\(\int \frac {x^4 (A+B x^3)}{a+b x^3} \, dx\) [65]

Optimal result

Integrand size = 20, antiderivative size = 167 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {(A b-a B) x^2}{2 b^2}+\frac {B x^5}{5 b}+\frac {a^{2/3} (A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}+\frac {a^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}-\frac {a^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}} \] Output:

1/2*(A*b-B*a)*x^2/b^2+1/5*B*x^5/b+1/3*a^(2/3)*(A*b-B*a)*arctan(1/3*(a^(1/3 
)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/b^(8/3)+1/3*a^(2/3)*(A*b-B*a)*ln(a 
^(1/3)+b^(1/3)*x)/b^(8/3)-1/6*a^(2/3)*(A*b-B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3) 
*x+b^(2/3)*x^2)/b^(8/3)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {15 b^{2/3} (A b-a B) x^2+6 b^{5/3} B x^5-10 \sqrt {3} a^{2/3} (-A b+a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-10 a^{2/3} (-A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+5 a^{2/3} (-A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{30 b^{8/3}} \] Input:

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

Output:

(15*b^(2/3)*(A*b - a*B)*x^2 + 6*b^(5/3)*B*x^5 - 10*Sqrt[3]*a^(2/3)*(-(A*b) 
 + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 10*a^(2/3)*(-(A*b) + 
 a*B)*Log[a^(1/3) + b^(1/3)*x] + 5*a^(2/3)*(-(A*b) + a*B)*Log[a^(2/3) - a^ 
(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(30*b^(8/3))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.94, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {959, 843, 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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(A b-a B) \int \frac {x^4}{b x^3+a}dx}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \int \frac {x}{b x^3+a}dx}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(A b-a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{b}+\frac {B x^5}{5 b}\)

Input:

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

Output:

(B*x^5)/(5*b) + ((A*b - a*B)*(x^2/(2*b) - (a*(-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))))/b))/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 843

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

rule 959

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

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 [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.81 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.39

method	result	size
risch	\(\frac {B \,x^{5}}{5 b}+\frac {A \,x^{2}}{2 b}-\frac {B a \,x^{2}}{2 b^{2}}+\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-A b +B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{3 b^{3}}\)	\(65\)
default	\(\frac {\frac {b B \,x^{5}}{5}+\frac {\left (A b -B a \right ) x^{2}}{2}}{b^{2}}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a \left (A b -B a \right )}{b^{2}}\)	\(131\)

Input:

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

Output:

1/5*B*x^5/b+1/2/b*A*x^2-1/2/b^2*B*a*x^2+1/3/b^3*a*sum((-A*b+B*a)/_R*ln(x-_ 
R),_R=RootOf(_Z^3*b+a))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {6 \, B b x^{5} - 15 \, {\left (B a - A b\right )} x^{2} + 10 \, \sqrt {3} {\left (B a - A b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 5 \, {\left (B a - A b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (B a - A b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{30 \, b^{2}} \] Input:

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

Output:

1/30*(6*B*b*x^5 - 15*(B*a - A*b)*x^2 + 10*sqrt(3)*(B*a - A*b)*(a^2/b^2)^(1 
/3)*arctan(1/3*(2*sqrt(3)*b*x*(a^2/b^2)^(1/3) - sqrt(3)*a)/a) + 5*(B*a - A 
*b)*(a^2/b^2)^(1/3)*log(a*x^2 - b*x*(a^2/b^2)^(2/3) + a*(a^2/b^2)^(1/3)) - 
 10*(B*a - A*b)*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)))/b^2
 

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.68 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {B x^{5}}{5 b} + x^{2} \left (\frac {A}{2 b} - \frac {B a}{2 b^{2}}\right ) + \operatorname {RootSum} {\left (27 t^{3} b^{8} - A^{3} a^{2} b^{3} + 3 A^{2} B a^{3} b^{2} - 3 A B^{2} a^{4} b + B^{3} a^{5}, \left ( t \mapsto t \log {\left (\frac {9 t^{2} b^{5}}{A^{2} a b^{2} - 2 A B a^{2} b + B^{2} a^{3}} + x \right )} \right )\right )} \] Input:

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

Output:

B*x**5/(5*b) + x**2*(A/(2*b) - B*a/(2*b**2)) + RootSum(27*_t**3*b**8 - A** 
3*a**2*b**3 + 3*A**2*B*a**3*b**2 - 3*A*B**2*a**4*b + B**3*a**5, Lambda(_t, 
 _t*log(9*_t**2*b**5/(A**2*a*b**2 - 2*A*B*a**2*b + B**2*a**3) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {\sqrt {3} {\left (B a^{2} - A a b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, B b x^{5} - 5 \, {\left (B a - A b\right )} x^{2}}{10 \, b^{2}} + \frac {{\left (B a^{2} - A a b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a^{2} - A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:

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

Output:

1/3*sqrt(3)*(B*a^2 - A*a*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^( 
1/3))/(b^3*(a/b)^(1/3)) + 1/10*(2*B*b*x^5 - 5*(B*a - A*b)*x^2)/b^2 + 1/6*( 
B*a^2 - A*a*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(1/3)) - 
1/3*(B*a^2 - A*a*b)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(1/3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.24 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=-\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{4}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{4}} - \frac {{\left (B a^{2} b^{3} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A a b^{4} \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{5}} + \frac {2 \, B b^{4} x^{5} - 5 \, B a b^{3} x^{2} + 5 \, A b^{4} x^{2}}{10 \, b^{5}} \] Input:

integrate(x^4*(B*x^3+A)/(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))/b^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))/b^4 - 1/3*(B*a^2*b^3 
*(-a/b)^(1/3) - A*a*b^4*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3 
)))/(a*b^5) + 1/10*(2*B*b^4*x^5 - 5*B*a*b^3*x^2 + 5*A*b^4*x^2)/b^5
 

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=x^2\,\left (\frac {A}{2\,b}-\frac {B\,a}{2\,b^2}\right )+\frac {B\,x^5}{5\,b}+\frac {a^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,b^{8/3}}+\frac {a^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,b^{8/3}}-\frac {a^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,b^{8/3}} \] Input:

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

Output:

x^2*(A/(2*b) - (B*a)/(2*b^2)) + (B*x^5)/(5*b) + (a^(2/3)*log(b^(1/3)*x + a 
^(1/3))*(A*b - B*a))/(3*b^(8/3)) + (a^(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*b^(8/3)) - (a^(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*b^(8/3))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {x^4 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {x^{5}}{5} \] Input:

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

Output:

x**5/5