \] Input:

Integrate[(A + B*x^3)/(x^4*(a + b*x^3)^(2/3)),x]
 

Output:

((-6*a^(2/3)*A*(a + b*x^3)^(1/3))/x^3 + 2*Sqrt[3]*(2*A*b - 3*a*B)*ArcTan[( 
1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 2*(-2*A*b + 3*a*B)*Log[-a^(1 
/3) + (a + b*x^3)^(1/3)] + (2*A*b - 3*a*B)*Log[a^(2/3) + a^(1/3)*(a + b*x^ 
3)^(1/3) + (a + b*x^3)^(2/3)])/(18*a^(5/3))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 87, 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.

\(\displaystyle \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {B x^3+A}{x^6 \left (b x^3+a\right )^{2/3}}dx^3\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(A + B*x^3)/(x^4*(a + b*x^3)^(2/3)),x]
 

Output:

(-((A*(a + b*x^3)^(1/3))/(a*x^3)) - ((2*A*b - 3*a*B)*(-((Sqrt[3]*ArcTan[(1 
 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x^3]/(2*a^(2/3) 
) + (3*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(2*a^(2/3))))/(3*a))/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 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 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 = 1.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-3 A \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}+\left (A b -\frac {3 B a}{2}\right ) \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )\right ) x^{3}}{9 a^{\frac {5}{3}} x^{3}}\)	\(112\)

Input:

int((B*x^3+A)/x^4/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

1/9*(-3*A*(b*x^3+a)^(1/3)*a^(2/3)+(A*b-3/2*B*a)*(2*arctan(1/3*(a^(1/3)+2*( 
b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)+ln((b*x^3+a)^(2/3)+a^(1/3)*(b*x^3 
+a)^(1/3)+a^(2/3))-2*ln((b*x^3+a)^(1/3)-a^(1/3)))*x^3)/a^(5/3)/x^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {6 \, \sqrt {\frac {1}{3}} {\left (3 \, B a^{2} - 2 \, A a b\right )} x^{3} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{a^{2}}\right ) + {\left (3 \, B a - 2 \, A b\right )} \left (-a^{2}\right )^{\frac {2}{3}} x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (3 \, B a - 2 \, A b\right )} \left (-a^{2}\right )^{\frac {2}{3}} x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} A a^{2}}{18 \, a^{3} x^{3}} \] Input:

integrate((B*x^3+A)/x^4/(b*x^3+a)^(2/3),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-1/18*(6*sqrt(1/3)*(3*B*a^2 - 2*A*a*b)*x^3*sqrt(-(-a^2)^(1/3))*arctan(-sqr 
t(1/3)*((-a^2)^(1/3)*a - 2*(b*x^3 + a)^(1/3)*(-a^2)^(2/3))*sqrt(-(-a^2)^(1 
/3))/a^2) + (3*B*a - 2*A*b)*(-a^2)^(2/3)*x^3*log((b*x^3 + a)^(2/3)*a - (-a 
^2)^(1/3)*a + (b*x^3 + a)^(1/3)*(-a^2)^(2/3)) - 2*(3*B*a - 2*A*b)*(-a^2)^( 
2/3)*x^3*log((b*x^3 + a)^(1/3)*a - (-a^2)^(2/3)) + 6*(b*x^3 + a)^(1/3)*A*a 
^2)/(a^3*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.61 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=- \frac {A \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{5} \Gamma \left (\frac {8}{3}\right )} - \frac {B \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{2} \Gamma \left (\frac {5}{3}\right )} \] Input:

integrate((B*x**3+A)/x**4/(b*x**3+a)**(2/3),x)
 

Output:

-A*gamma(5/3)*hyper((2/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**3))/(3*b** 
(2/3)*x**5*gamma(8/3)) - B*gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_pola 
r(I*pi)/(b*x**3))/(3*b**(2/3)*x**2*gamma(5/3))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.55 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {1}{6} \, B {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {2 \, \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{a^{\frac {2}{3}}}\right )} + \frac {1}{9} \, A {\left (\frac {2 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{{\left (b x^{3} + a\right )} a - a^{2}} + \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{a^{\frac {5}{3}}}\right )} \] Input:

integrate((B*x^3+A)/x^4/(b*x^3+a)^(2/3),x, algorithm="maxima")
 

Output:

-1/6*B*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/ 
3))/a^(2/3) + log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) 
/a^(2/3) - 2*log((b*x^3 + a)^(1/3) - a^(1/3))/a^(2/3)) + 1/9*A*(2*sqrt(3)* 
b*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a^(5/3) - 3* 
(b*x^3 + a)^(1/3)*b/((b*x^3 + a)*a - a^2) + b*log((b*x^3 + a)^(2/3) + (b*x 
^3 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 2*b*log((b*x^3 + a)^(1/3) - a^( 
1/3))/a^(5/3))
 

