\(\int \frac {c+d x^3+e x^6+f x^9}{x^{12} (a+b x^3)} \, dx\) [247]

Optimal result

Integrand size = 30, antiderivative size = 280 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=-\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}-\frac {b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{14/3}}+\frac {b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{14/3}}-\frac {b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{14/3}} \]

[Out]

-1/11*c/a/x^11+1/8*(-a*d+b*c)/a^2/x^8+1/5*(-a^2*e+a*b*d-b^2*c)/a^3/x^5+1/2*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)/a^4/
x^2+1/3*b^(2/3)*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*ln(a^(1/3)+b^(1/3)*x)/a^(14/3)-1/6*b^(2/3)*(-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^(14/3)-1/3*b^(2/3)*(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*arcta
n(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(14/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1848, 206, 31, 648, 631, 210, 642} \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\frac {b c-a d}{8 a^2 x^8}-\frac {a^2 e-a b d+b^2 c}{5 a^3 x^5}-\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{\sqrt {3} a^{14/3}}-\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{6 a^{14/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^{14/3}}+\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{2 a^4 x^2}-\frac {c}{11 a x^{11}} \]

[In]

Int[(c + d*x^3 + e*x^6 + f*x^9)/(x^12*(a + b*x^3)),x]

[Out]

-1/11*c/(a*x^11) + (b*c - a*d)/(8*a^2*x^8) - (b^2*c - a*b*d + a^2*e)/(5*a^3*x^5) + (b^3*c - a*b^2*d + a^2*b*e
- a^3*f)/(2*a^4*x^2) - (b^(2/3)*(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^(14/3)) + (b^(2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(1/3) + b^(1/3)*x])/(3*a^(14/
3)) - (b^(2/3)*(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^(14/3)
)

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 1848

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(Pq/(a + b*x
^n)), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c}{a x^{12}}+\frac {-b c+a d}{a^2 x^9}+\frac {b^2 c-a b d+a^2 e}{a^3 x^6}+\frac {-b^3 c+a b^2 d-a^2 b e+a^3 f}{a^4 x^3}-\frac {b \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right )}{a^4 \left (a+b x^3\right )}\right ) \, dx \\ & = -\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}+\frac {\left (b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\right ) \int \frac {1}{a+b x^3} \, dx}{a^4} \\ & = -\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}+\frac {\left (b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{14/3}}+\frac {\left (b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\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}{3 a^{14/3}} \\ & = -\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}+\frac {b^{2/3} \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 a^{14/3}}-\frac {\left (b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\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 a^{14/3}}+\frac {\left (b \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 a^{13/3}} \\ & = -\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}+\frac {b^{2/3} \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 a^{14/3}}-\frac {b^{2/3} \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 a^{14/3}}+\frac {\left (b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{14/3}} \\ & = -\frac {c}{11 a x^{11}}+\frac {b c-a d}{8 a^2 x^8}-\frac {b^2 c-a b d+a^2 e}{5 a^3 x^5}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{2 a^4 x^2}-\frac {b^{2/3} \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} a^{14/3}}+\frac {b^{2/3} \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 a^{14/3}}-\frac {b^{2/3} \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 a^{14/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\frac {-\frac {120 a^{11/3} c}{x^{11}}+\frac {165 a^{8/3} (b c-a d)}{x^8}-\frac {264 a^{5/3} \left (b^2 c-a b d+a^2 e\right )}{x^5}+\frac {660 a^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right )}{x^2}-440 \sqrt {3} b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+440 b^{2/3} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+220 b^{2/3} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{1320 a^{14/3}} \]

[In]

Integrate[(c + d*x^3 + e*x^6 + f*x^9)/(x^12*(a + b*x^3)),x]

[Out]

((-120*a^(11/3)*c)/x^11 + (165*a^(8/3)*(b*c - a*d))/x^8 - (264*a^(5/3)*(b^2*c - a*b*d + a^2*e))/x^5 + (660*a^(
2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f))/x^2 - 440*Sqrt[3]*b^(2/3)*(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]] + 440*b^(2/3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(1/3) + b^(1/3)*
x] + 220*b^(2/3)*(-(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])/(1320*
a^(14/3))

