\(\int \frac {c+d x^3+e x^6+f x^9}{x^6 (a+b x^3)^2} \, dx\) [270]

Optimal result

Integrand size = 30, antiderivative size = 270 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=-\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}-\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{11/3} b^{4/3}}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3} b^{4/3}}-\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{11/3} b^{4/3}} \]

[Out]

-1/5*c/a^2/x^5+1/2*(-a*d+2*b*c)/a^3/x^2+1/3*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*x/a^3/b/(b*x^3+a)+1/9*(a^3*f+2*a^2*
b*e-5*a*b^2*d+8*b^3*c)*ln(a^(1/3)+b^(1/3)*x)/a^(11/3)/b^(4/3)-1/18*(a^3*f+2*a^2*b*e-5*a*b^2*d+8*b^3*c)*ln(a^(2
/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(11/3)/b^(4/3)-1/9*(a^3*f+2*a^2*b*e-5*a*b^2*d+8*b^3*c)*arctan(1/3*(a^(1/3
)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(11/3)/b^(4/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1843, 1502, 206, 31, 648, 631, 210, 642} \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\frac {2 b c-a d}{2 a^3 x^2}-\frac {c}{5 a^2 x^5}+\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^3 b \left (a+b x^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 f+2 a^2 b e-5 a b^2 d+8 b^3 c\right )}{3 \sqrt {3} a^{11/3} b^{4/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 f+2 a^2 b e-5 a b^2 d+8 b^3 c\right )}{18 a^{11/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 f+2 a^2 b e-5 a b^2 d+8 b^3 c\right )}{9 a^{11/3} b^{4/3}} \]

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^6*(a + b*x^3)^2),x]

[Out]

-1/5*c/(a^2*x^5) + (2*b*c - a*d)/(2*a^3*x^2) + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/(3*a^3*b*(a + b*x^3)) -
((8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(1
1/3)*b^(4/3)) + ((8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(11/3)*b^(4/3)) - ((
8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(11/3)*b^(4/3))

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 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 1843

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x] + S
imp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x]]] /; FreeQ[{a, b}, x] && PolyQ[P
q, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}-\frac {\int \frac {-3 b^3 c+3 b^3 \left (\frac {b c}{a}-d\right ) x^3-b^2 \left (\frac {2 b^3 c}{a^2}-\frac {2 b^2 d}{a}+2 b e+a f\right ) x^6}{x^6 \left (a+b x^3\right )} \, dx}{3 a b^3} \\ & = \frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}-\frac {\int \left (-\frac {3 b^3 c}{a x^6}-\frac {3 b^3 (-2 b c+a d)}{a^2 x^3}-\frac {b^2 \left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right )}{a^2 \left (a+b x^3\right )}\right ) \, dx}{3 a b^3} \\ & = -\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \int \frac {1}{a+b x^3} \, dx}{3 a^3 b} \\ & = -\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 a^{11/3} b}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{9 a^{11/3} b} \\ & = -\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3} b^{4/3}}-\frac {\left (8 b^3 c-5 a b^2 d+2 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}{18 a^{11/3} b^{4/3}}+\frac {\left (8 b^3 c-5 a b^2 d+2 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}{6 a^{10/3} b} \\ & = -\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3} b^{4/3}}-\frac {\left (8 b^3 c-5 a b^2 d+2 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 )}{18 a^{11/3} b^{4/3}}+\frac {\left (8 b^3 c-5 a b^2 d+2 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 )}{3 a^{11/3} b^{4/3}} \\ & = -\frac {c}{5 a^2 x^5}+\frac {2 b c-a d}{2 a^3 x^2}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{3 a^3 b \left (a+b x^3\right )}-\frac {\left (8 b^3 c-5 a b^2 d+2 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 )}{3 \sqrt {3} a^{11/3} b^{4/3}}+\frac {\left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{11/3} b^{4/3}}-\frac {\left (8 b^3 c-5 a b^2 d+2 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 )}{18 a^{11/3} b^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\frac {-\frac {18 a^{5/3} c}{x^5}-\frac {45 a^{2/3} (-2 b c+a d)}{x^2}-\frac {30 a^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{b \left (a+b x^3\right )}-\frac {10 \sqrt {3} \left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{4/3}}+\frac {10 \left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{4/3}}-\frac {5 \left (8 b^3 c-5 a b^2 d+2 a^2 b e+a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{4/3}}}{90 a^{11/3}} \]

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^6*(a + b*x^3)^2),x]

