\(\int \frac {A+B x^3}{x^2 (a+b x^3)^2} \, dx\) [87]

Optimal result

Integrand size = 20, antiderivative size = 180 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=-\frac {A}{a^2 x}-\frac {(A b-a B) x^2}{3 a^2 \left (a+b x^3\right )}+\frac {(4 A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} b^{2/3}}+\frac {(4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{7/3} b^{2/3}}-\frac {(4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{7/3} b^{2/3}} \] Output:

-A/a^2/x-1/3*(A*b-B*a)*x^2/a^2/(b*x^3+a)+1/9*(4*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^(7/3)/b^(2/3)+1/9*(4*A*b-B*a)*l 
n(a^(1/3)+b^(1/3)*x)/a^(7/3)/b^(2/3)-1/18*(4*A*b-B*a)*ln(a^(2/3)-a^(1/3)*b 
^(1/3)*x+b^(2/3)*x^2)/a^(7/3)/b^(2/3)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {-\frac {18 \sqrt [3]{a} A}{x}+\frac {6 \sqrt [3]{a} (-A b+a B) x^2}{a+b x^3}+\frac {2 \sqrt {3} (4 A b-a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {2 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {(-4 A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}}{18 a^{7/3}} \] Input:

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

Output:

((-18*a^(1/3)*A)/x + (6*a^(1/3)*(-(A*b) + a*B)*x^2)/(a + b*x^3) + (2*Sqrt[ 
3]*(4*A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(2/3) + (2 
*(4*A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) + ((-4*A*b + a*B)*Log[a^( 
2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(2/3))/(18*a^(7/3))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 847, 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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {(4 A b-a B) \int \frac {1}{x^2 \left (b x^3+a\right )}dx}{3 a b}+\frac {A b-a B}{3 a b x \left (a+b x^3\right )}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {(4 A b-a B) \left (-\frac {b \int \frac {x}{b x^3+a}dx}{a}-\frac {1}{a x}\right )}{3 a b}+\frac {A b-a B}{3 a b x \left (a+b x^3\right )}\)

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(A*b - a*B)/(3*a*b*x*(a + b*x^3)) + ((4*A*b - a*B)*(-(1/(a*x)) - (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))/(3*a*b)
 

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 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(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 = 0.83 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {A}{a^{2} x}-\frac {\frac {\left (\frac {A b}{3}-\frac {B a}{3}\right ) x^{2}}{b \,x^{3}+a}+\left (\frac {4 A b}{3}-\frac {B a}{3}\right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a^{2}}\)	\(139\)
risch	\(\frac {-\frac {\left (4 A b -B a \right ) x^{3}}{3 a^{2}}-\frac {A}{a}}{x \left (b \,x^{3}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} b^{2} \textit {\_Z}^{3}-64 A^{3} b^{3}+48 A^{2} B a \,b^{2}-12 A \,B^{2} a^{2} b +B^{3} a^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{7} b^{2}+192 A^{3} b^{3}-144 A^{2} B a \,b^{2}+36 A \,B^{2} a^{2} b -3 B^{3} a^{3}\right ) x +\left (-4 A \,a^{5} b^{2}+B \,a^{6} b \right ) \textit {\_R}^{2}\right )\right )}{9}\)	\(165\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.17 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/18*(18*A*a^2*b^2 - 6*(B*a^2*b^2 - 4*A*a*b^3)*x^3 + 3*sqrt(1/3)*((B*a^2 
*b^2 - 4*A*a*b^3)*x^4 + (B*a^3*b - 4*A*a^2*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)) - ((B*a* 
b - 4*A*b^2)*x^4 + (B*a^2 - 4*A*a*b)*x)*(a*b^2)^(2/3)*log(b^2*x^2 - (a*b^2 
)^(1/3)*b*x + (a*b^2)^(2/3)) + 2*((B*a*b - 4*A*b^2)*x^4 + (B*a^2 - 4*A*a*b 
)*x)*(a*b^2)^(2/3)*log(b*x + (a*b^2)^(1/3)))/(a^3*b^3*x^4 + a^4*b^2*x), -1 
/18*(18*A*a^2*b^2 - 6*(B*a^2*b^2 - 4*A*a*b^3)*x^3 + 6*sqrt(1/3)*((B*a^2*b^ 
2 - 4*A*a*b^3)*x^4 + (B*a^3*b - 4*A*a^2*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) - ((B*a*b - 
 4*A*b^2)*x^4 + (B*a^2 - 4*A*a*b)*x)*(a*b^2)^(2/3)*log(b^2*x^2 - (a*b^2)^( 
1/3)*b*x + (a*b^2)^(2/3)) + 2*((B*a*b - 4*A*b^2)*x^4 + (B*a^2 - 4*A*a*b)*x 
)*(a*b^2)^(2/3)*log(b*x + (a*b^2)^(1/3)))/(a^3*b^3*x^4 + a^4*b^2*x)]
 