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {1}{18} \, b {\left (\frac {2 \, \sqrt {3} {\left (3 \, B a - 2 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}} b} + \frac {{\left (3 \, B a - 2 \, A b\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}} b} - \frac {2 \, {\left (3 \, B a - 2 \, A b\right )} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}} b} + \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} A}{a b x^{3}}\right )} \] Input:

integrate((B*x^3+A)/x^4/(b*x^3+a)^(2/3),x, algorithm="giac")
 

Output:

-1/18*b*(2*sqrt(3)*(3*B*a - 2*A*b)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) 
 + a^(1/3))/a^(1/3))/(a^(5/3)*b) + (3*B*a - 2*A*b)*log((b*x^3 + a)^(2/3) + 
 (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/(a^(5/3)*b) - 2*(3*B*a - 2*A*b)*log( 
abs((b*x^3 + a)^(1/3) - a^(1/3)))/(a^(5/3)*b) + 6*(b*x^3 + a)^(1/3)*A/(a*b 
*x^3))
 

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.88 \[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\frac {\ln \left (\frac {A\,b-\sqrt {3}\,A\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,A\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (A\,b-\sqrt {3}\,A\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}+\frac {\ln \left (\frac {A\,b+\sqrt {3}\,A\,b\,1{}\mathrm {i}}{a^{2/3}}+\frac {2\,A\,b\,{\left (b\,x^3+a\right )}^{1/3}}{a}\right )\,\left (A\,b+\sqrt {3}\,A\,b\,1{}\mathrm {i}\right )}{9\,a^{5/3}}-\frac {\ln \left (3\,B\,{\left (b\,x^3+a\right )}^{1/3}+\frac {3\,a^{1/3}\,\left (B-\sqrt {3}\,B\,1{}\mathrm {i}\right )}{2}\right )\,\left (B-\sqrt {3}\,B\,1{}\mathrm {i}\right )}{6\,a^{2/3}}-\frac {\ln \left (3\,B\,{\left (b\,x^3+a\right )}^{1/3}+\frac {3\,a^{1/3}\,\left (B+\sqrt {3}\,B\,1{}\mathrm {i}\right )}{2}\right )\,\left (B+\sqrt {3}\,B\,1{}\mathrm {i}\right )}{6\,a^{2/3}}+\frac {B\,\ln \left (3\,B\,a^{1/3}-3\,B\,{\left (b\,x^3+a\right )}^{1/3}\right )}{3\,a^{2/3}}-\frac {2\,A\,b\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}-a^{1/3}\right )}{9\,a^{5/3}}-\frac {A\,{\left (b\,x^3+a\right )}^{1/3}}{3\,a\,x^3} \] Input:

int((A + B*x^3)/(x^4*(a + b*x^3)^(2/3)),x)
 

Output:

(log((A*b - 3^(1/2)*A*b*1i)/a^(2/3) + (2*A*b*(a + b*x^3)^(1/3))/a)*(A*b - 
3^(1/2)*A*b*1i))/(9*a^(5/3)) + (log((A*b + 3^(1/2)*A*b*1i)/a^(2/3) + (2*A* 
b*(a + b*x^3)^(1/3))/a)*(A*b + 3^(1/2)*A*b*1i))/(9*a^(5/3)) - (log(3*B*(a 
+ b*x^3)^(1/3) + (3*a^(1/3)*(B - 3^(1/2)*B*1i))/2)*(B - 3^(1/2)*B*1i))/(6* 
a^(2/3)) - (log(3*B*(a + b*x^3)^(1/3) + (3*a^(1/3)*(B + 3^(1/2)*B*1i))/2)* 
(B + 3^(1/2)*B*1i))/(6*a^(2/3)) + (B*log(3*B*a^(1/3) - 3*B*(a + b*x^3)^(1/ 
3)))/(3*a^(2/3)) - (2*A*b*log((a + b*x^3)^(1/3) - a^(1/3)))/(9*a^(5/3)) - 
(A*(a + b*x^3)^(1/3))/(3*a*x^3)
 

Reduce [F]

\[ \int \frac {A+B x^3}{x^4 \left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{4}}d x \right ) a +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x}d x \right ) b \] Input:

int((B*x^3+A)/x^4/(b*x^3+a)^(2/3),x)
 

Output:

int(1/((a + b*x**3)**(2/3)*x**4),x)*a + int(1/((a + b*x**3)**(2/3)*x),x)*b