\(\int \frac {A+B x^3}{x^8 (a+b x^3)} \, dx\) [73]

Optimal result

Integrand size = 20, antiderivative size = 184 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}+\frac {b^{4/3} (A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{10/3}}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}-\frac {b^{4/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{10/3}} \] Output:

-1/7*A/a/x^7+1/4*(A*b-B*a)/a^2/x^4-b*(A*b-B*a)/a^3/x+1/3*b^(4/3)*(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^(10/3)+1/3*b^ 
(4/3)*(A*b-B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(10/3)-1/6*b^(4/3)*(A*b-B*a)*ln(a^ 
(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(10/3)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=\frac {-\frac {12 a^{7/3} A}{x^7}+\frac {21 a^{4/3} (A b-a B)}{x^4}+\frac {84 \sqrt [3]{a} b (-A b+a B)}{x}+28 \sqrt {3} b^{4/3} (A b-a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+28 b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+14 b^{4/3} (-A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{84 a^{10/3}} \] Input:

Integrate[(A + B*x^3)/(x^8*(a + b*x^3)),x]
 

Output:

((-12*a^(7/3)*A)/x^7 + (21*a^(4/3)*(A*b - a*B))/x^4 + (84*a^(1/3)*b*(-(A*b 
) + a*B))/x + 28*Sqrt[3]*b^(4/3)*(A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^( 
1/3))/Sqrt[3]] + 28*b^(4/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x] + 14*b^(4 
/3)*(-(A*b) + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(84*a^( 
10/3))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {955, 847, 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.

\(\displaystyle \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx\)

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(A b-a B) \int \frac {1}{x^5 \left (b x^3+a\right )}dx}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \int \frac {1}{x^2 \left (b x^3+a\right )}dx}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \int \frac {x}{b x^3+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 821

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 1142

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{4 a x^4}\right )}{a}-\frac {A}{7 a x^7}\)

Input:

Int[(A + B*x^3)/(x^8*(a + b*x^3)),x]
 

Output:

-1/7*A/(a*x^7) - ((A*b - a*B)*(-1/4*1/(a*x^4) - (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))/a))/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 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 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 = 0.82 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {A}{7 a \,x^{7}}-\frac {-A b +B a}{4 a^{2} x^{4}}-\frac {b \left (A b -B a \right )}{a^{3} x}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) b^{2} \left (A b -B a \right )}{a^{3}}\)	\(150\)
risch	\(\frac {-\frac {\left (A b -B a \right ) b \,x^{6}}{a^{3}}+\frac {\left (A b -B a \right ) x^{3}}{4 a^{2}}-\frac {A}{7 a}}{x^{7}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{10} \textit {\_Z}^{3}-A^{3} b^{7}+3 A^{2} B a \,b^{6}-3 A \,B^{2} a^{2} b^{5}+B^{3} a^{3} b^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{10} \textit {\_R}^{3}+3 A^{3} b^{7}-9 A^{2} B a \,b^{6}+9 A \,B^{2} a^{2} b^{5}-3 B^{3} a^{3} b^{4}\right ) x +\left (-A \,a^{7} b^{2}+B \,a^{8} b \right ) \textit {\_R}^{2}\right )\right )}{3}\)	\(176\)

Input:

int((B*x^3+A)/x^8/(b*x^3+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=\frac {28 \, \sqrt {3} {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 14 \, {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 28 \, {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 84 \, {\left (B a b - A b^{2}\right )} x^{6} - 21 \, {\left (B a^{2} - A a b\right )} x^{3} - 12 \, A a^{2}}{84 \, a^{3} x^{7}} \] Input:

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="fricas")
 

Output:

1/84*(28*sqrt(3)*(B*a*b - A*b^2)*x^7*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(b/a 
)^(1/3) - 1/3*sqrt(3)) + 14*(B*a*b - A*b^2)*x^7*(b/a)^(1/3)*log(b*x^2 - a* 
x*(b/a)^(2/3) + a*(b/a)^(1/3)) - 28*(B*a*b - A*b^2)*x^7*(b/a)^(1/3)*log(b* 
x + a*(b/a)^(2/3)) + 84*(B*a*b - A*b^2)*x^6 - 21*(B*a^2 - A*a*b)*x^3 - 12* 
A*a^2)/(a^3*x^7)
 

