\(\int \frac {A+B x^3}{x^2 (a+b x^3)} \, dx\) [69]

Optimal result

Integrand size = 20, antiderivative size = 147 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=-\frac {A}{a x}+\frac {(A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} b^{2/3}}+\frac {(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}-\frac {(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3}} \] Output:

-A/a/x+1/3*(A*b-B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^( 
1/2)/a^(4/3)/b^(2/3)+1/3*(A*b-B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)/b^(2/3)-1 
/6*(A*b-B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(4/3)/b^(2/3)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=\frac {-6 \sqrt [3]{a} A b^{2/3}+2 \sqrt {3} (A b-a B) x \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 (A b-a B) x \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-(A b-a B) x \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3} x} \] Input:

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

Output:

(-6*a^(1/3)*A*b^(2/3) + 2*Sqrt[3]*(A*b - a*B)*x*ArcTan[(1 - (2*b^(1/3)*x)/ 
a^(1/3))/Sqrt[3]] + 2*(A*b - a*B)*x*Log[a^(1/3) + b^(1/3)*x] - (A*b - a*B) 
*x*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(4/3)*b^(2/3)*x)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {955, 821, 16, 1142, 25, 27, 1082, 217, 1103}

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.

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

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(A b-a B) \int \frac {x}{b x^3+a}dx}{a}-\frac {A}{a x}\)

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

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]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {A}{a x}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \left (A b -B a \right )}{a}\)	\(114\)
risch	\(-\frac {A}{a x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b^{2} a^{4}-A^{3} b^{3}+3 A^{2} B a \,b^{2}-3 A \,B^{2} a^{2} b +B^{3} a^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{4} b^{2}+3 A^{3} b^{3}-9 A^{2} B a \,b^{2}+9 A \,B^{2} a^{2} b -3 B^{3} a^{3}\right ) x +\left (-a^{3} b^{2} A +a^{4} b B \right ) \textit {\_R}^{2}\right )\right )}{3}\)	\(137\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.53 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=\left [-\frac {6 \, A a b^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (B a^{2} b - A a b^{2}\right )} x \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b - 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) - \left (a b^{2}\right )^{\frac {2}{3}} {\left (B a - A b\right )} x \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} {\left (B a - A b\right )} x \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a^{2} b^{2} x}, -\frac {6 \, A a b^{2} + 6 \, \sqrt {\frac {1}{3}} {\left (B a^{2} b - A a b^{2}\right )} x \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x - \left (a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) - \left (a b^{2}\right )^{\frac {2}{3}} {\left (B a - A b\right )} x \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} {\left (B a - A b\right )} x \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a^{2} b^{2} x}\right ] \] Input:

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

Output:

[-1/6*(6*A*a*b^2 + 3*sqrt(1/3)*(B*a^2*b - A*a*b^2)*x*sqrt(-(a*b^2)^(1/3)/a 
)*log((2*b^2*x^3 - a*b - 3*sqrt(1/3)*(a*b*x + 2*(a*b^2)^(2/3)*x^2 - (a*b^2 
)^(1/3)*a)*sqrt(-(a*b^2)^(1/3)/a) - 3*(a*b^2)^(2/3)*x)/(b*x^3 + a)) - (a*b 
^2)^(2/3)*(B*a - A*b)*x*log(b^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) + 
 2*(a*b^2)^(2/3)*(B*a - A*b)*x*log(b*x + (a*b^2)^(1/3)))/(a^2*b^2*x), -1/6 
*(6*A*a*b^2 + 6*sqrt(1/3)*(B*a^2*b - A*a*b^2)*x*sqrt((a*b^2)^(1/3)/a)*arct 
an(-sqrt(1/3)*(2*b*x - (a*b^2)^(1/3))*sqrt((a*b^2)^(1/3)/a)/b) - (a*b^2)^( 
2/3)*(B*a - A*b)*x*log(b^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) + 2*(a 
*b^2)^(2/3)*(B*a - A*b)*x*log(b*x + (a*b^2)^(1/3)))/(a^2*b^2*x)]
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=- \frac {A}{a x} + \operatorname {RootSum} {\left (27 t^{3} a^{4} b^{2} - A^{3} b^{3} + 3 A^{2} B a b^{2} - 3 A B^{2} a^{2} b + B^{3} a^{3}, \left ( t \mapsto t \log {\left (\frac {9 t^{2} a^{3} b}{A^{2} b^{2} - 2 A B a b + B^{2} a^{2}} + x \right )} \right )\right )} \] Input:

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

Output:

-A/(a*x) + RootSum(27*_t**3*a**4*b**2 - A**3*b**3 + 3*A**2*B*a*b**2 - 3*A* 
B**2*a**2*b + B**3*a**3, Lambda(_t, _t*log(9*_t**2*a**3*b/(A**2*b**2 - 2*A 
*B*a*b + B**2*a**2) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=\frac {\sqrt {3} {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a - A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a - A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {A}{a x} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=\frac {\sqrt {3} {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {1}{3}} a} - \frac {{\left (B a - A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {1}{3}} a} - \frac {{\left (B a \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{2}} - \frac {A}{a x} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,a^{4/3}\,b^{2/3}}-\frac {A}{a\,x}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{4/3}\,b^{2/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{4/3}\,b^{2/3}} \] Input:

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

Output:

(log(b^(1/3)*x + a^(1/3))*(A*b - B*a))/(3*a^(4/3)*b^(2/3)) - A/(a*x) + (lo 
g(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(A*b 
- B*a))/(3*a^(4/3)*b^(2/3)) - (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1 
/3))*((3^(1/2)*1i)/2 + 1/2)*(A*b - B*a))/(3*a^(4/3)*b^(2/3))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )} \, dx=-\frac {1}{x} \] Input:

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

Output:

( - 1)/x