\(\int \frac {x (c+d x^3+e x^6+f x^9)}{a+b x^3} \, dx\) [238]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 28, antiderivative size = 245 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{11/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{11/3}} \]

[Out]

1/2*(a^2*f-a*b*e+b^2*d)*x^2/b^3+1/5*(-a*f+b*e)*x^5/b^2+1/8*f*x^8/b-1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(a^(1/
3)+b^(1/3)*x)/a^(1/3)/b^(11/3)+1/6*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^
(1/3)/b^(11/3)-1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(1/3)/b^
(11/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1850, 1502, 298, 31, 648, 631, 210, 642} \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {x^2 \left (a^2 f-a b e+b^2 d\right )}{2 b^3}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{\sqrt {3} \sqrt [3]{a} b^{11/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 \sqrt [3]{a} b^{11/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{6 \sqrt [3]{a} b^{11/3}}+\frac {x^5 (b e-a f)}{5 b^2}+\frac {f x^8}{8 b} \]

[In]

Int[(x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3),x]

[Out]

((b^2*d - a*b*e + a^2*f)*x^2)/(2*b^3) + ((b*e - a*f)*x^5)/(5*b^2) + (f*x^8)/(8*b) - ((b^3*c - a*b^2*d + a^2*b*
e - a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(1/3)*b^(11/3)) - ((b^3*c - a*b^2*d +
 a^2*b*e - a^3*f)*Log[a^(1/3) + b^(1/3)*x])/(3*a^(1/3)*b^(11/3)) + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^
(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(1/3)*b^(11/3))

Rule 31

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

Rule 210

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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1502

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 1850

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq =
Coeff[Pq, x, q]}, Dist[1/(b*(m + q + n*p + 1)), Int[(c*x)^m*ExpandToSum[b*(m + q + n*p + 1)*(Pq - Pqq*x^q) - a
*Pqq*(m + q - n + 1)*x^(q - n), x]*(a + b*x^n)^p, x], x] + Simp[Pqq*(c*x)^(m + q - n + 1)*((a + b*x^n)^(p + 1)
/(b*c^(q - n + 1)*(m + q + n*p + 1))), x]] /; NeQ[m + q + n*p + 1, 0] && q - n >= 0 && (IntegerQ[2*p] || Integ
erQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {f x^8}{8 b}+\frac {\int \frac {x \left (8 b c+8 b d x^3+8 (b e-a f) x^6\right )}{a+b x^3} \, dx}{8 b} \\ & = \frac {f x^8}{8 b}+\frac {\int \left (\frac {8 \left (b^2 d-a b e+a^2 f\right ) x}{b^2}+\frac {8 (b e-a f) x^4}{b}+\frac {8 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{b^2 \left (a+b x^3\right )}\right ) \, dx}{8 b} \\ & = \frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {x}{a+b x^3} \, dx}{b^3} \\ & = \frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 \sqrt [3]{a} b^{10/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 \sqrt [3]{a} b^{10/3}} \\ & = \frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b^{10/3}} \\ & = \frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{a} b^{11/3}} \\ & = \frac {\left (b^2 d-a b e+a^2 f\right ) x^2}{2 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^8}{8 b}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{11/3}}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{11/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{11/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {60 b^{2/3} \left (b^2 d-a b e+a^2 f\right ) x^2+24 b^{5/3} (b e-a f) x^5+15 b^{8/3} f x^8+\frac {40 \sqrt {3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {40 \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {20 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}}{120 b^{11/3}} \]

[In]

Integrate[(x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3),x]

[Out]

(60*b^(2/3)*(b^2*d - a*b*e + a^2*f)*x^2 + 24*b^(5/3)*(b*e - a*f)*x^5 + 15*b^(8/3)*f*x^8 + (40*Sqrt[3]*(-(b^3*c
) + a*b^2*d - a^2*b*e + a^3*f)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(1/3) + (40*(-(b^3*c) + a*b^2*d
- a^2*b*e + a^3*f)*Log[a^(1/3) + b^(1/3)*x])/a^(1/3) + (20*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(2/3) - a
^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(1/3))/(120*b^(11/3))

Maple [C] (verified)

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

Time = 1.52 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.46

method result size
risch \(\frac {f \,x^{8}}{8 b}-\frac {x^{5} a f}{5 b^{2}}+\frac {x^{5} e}{5 b}+\frac {a^{2} f \,x^{2}}{2 b^{3}}-\frac {a e \,x^{2}}{2 b^{2}}+\frac {d \,x^{2}}{2 b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-f \,a^{3}+a^{2} b e -a \,b^{2} d +b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{3 b^{4}}\) \(112\)
default \(\frac {\frac {b^{2} f \,x^{8}}{8}+\frac {\left (-a f b +b^{2} e \right ) x^{5}}{5}+\frac {\left (a^{2} f -a e b +b^{2} d \right ) x^{2}}{2}}{b^{3}}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right )}{b^{3}}\) \(173\)