[Out]

((-18*a^(5/3)*c)/x^5 - (45*a^(2/3)*(-2*b*c + a*d))/x^2 - (30*a^(2/3)*(-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*x)
/(b*(a + b*x^3)) - (10*Sqrt[3]*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sq
rt[3]])/b^(4/3) + (10*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*Log[a^(1/3) + b^(1/3)*x])/b^(4/3) - (5*(8*b^3*
c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(4/3))/(90*a^(11/3))

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.71

method	result	size
default	\(-\frac {c}{5 a^{2} x^{5}}-\frac {a d -2 b c}{2 a^{3} x^{2}}+\frac {-\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) x}{3 b \left (b \,x^{3}+a \right )}+\frac {\left (f \,a^{3}+2 a^{2} b e -5 a \,b^{2} d +8 b^{3} c \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3 b}}{a^{3}}\)	\(193\)
risch	\(\frac {-\frac {\left (2 f \,a^{3}-2 a^{2} b e +5 a \,b^{2} d -8 b^{3} c \right ) x^{6}}{6 a^{3} b}-\frac {\left (5 a d -8 b c \right ) x^{3}}{10 a^{2}}-\frac {c}{5 a}}{x^{5} \left (b \,x^{3}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{11} b^{4} \textit {\_Z}^{3}-a^{9} f^{3}-6 a^{8} b e \,f^{2}+15 a^{7} b^{2} d \,f^{2}-12 a^{7} b^{2} e^{2} f -24 a^{6} b^{3} c \,f^{2}+60 a^{6} b^{3} d e f -8 a^{6} b^{3} e^{3}-96 a^{5} b^{4} c e f -75 a^{5} b^{4} d^{2} f +60 a^{5} b^{4} d \,e^{2}+240 a^{4} b^{5} c d f -96 a^{4} b^{5} c \,e^{2}-150 a^{4} b^{5} d^{2} e -192 a^{3} b^{6} c^{2} f +480 a^{3} b^{6} c d e +125 a^{3} b^{6} d^{3}-384 a^{2} b^{7} c^{2} e -600 a^{2} b^{7} c \,d^{2}+960 a \,b^{8} c^{2} d -512 c^{3} b^{9}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{11} b^{4}+3 a^{9} f^{3}+18 a^{8} b e \,f^{2}-45 a^{7} b^{2} d \,f^{2}+36 a^{7} b^{2} e^{2} f +72 a^{6} b^{3} c \,f^{2}-180 a^{6} b^{3} d e f +24 a^{6} b^{3} e^{3}+288 a^{5} b^{4} c e f +225 a^{5} b^{4} d^{2} f -180 a^{5} b^{4} d \,e^{2}-720 a^{4} b^{5} c d f +288 a^{4} b^{5} c \,e^{2}+450 a^{4} b^{5} d^{2} e +576 a^{3} b^{6} c^{2} f -1440 a^{3} b^{6} c d e -375 a^{3} b^{6} d^{3}+1152 a^{2} b^{7} c^{2} e +1800 a^{2} b^{7} c \,d^{2}-2880 a \,b^{8} c^{2} d +1536 c^{3} b^{9}\right ) x +\left (-a^{10} f^{2} b -4 b^{2} e f \,a^{9}+10 b^{3} d f \,a^{8}-4 b^{3} e^{2} a^{8}-16 b^{4} c f \,a^{7}+20 b^{4} d e \,a^{7}-32 b^{5} c e \,a^{6}-25 b^{5} d^{2} a^{6}+80 b^{6} c d \,a^{5}-64 b^{7} c^{2} a^{4}\right ) \textit {\_R} \right )\right )}{9}\)	\(661\)

[In]