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.73

method	result	size
default	\(-\frac {c}{11 a \,x^{11}}-\frac {a d -b c}{8 a^{2} x^{8}}-\frac {a^{2} e -a b d +b^{2} c}{5 a^{3} x^{5}}-\frac {f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c}{2 a^{4} x^{2}}-\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) b}{a^{4}}\)	\(205\)
risch	\(\frac {-\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) x^{9}}{2 a^{4}}-\frac {\left (a^{2} e -a b d +b^{2} c \right ) x^{6}}{5 a^{3}}-\frac {\left (a d -b c \right ) x^{3}}{8 a^{2}}-\frac {c}{11 a}}{x^{11}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{14} \textit {\_Z}^{3}+a^{9} b^{2} f^{3}-3 a^{8} b^{3} e \,f^{2}+3 a^{7} b^{4} d \,f^{2}+3 a^{7} b^{4} e^{2} f -3 a^{6} b^{5} c \,f^{2}-6 a^{6} b^{5} d e f -a^{6} b^{5} e^{3}+6 a^{5} b^{6} c e f +3 a^{5} b^{6} d^{2} f +3 a^{5} b^{6} d \,e^{2}-6 a^{4} b^{7} c d f -3 a^{4} b^{7} c \,e^{2}-3 a^{4} b^{7} d^{2} e +3 a^{3} b^{8} c^{2} f +6 a^{3} b^{8} c d e +a^{3} b^{8} d^{3}-3 a^{2} b^{9} c^{2} e -3 a^{2} b^{9} c \,d^{2}+3 a \,b^{10} c^{2} d -b^{11} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{14}-3 a^{9} b^{2} f^{3}+9 a^{8} b^{3} e \,f^{2}-9 a^{7} b^{4} d \,f^{2}-9 a^{7} b^{4} e^{2} f +9 a^{6} b^{5} c \,f^{2}+18 a^{6} b^{5} d e f +3 a^{6} b^{5} e^{3}-18 a^{5} b^{6} c e f -9 a^{5} b^{6} d^{2} f -9 a^{5} b^{6} d \,e^{2}+18 a^{4} b^{7} c d f +9 a^{4} b^{7} c \,e^{2}+9 a^{4} b^{7} d^{2} e -9 a^{3} b^{8} c^{2} f -18 a^{3} b^{8} c d e -3 a^{3} b^{8} d^{3}+9 a^{2} b^{9} c^{2} e +9 a^{2} b^{9} c \,d^{2}-9 a \,b^{10} c^{2} d +3 b^{11} c^{3}\right ) x +\left (-a^{11} b \,f^{2}+2 a^{10} b^{2} e f -2 a^{9} b^{3} d f -a^{9} b^{3} e^{2}+2 a^{8} b^{4} c f +2 a^{8} b^{4} d e -2 a^{7} b^{5} c e -a^{7} b^{5} d^{2}+2 a^{6} b^{6} c d -a^{5} b^{7} c^{2}\right ) \textit {\_R} \right )\right )}{3}\)	\(672\)

[In]

int((f*x^9+e*x^6+d*x^3+c)/x^12/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=-\frac {440 \, \sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{11} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 220 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{11} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} + a b x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 440 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{11} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 660 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} + 264 \, {\left (a b^{2} c - a^{2} b d + a^{3} e\right )} x^{6} + 120 \, a^{3} c - 165 \, {\left (a^{2} b c - a^{3} d\right )} x^{3}}{1320 \, a^{4} x^{11}} \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^12/(b*x^3+a),x, algorithm="fricas")

[Out]

-1/1320*(440*sqrt(3)*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^11*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(-b^2
/a^2)^(2/3) - sqrt(3)*b)/b) - 220*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^11*(-b^2/a^2)^(1/3)*log(b^2*x^2 + a*b*
x*(-b^2/a^2)^(1/3) + a^2*(-b^2/a^2)^(2/3)) + 440*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^11*(-b^2/a^2)^(1/3)*log
(b*x - a*(-b^2/a^2)^(1/3)) - 660*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x^9 + 264*(a*b^2*c - a^2*b*d + a^3*e)*x^6
+ 120*a^3*c - 165*(a^2*b*c - a^3*d)*x^3)/(a^4*x^11)

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\text {Timed out} \]