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {- 3 A a + x^{3} \left (- 4 A b + B a\right )}{3 a^{3} x + 3 a^{2} b x^{4}} + \operatorname {RootSum} {\left (729 t^{3} a^{7} b^{2} - 64 A^{3} b^{3} + 48 A^{2} B a b^{2} - 12 A B^{2} a^{2} b + B^{3} a^{3}, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a^{5} b}{16 A^{2} b^{2} - 8 A B a b + B^{2} a^{2}} + x \right )} \right )\right )} \] Input:

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

Output:

(-3*A*a + x**3*(-4*A*b + B*a))/(3*a**3*x + 3*a**2*b*x**4) + RootSum(729*_t 
**3*a**7*b**2 - 64*A**3*b**3 + 48*A**2*B*a*b**2 - 12*A*B**2*a**2*b + B**3* 
a**3, Lambda(_t, _t*log(81*_t**2*a**5*b/(16*A**2*b**2 - 8*A*B*a*b + B**2*a 
**2) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {{\left (B a - 4 \, A b\right )} x^{3} - 3 \, A a}{3 \, {\left (a^{2} b x^{4} + a^{3} x\right )}} + \frac {\sqrt {3} {\left (B a - 4 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a - 4 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a - 4 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:

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

Output:

1/3*((B*a - 4*A*b)*x^3 - 3*A*a)/(a^2*b*x^4 + a^3*x) + 1/9*sqrt(3)*(B*a - 4 
*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^2*b*(a/b)^(1/ 
3)) + 1/18*(B*a - 4*A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*(a/ 
b)^(1/3)) - 1/9*(B*a - 4*A*b)*log(x + (a/b)^(1/3))/(a^2*b*(a/b)^(1/3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {\sqrt {3} {\left (B a - 4 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2}} - \frac {{\left (B a - 4 \, A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2}} - \frac {{\left (B a \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, A b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} + \frac {B a x^{3} - 4 \, A b x^{3} - 3 \, A a}{3 \, {\left (b x^{4} + a x\right )} a^{2}} \] Input:

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

Output:

1/9*sqrt(3)*(B*a - 4*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^( 
1/3))/((-a*b^2)^(1/3)*a^2) - 1/18*(B*a - 4*A*b)*log(x^2 + x*(-a/b)^(1/3) + 
 (-a/b)^(2/3))/((-a*b^2)^(1/3)*a^2) - 1/9*(B*a*(-a/b)^(1/3) - 4*A*b*(-a/b) 
^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^3 + 1/3*(B*a*x^3 - 4*A*b 
*x^3 - 3*A*a)/((b*x^4 + a*x)*a^2)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.00 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (4\,A\,b-B\,a\right )}{9\,a^{7/3}\,b^{2/3}}-\frac {\frac {A}{a}+\frac {x^3\,\left (4\,A\,b-B\,a\right )}{3\,a^2}}{b\,x^4+a\,x}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (4\,A\,b-B\,a\right )}{9\,a^{7/3}\,b^{2/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (4\,A\,b-B\,a\right )}{9\,a^{7/3}\,b^{2/3}} \] Input:

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

Output:

(log(b^(1/3)*x + a^(1/3))*(4*A*b - B*a))/(9*a^(7/3)*b^(2/3)) - (A/a + (x^3 
*(4*A*b - B*a))/(3*a^2))/(a*x + b*x^4) + (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/ 
3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(4*A*b - B*a))/(9*a^(7/3)*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)* 
(4*A*b - B*a))/(9*a^(7/3)*b^(2/3))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.47 \[ \int \frac {A+B x^3}{x^2 \left (a+b x^3\right )^2} \, dx=\frac {2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b x -6 b^{\frac {2}{3}} a^{\frac {1}{3}}-\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b x +2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b x}{6 b^{\frac {2}{3}} a^{\frac {4}{3}} x} \] Input:

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

Output:

(2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b*x - 6*b**( 
2/3)*a**(1/3) - log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b*x + 
2*log(a**(1/3) + b**(1/3)*x)*b*x)/(6*b**(2/3)*a**(1/3)*a*x)