int((f*x^9+e*x^6+d*x^3+c)/x^6/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 897, normalized size of antiderivative = 3.32 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\left [-\frac {18 \, a^{4} b^{2} c - 15 \, {\left (8 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 2 \, a^{4} b^{2} e - 2 \, a^{5} b f\right )} x^{6} - 9 \, {\left (8 \, a^{3} b^{3} c - 5 \, a^{4} b^{2} d\right )} x^{3} - 15 \, \sqrt {\frac {1}{3}} {\left ({\left (8 \, a b^{5} c - 5 \, a^{2} b^{4} d + 2 \, a^{3} b^{3} e + a^{4} b^{2} f\right )} x^{8} + {\left (8 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 2 \, a^{4} b^{2} e + a^{5} b f\right )} x^{5}\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) + 5 \, {\left ({\left (8 \, b^{4} c - 5 \, a b^{3} d + 2 \, a^{2} b^{2} e + a^{3} b f\right )} x^{8} + {\left (8 \, a b^{3} c - 5 \, a^{2} b^{2} d + 2 \, a^{3} b e + a^{4} f\right )} x^{5}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 10 \, {\left ({\left (8 \, b^{4} c - 5 \, a b^{3} d + 2 \, a^{2} b^{2} e + a^{3} b f\right )} x^{8} + {\left (8 \, a b^{3} c - 5 \, a^{2} b^{2} d + 2 \, a^{3} b e + a^{4} f\right )} x^{5}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{90 \, {\left (a^{5} b^{3} x^{8} + a^{6} b^{2} x^{5}\right )}}, -\frac {18 \, a^{4} b^{2} c - 15 \, {\left (8 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 2 \, a^{4} b^{2} e - 2 \, a^{5} b f\right )} x^{6} - 9 \, {\left (8 \, a^{3} b^{3} c - 5 \, a^{4} b^{2} d\right )} x^{3} - 30 \, \sqrt {\frac {1}{3}} {\left ({\left (8 \, a b^{5} c - 5 \, a^{2} b^{4} d + 2 \, a^{3} b^{3} e + a^{4} b^{2} f\right )} x^{8} + {\left (8 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + 2 \, a^{4} b^{2} e + a^{5} b f\right )} x^{5}\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) + 5 \, {\left ({\left (8 \, b^{4} c - 5 \, a b^{3} d + 2 \, a^{2} b^{2} e + a^{3} b f\right )} x^{8} + {\left (8 \, a b^{3} c - 5 \, a^{2} b^{2} d + 2 \, a^{3} b e + a^{4} f\right )} x^{5}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 10 \, {\left ({\left (8 \, b^{4} c - 5 \, a b^{3} d + 2 \, a^{2} b^{2} e + a^{3} b f\right )} x^{8} + {\left (8 \, a b^{3} c - 5 \, a^{2} b^{2} d + 2 \, a^{3} b e + a^{4} f\right )} x^{5}\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{90 \, {\left (a^{5} b^{3} x^{8} + a^{6} b^{2} x^{5}\right )}}\right ] \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^6/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[-1/90*(18*a^4*b^2*c - 15*(8*a^2*b^4*c - 5*a^3*b^3*d + 2*a^4*b^2*e - 2*a^5*b*f)*x^6 - 9*(8*a^3*b^3*c - 5*a^4*b
^2*d)*x^3 - 15*sqrt(1/3)*((8*a*b^5*c - 5*a^2*b^4*d + 2*a^3*b^3*e + a^4*b^2*f)*x^8 + (8*a^2*b^4*c - 5*a^3*b^3*d
+ 2*a^4*b^2*e + a^5*b*f)*x^5)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)
*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) + 5*((8*b^4*c - 5*a*b^3*
d + 2*a^2*b^2*e + a^3*b*f)*x^8 + (8*a*b^3*c - 5*a^2*b^2*d + 2*a^3*b*e + a^4*f)*x^5)*(a^2*b)^(2/3)*log(a*b*x^2
- (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) - 10*((8*b^4*c - 5*a*b^3*d + 2*a^2*b^2*e + a^3*b*f)*x^8 + (8*a*b^3*c - 5*
a^2*b^2*d + 2*a^3*b*e + a^4*f)*x^5)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^5*b^3*x^8 + a^6*b^2*x^5), -1/
90*(18*a^4*b^2*c - 15*(8*a^2*b^4*c - 5*a^3*b^3*d + 2*a^4*b^2*e - 2*a^5*b*f)*x^6 - 9*(8*a^3*b^3*c - 5*a^4*b^2*d
)*x^3 - 30*sqrt(1/3)*((8*a*b^5*c - 5*a^2*b^4*d + 2*a^3*b^3*e + a^4*b^2*f)*x^8 + (8*a^2*b^4*c - 5*a^3*b^3*d + 2
*a^4*b^2*e + a^5*b*f)*x^5)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((
a^2*b)^(1/3)/b)/a^2) + 5*((8*b^4*c - 5*a*b^3*d + 2*a^2*b^2*e + a^3*b*f)*x^8 + (8*a*b^3*c - 5*a^2*b^2*d + 2*a^3
*b*e + a^4*f)*x^5)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) - 10*((8*b^4*c - 5*a*b^3*d +
2*a^2*b^2*e + a^3*b*f)*x^8 + (8*a*b^3*c - 5*a^2*b^2*d + 2*a^3*b*e + a^4*f)*x^5)*(a^2*b)^(2/3)*log(a*b*x + (a^
2*b)^(2/3)))/(a^5*b^3*x^8 + a^6*b^2*x^5)]