[In]

integrate((f*x**9+e*x**6+d*x**3+c)/x**12/(b*x**3+a),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\frac {\sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {220 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x^{9} - 88 \, {\left (a b^{2} c - a^{2} b d + a^{3} e\right )} x^{6} - 40 \, a^{3} c + 55 \, {\left (a^{2} b c - a^{3} d\right )} x^{3}}{440 \, a^{4} x^{11}} \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^12/(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))/(a^4*(a/b)
^(2/3)) - 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^4*(a/b)^(2/3)) + 1
/3*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*log(x + (a/b)^(1/3))/(a^4*(a/b)^(2/3)) + 1/440*(220*(b^3*c - a*b^2*d +
a^2*b*e - a^3*f)*x^9 - 88*(a*b^2*c - a^2*b*d + a^3*e)*x^6 - 40*a^3*c + 55*(a^2*b*c - a^3*d)*x^3)/(a^4*x^11)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.19 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{3} c - \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2} d + \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b e - \left (-a b^{2}\right )^{\frac {1}{3}} a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{5}} - \frac {{\left (b^{4} c - a b^{3} d + a^{2} b^{2} e - a^{3} b f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{5}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b^{3} c - \left (-a b^{2}\right )^{\frac {1}{3}} a b^{2} d + \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b e - \left (-a b^{2}\right )^{\frac {1}{3}} a^{3} f\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{5}} + \frac {220 \, b^{3} c x^{9} - 220 \, a b^{2} d x^{9} + 220 \, a^{2} b e x^{9} - 220 \, a^{3} f x^{9} - 88 \, a b^{2} c x^{6} + 88 \, a^{2} b d x^{6} - 88 \, a^{3} e x^{6} + 55 \, a^{2} b c x^{3} - 55 \, a^{3} d x^{3} - 40 \, a^{3} c}{440 \, a^{4} x^{11}} \]

[In]

integrate((f*x^9+e*x^6+d*x^3+c)/x^12/(b*x^3+a),x, algorithm="giac")

[Out]

1/3*sqrt(3)*((-a*b^2)^(1/3)*b^3*c - (-a*b^2)^(1/3)*a*b^2*d + (-a*b^2)^(1/3)*a^2*b*e - (-a*b^2)^(1/3)*a^3*f)*ar
ctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/a^5 - 1/3*(b^4*c - a*b^3*d + a^2*b^2*e - a^3*b*f)*(-a/b)^(
1/3)*log(abs(x - (-a/b)^(1/3)))/a^5 + 1/6*((-a*b^2)^(1/3)*b^3*c - (-a*b^2)^(1/3)*a*b^2*d + (-a*b^2)^(1/3)*a^2*
b*e - (-a*b^2)^(1/3)*a^3*f)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^5 + 1/440*(220*b^3*c*x^9 - 220*a*b^2*d*
x^9 + 220*a^2*b*e*x^9 - 220*a^3*f*x^9 - 88*a*b^2*c*x^6 + 88*a^2*b*d*x^6 - 88*a^3*e*x^6 + 55*a^2*b*c*x^3 - 55*a
^3*d*x^3 - 40*a^3*c)/(a^4*x^11)

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.90 \[ \int \frac {c+d x^3+e x^6+f x^9}{x^{12} \left (a+b x^3\right )} \, dx=\frac {b^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{14/3}}-\frac {\frac {c}{11\,a}-\frac {x^9\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{2\,a^4}+\frac {x^3\,\left (a\,d-b\,c\right )}{8\,a^2}+\frac {x^6\,\left (e\,a^2-d\,a\,b+c\,b^2\right )}{5\,a^3}}{x^{11}}+\frac {b^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{14/3}}-\frac {b^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{14/3}} \]

[In]

int((c + d*x^3 + e*x^6 + f*x^9)/(x^12*(a + b*x^3)),x)

[Out]

(b^(2/3)*log(b^(1/3)*x + a^(1/3))*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(14/3)) - (c/(11*a) - (x^9*(b^3*c
- a^3*f - a*b^2*d + a^2*b*e))/(2*a^4) + (x^3*(a*d - b*c))/(8*a^2) + (x^6*(b^2*c + a^2*e - a*b*d))/(5*a^3))/x^1
1 + (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)*(b^3*c - a^3*f - a*b^2*d +
a^2*b*e))/(3*a^(14/3)) - (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)*(b^3
*c - a^3*f - a*b^2*d + a^2*b*e))/(3*a^(14/3))