[In]

int(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

1/8*f*x^8/b-1/5/b^2*x^5*a*f+1/5/b*x^5*e+1/2/b^3*a^2*f*x^2-1/2/b^2*a*e*x^2+1/2*d*x^2/b+1/3/b^4*sum((-a^3*f+a^2*
b*e-a*b^2*d+b^3*c)/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 568, normalized size of antiderivative = 2.32 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\left [\frac {15 \, a b^{4} f x^{8} + 24 \, {\left (a b^{4} e - a^{2} b^{3} f\right )} x^{5} + 60 \, {\left (a b^{4} d - a^{2} b^{3} e + a^{3} b^{2} f\right )} x^{2} - 60 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} \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 ) + 20 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (a b^{2}\right )^{\frac {2}{3}} \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 ) - 40 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{120 \, a b^{5}}, \frac {15 \, a b^{4} f x^{8} + 24 \, {\left (a b^{4} e - a^{2} b^{3} f\right )} x^{5} + 60 \, {\left (a b^{4} d - a^{2} b^{3} e + a^{3} b^{2} f\right )} x^{2} - 120 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} \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 ) + 20 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (a b^{2}\right )^{\frac {2}{3}} \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 ) - 40 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{120 \, a b^{5}}\right ] \]

[In]

integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a),x, algorithm="fricas")

[Out]

[1/120*(15*a*b^4*f*x^8 + 24*(a*b^4*e - a^2*b^3*f)*x^5 + 60*(a*b^4*d - a^2*b^3*e + a^3*b^2*f)*x^2 - 60*sqrt(1/3
)*(a*b^4*c - a^2*b^3*d + a^3*b^2*e - a^4*b*f)*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)) + 20*(b^3*c
 - a*b^2*d + a^2*b*e - a^3*f)*(a*b^2)^(2/3)*log(b^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) - 40*(b^3*c - a*b
^2*d + a^2*b*e - a^3*f)*(a*b^2)^(2/3)*log(b*x + (a*b^2)^(1/3)))/(a*b^5), 1/120*(15*a*b^4*f*x^8 + 24*(a*b^4*e -
 a^2*b^3*f)*x^5 + 60*(a*b^4*d - a^2*b^3*e + a^3*b^2*f)*x^2 - 120*sqrt(1/3)*(a*b^4*c - a^2*b^3*d + a^3*b^2*e -
a^4*b*f)*sqrt((a*b^2)^(1/3)/a)*arctan(-sqrt(1/3)*(2*b*x - (a*b^2)^(1/3))*sqrt((a*b^2)^(1/3)/a)/b) + 20*(b^3*c
- a*b^2*d + a^2*b*e - a^3*f)*(a*b^2)^(2/3)*log(b^2*x^2 - (a*b^2)^(1/3)*b*x + (a*b^2)^(2/3)) - 40*(b^3*c - a*b^
2*d + a^2*b*e - a^3*f)*(a*b^2)^(2/3)*log(b*x + (a*b^2)^(1/3)))/(a*b^5)]

Sympy [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.74 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=x^{5} \left (- \frac {a f}{5 b^{2}} + \frac {e}{5 b}\right ) + x^{2} \left (\frac {a^{2} f}{2 b^{3}} - \frac {a e}{2 b^{2}} + \frac {d}{2 b}\right ) + \operatorname {RootSum} {\left (27 t^{3} a b^{11} - a^{9} f^{3} + 3 a^{8} b e f^{2} - 3 a^{7} b^{2} d f^{2} - 3 a^{7} b^{2} e^{2} f + 3 a^{6} b^{3} c f^{2} + 6 a^{6} b^{3} d e f + a^{6} b^{3} e^{3} - 6 a^{5} b^{4} c e f - 3 a^{5} b^{4} d^{2} f - 3 a^{5} b^{4} d e^{2} + 6 a^{4} b^{5} c d f + 3 a^{4} b^{5} c e^{2} + 3 a^{4} b^{5} d^{2} e - 3 a^{3} b^{6} c^{2} f - 6 a^{3} b^{6} c d e - a^{3} b^{6} d^{3} + 3 a^{2} b^{7} c^{2} e + 3 a^{2} b^{7} c d^{2} - 3 a b^{8} c^{2} d + b^{9} c^{3}, \left ( t \mapsto t \log {\left (\frac {9 t^{2} a b^{7}}{a^{6} f^{2} - 2 a^{5} b e f + 2 a^{4} b^{2} d f + a^{4} b^{2} e^{2} - 2 a^{3} b^{3} c f - 2 a^{3} b^{3} d e + 2 a^{2} b^{4} c e + a^{2} b^{4} d^{2} - 2 a b^{5} c d + b^{6} c^{2}} + x \right )} \right )\right )} + \frac {f x^{8}}{8 b} \]