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=\operatorname {RootSum} {\left (27 t^{3} a^{10} - A^{3} b^{7} + 3 A^{2} B a b^{6} - 3 A B^{2} a^{2} b^{5} + B^{3} a^{3} b^{4}, \left ( t \mapsto t \log {\left (\frac {9 t^{2} a^{7}}{A^{2} b^{5} - 2 A B a b^{4} + B^{2} a^{2} b^{3}} + x \right )} \right )\right )} + \frac {- 4 A a^{2} + x^{6} \left (- 28 A b^{2} + 28 B a b\right ) + x^{3} \cdot \left (7 A a b - 7 B a^{2}\right )}{28 a^{3} x^{7}} \] Input:

integrate((B*x**3+A)/x**8/(b*x**3+a),x)
 

Output:

RootSum(27*_t**3*a**10 - A**3*b**7 + 3*A**2*B*a*b**6 - 3*A*B**2*a**2*b**5 
+ B**3*a**3*b**4, Lambda(_t, _t*log(9*_t**2*a**7/(A**2*b**5 - 2*A*B*a*b**4 
 + B**2*a**2*b**3) + x))) + (-4*A*a**2 + x**6*(-28*A*b**2 + 28*B*a*b) + x* 
*3*(7*A*a*b - 7*B*a**2))/(28*a**3*x**7)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=\frac {\sqrt {3} {\left (B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a b - A b^{2}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a b - A b^{2}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {28 \, {\left (B a b - A b^{2}\right )} x^{6} - 7 \, {\left (B a^{2} - A a b\right )} x^{3} - 4 \, A a^{2}}{28 \, a^{3} x^{7}} \] Input:

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="maxima")
 

Output:

1/3*sqrt(3)*(B*a*b - A*b^2)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^( 
1/3))/(a^3*(a/b)^(1/3)) + 1/6*(B*a*b - A*b^2)*log(x^2 - x*(a/b)^(1/3) + (a 
/b)^(2/3))/(a^3*(a/b)^(1/3)) - 1/3*(B*a*b - A*b^2)*log(x + (a/b)^(1/3))/(a 
^3*(a/b)^(1/3)) + 1/28*(28*(B*a*b - A*b^2)*x^6 - 7*(B*a^2 - A*a*b)*x^3 - 4 
*A*a^2)/(a^3*x^7)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=-\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{4}} - \frac {{\left (B a b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b^{3} \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{4}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{4}} + \frac {28 \, B a b x^{6} - 28 \, A b^{2} x^{6} - 7 \, B a^{2} x^{3} + 7 \, A a b x^{3} - 4 \, A a^{2}}{28 \, a^{3} x^{7}} \] Input:

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="giac")
 

Output:

-1/3*sqrt(3)*((-a*b^2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*arctan(1/3*sqrt(3)* 
(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/a^4 - 1/3*(B*a*b^2*(-a/b)^(1/3) - A*b^3 
*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^4 + 1/6*((-a*b^2) 
^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/ 
a^4 + 1/28*(28*B*a*b*x^6 - 28*A*b^2*x^6 - 7*B*a^2*x^3 + 7*A*a*b*x^3 - 4*A* 
a^2)/(a^3*x^7)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=\frac {b^{4/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,a^{10/3}}-\frac {\frac {A}{7\,a}-\frac {x^3\,\left (A\,b-B\,a\right )}{4\,a^2}+\frac {b\,x^6\,\left (A\,b-B\,a\right )}{a^3}}{x^7}+\frac {b^{4/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{10/3}}-\frac {b^{4/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{10/3}} \] Input:

int((A + B*x^3)/(x^8*(a + b*x^3)),x)
 

Output:

(b^(4/3)*log(b^(1/3)*x + a^(1/3))*(A*b - B*a))/(3*a^(10/3)) - (A/(7*a) - ( 
x^3*(A*b - B*a))/(4*a^2) + (b*x^6*(A*b - B*a))/a^3)/x^7 + (b^(4/3)*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^(10/3)) - (b^(4/3)*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^(10/3))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx=-\frac {1}{7 x^{7}} \] Input:

int((B*x^3+A)/x^8/(b*x^3+a),x)
 

Output:

( - 1)/(7*x**7)