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((f*x**9+e*x**6+d*x**3+c)/x**6/(b*x**3+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\frac {5 \, {\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e - 2 \, a^{3} f\right )} x^{6} - 6 \, a^{2} b c + 3 \, {\left (8 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{3}}{30 \, {\left (a^{3} b^{2} x^{8} + a^{4} b x^{5}\right )}} + \frac {\sqrt {3} {\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{3} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{3} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^6/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/30*(5*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e - 2*a^3*f)*x^6 - 6*a^2*b*c + 3*(8*a*b^2*c - 5*a^2*b*d)*x^3)/(a^3*b^2*
x^8 + a^4*b*x^5) + 1/9*sqrt(3)*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3)
)/(a/b)^(1/3))/(a^3*b^2*(a/b)^(2/3)) - 1/18*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*log(x^2 - x*(a/b)^(1/3)
+ (a/b)^(2/3))/(a^3*b^2*(a/b)^(2/3)) + 1/9*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*log(x + (a/b)^(1/3))/(a^3
*b^2*(a/b)^(2/3))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=-\frac {\sqrt {3} {\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{3}} - \frac {{\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{3}} - \frac {{\left (8 \, b^{3} c - 5 \, a b^{2} d + 2 \, a^{2} b e + a^{3} f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{4} b} + \frac {b^{3} c x - a b^{2} d x + a^{2} b e x - a^{3} f x}{3 \, {\left (b x^{3} + a\right )} a^{3} b} + \frac {10 \, b c x^{3} - 5 \, a d x^{3} - 2 \, a c}{10 \, a^{3} x^{5}} \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^6/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/9*sqrt(3)*(8*b^3*c - 5*a*b^2*d + 2*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)^(2/3)*a^3) - 1/18*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/
((-a*b^2)^(2/3)*a^3) - 1/9*(8*b^3*c - 5*a*b^2*d + 2*a^2*b*e + a^3*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(
a^4*b) + 1/3*(b^3*c*x - a*b^2*d*x + a^2*b*e*x - a^3*f*x)/((b*x^3 + a)*a^3*b) + 1/10*(10*b*c*x^3 - 5*a*d*x^3 -
2*a*c)/(a^3*x^5)

Mupad [B] (verification not implemented)

Time = 9.44 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )^2} \, dx=\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (f\,a^3+2\,e\,a^2\,b-5\,d\,a\,b^2+8\,c\,b^3\right )}{9\,a^{11/3}\,b^{4/3}}-\frac {\frac {c}{5\,a}+\frac {x^3\,\left (5\,a\,d-8\,b\,c\right )}{10\,a^2}-\frac {x^6\,\left (-2\,f\,a^3+2\,e\,a^2\,b-5\,d\,a\,b^2+8\,c\,b^3\right )}{6\,a^3\,b}}{b\,x^8+a\,x^5}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (f\,a^3+2\,e\,a^2\,b-5\,d\,a\,b^2+8\,c\,b^3\right )}{9\,a^{11/3}\,b^{4/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (f\,a^3+2\,e\,a^2\,b-5\,d\,a\,b^2+8\,c\,b^3\right )}{9\,a^{11/3}\,b^{4/3}} \]

[In]

int((c + d*x^3 + e*x^6 + f*x^9)/(x^6*(a + b*x^3)^2),x)

[Out]

(log(b^(1/3)*x + a^(1/3))*(8*b^3*c + a^3*f - 5*a*b^2*d + 2*a^2*b*e))/(9*a^(11/3)*b^(4/3)) - (c/(5*a) + (x^3*(5
*a*d - 8*b*c))/(10*a^2) - (x^6*(8*b^3*c - 2*a^3*f - 5*a*b^2*d + 2*a^2*b*e))/(6*a^3*b))/(a*x^5 + b*x^8) + (log(
3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(8*b^3*c + a^3*f - 5*a*b^2*d + 2*a^2*b*e))/
(9*a^(11/3)*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)*(8*b^3*c + a^3*
f - 5*a*b^2*d + 2*a^2*b*e))/(9*a^(11/3)*b^(4/3))