[In]

integrate(x*(f*x**9+e*x**6+d*x**3+c)/(b*x**3+a),x)

[Out]

x**5*(-a*f/(5*b**2) + e/(5*b)) + x**2*(a**2*f/(2*b**3) - a*e/(2*b**2) + d/(2*b)) + RootSum(27*_t**3*a*b**11 -
a**9*f**3 + 3*a**8*b*e*f**2 - 3*a**7*b**2*d*f**2 - 3*a**7*b**2*e**2*f + 3*a**6*b**3*c*f**2 + 6*a**6*b**3*d*e*f
 + a**6*b**3*e**3 - 6*a**5*b**4*c*e*f - 3*a**5*b**4*d**2*f - 3*a**5*b**4*d*e**2 + 6*a**4*b**5*c*d*f + 3*a**4*b
**5*c*e**2 + 3*a**4*b**5*d**2*e - 3*a**3*b**6*c**2*f - 6*a**3*b**6*c*d*e - a**3*b**6*d**3 + 3*a**2*b**7*c**2*e
 + 3*a**2*b**7*c*d**2 - 3*a*b**8*c**2*d + b**9*c**3, Lambda(_t, _t*log(9*_t**2*a*b**7/(a**6*f**2 - 2*a**5*b*e*
f + 2*a**4*b**2*d*f + a**4*b**2*e**2 - 2*a**3*b**3*c*f - 2*a**3*b**3*d*e + 2*a**2*b**4*c*e + a**2*b**4*d**2 -
2*a*b**5*c*d + b**6*c**2) + x))) + f*x**8/(8*b)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.92 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {\sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, b^{2} f x^{8} + 8 \, {\left (b^{2} e - a b f\right )} x^{5} + 20 \, {\left (b^{2} d - a b e + a^{2} f\right )} x^{2}}{40 \, b^{3}} + \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

[In]

integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^4*(a/b)
^(1/3)) + 1/40*(5*b^2*f*x^8 + 8*(b^2*e - a*b*f)*x^5 + 20*(b^2*d - a*b*e + a^2*f)*x^2)/b^3 + 1/6*(b^3*c - a*b^2
*d + a^2*b*e - a^3*f)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^4*(a/b)^(1/3)) - 1/3*(b^3*c - a*b^2*d + a^2*b*
e - a^3*f)*log(x + (a/b)^(1/3))/(b^4*(a/b)^(1/3))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.17 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=\frac {\sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3}} - \frac {{\left (b^{8} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a b^{7} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} b^{6} e \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} b^{5} f \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{8}} + \frac {5 \, b^{7} f x^{8} + 8 \, b^{7} e x^{5} - 8 \, a b^{6} f x^{5} + 20 \, b^{7} d x^{2} - 20 \, a b^{6} e x^{2} + 20 \, a^{2} b^{5} f x^{2}}{40 \, b^{8}} \]

[In]

integrate(x*(f*x^9+e*x^6+d*x^3+c)/(b*x^3+a),x, algorithm="giac")

[Out]

1/3*sqrt(3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2
)^(1/3)*b^3) - 1/6*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(1/3
)*b^3) - 1/3*(b^8*c*(-a/b)^(1/3) - a*b^7*d*(-a/b)^(1/3) + a^2*b^6*e*(-a/b)^(1/3) - a^3*b^5*f*(-a/b)^(1/3))*(-a
/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^8) + 1/40*(5*b^7*f*x^8 + 8*b^7*e*x^5 - 8*a*b^6*f*x^5 + 20*b^7*d*x^2
- 20*a*b^6*e*x^2 + 20*a^2*b^5*f*x^2)/b^8

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.92 \[ \int \frac {x \left (c+d x^3+e x^6+f x^9\right )}{a+b x^3} \, dx=x^5\,\left (\frac {e}{5\,b}-\frac {a\,f}{5\,b^2}\right )+x^2\,\left (\frac {d}{2\,b}-\frac {a\,\left (\frac {e}{b}-\frac {a\,f}{b^2}\right )}{2\,b}\right )+\frac {f\,x^8}{8\,b}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{1/3}\,b^{11/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{1/3}\,b^{11/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{1/3}\,b^{11/3}} \]

[In]

int((x*(c + d*x^3 + e*x^6 + f*x^9))/(a + b*x^3),x)

[Out]

x^5*(e/(5*b) - (a*f)/(5*b^2)) + x^2*(d/(2*b) - (a*(e/b - (a*f)/b^2))/(2*b)) + (f*x^8)/(8*b) - (log(b^(1/3)*x +
 a^(1/3))*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(1/3)*b^(11/3)) + (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x -
a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(1/3)*b^(11/3)) - (log(3^(1/2)*a^(1/
3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(1/3)*b^(11/3)
)