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

Optimal. Leaf size=225 \[ \frac {b c-a d}{2 a^2 x^2}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{6 a^{8/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^{8/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{\sqrt {3} a^{8/3} b^{4/3}}-\frac {c}{5 a x^5}+\frac {f x}{b} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1834, 200, 31, 634, 617, 204, 628} \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{6 a^{8/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{3 a^{8/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{\sqrt {3} a^{8/3} b^{4/3}}+\frac {b c-a d}{2 a^2 x^2}-\frac {c}{5 a x^5}+\frac {f x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(5*a*x^5) + (b*c - a*d)/(2*a^2*x^2) + (f*x)/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^(8/3)*b^(4/3)) + ((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Log[a^(1/3) + b^
(1/3)*x])/(3*a^(8/3)*b^(4/3)) - ((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^(8/3)*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 200

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

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 628

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 634

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 1834

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*} \int \frac {c+d x^3+e x^6+f x^9}{x^6 \left (a+b x^3\right )} \, dx &=\int \left (\frac {f}{b}+\frac {c}{a x^6}+\frac {-b c+a d}{a^2 x^3}+\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{a^2 b \left (a+b x^3\right )}\right ) \, dx\\ &=-\frac {c}{5 a x^5}+\frac {b c-a d}{2 a^2 x^2}+\frac {f x}{b}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \int \frac {1}{a+b x^3} \, dx}{a^2 b}\\ &=-\frac {c}{5 a x^5}+\frac {b c-a d}{2 a^2 x^2}+\frac {f x}{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 a^{8/3} b}+\frac {\left (b^3 c-a b^2 d+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}{3 a^{8/3} b}\\ &=-\frac {c}{5 a x^5}+\frac {b c-a d}{2 a^2 x^2}+\frac {f x}{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 a^{8/3} b^{4/3}}-\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 a^{8/3} b^{4/3}}+\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 a^{7/3} b}\\ &=-\frac {c}{5 a x^5}+\frac {b c-a d}{2 a^2 x^2}+\frac {f x}{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 a^{8/3} b^{4/3}}-\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 a^{8/3} b^{4/3}}+\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{8/3} b^{4/3}}\\ &=-\frac {c}{5 a x^5}+\frac {b c-a d}{2 a^2 x^2}+\frac {f x}{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} a^{8/3} b^{4/3}}+\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 a^{8/3} b^{4/3}}-\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 a^{8/3} b^{4/3}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 220, normalized size = 0.98 \[ \frac {b c-a d}{2 a^2 x^2}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^3 f-a^2 b e+a b^2 d-b^3 c\right )}{6 a^{8/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{3 a^{8/3} b^{4/3}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right ) \left (a^3 f-a^2 b e+a b^2 d-b^3 c\right )}{\sqrt {3} a^{8/3} b^{4/3}}-\frac {c}{5 a x^5}+\frac {f x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 584, normalized size = 2.60 \[ \left [\frac {30 \, a^{4} b f x^{6} - 15 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{5} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} + 3 \, \left (-a^{2} b\right )^{\frac {1}{3}} a x - a^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (-a^{2} b\right )^{\frac {2}{3}} x + \left (-a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) - 5 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a^{2} b\right )^{\frac {2}{3}} x^{5} \log \left (a b x^{2} - \left (-a^{2} b\right )^{\frac {2}{3}} x - \left (-a^{2} b\right )^{\frac {1}{3}} a\right ) + 10 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a^{2} b\right )^{\frac {2}{3}} x^{5} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) - 6 \, a^{3} b^{2} c + 15 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}}{30 \, a^{4} b^{2} x^{5}}, \frac {30 \, a^{4} b f x^{6} + 30 \, \sqrt {\frac {1}{3}} {\left (a b^{4} c - a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{5} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-a^{2} b\right )^{\frac {2}{3}} x + \left (-a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - 5 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a^{2} b\right )^{\frac {2}{3}} x^{5} \log \left (a b x^{2} - \left (-a^{2} b\right )^{\frac {2}{3}} x - \left (-a^{2} b\right )^{\frac {1}{3}} a\right ) + 10 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \left (-a^{2} b\right )^{\frac {2}{3}} x^{5} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) - 6 \, a^{3} b^{2} c + 15 \, {\left (a^{2} b^{3} c - a^{3} b^{2} d\right )} x^{3}}{30 \, a^{4} b^{2} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.48, size = 220, normalized size = 0.98 \[ \frac {f x}{b} - \frac {\sqrt {3} {\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (b^{3} c - a b^{2} d - a^{3} f + a^{2} b e\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{3} b} + \frac {5 \, b c x^{3} - 5 \, a d x^{3} - 2 \, a c}{10 \, a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*x/b - 1/3*sqrt(3)*(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/
((-a*b^2)^(2/3)*a^2) - 1/6*(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b
^2)^(2/3)*a^2) - 1/3*(b^3*c - a*b^2*d - a^3*f + a^2*b*e)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b) + 1/1
0*(5*b*c*x^3 - 5*a*d*x^3 - 2*a*c)/(a^2*x^5)

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maple [B]  time = 0.05, size = 410, normalized size = 1.82 \[ -\frac {\sqrt {3}\, a f \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {a f \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}+\frac {a f \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {\sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\frac {d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\frac {\sqrt {3}\, b c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}+\frac {b c \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}-\frac {b c \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}+\frac {\sqrt {3}\, e \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {e \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}-\frac {e \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {f x}{b}-\frac {d}{2 a \,x^{2}}+\frac {b c}{2 a^{2} x^{2}}-\frac {c}{5 a \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.08, size = 214, normalized size = 0.95 \[ \frac {f x}{b} + \frac {\sqrt {3} {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {5 \, {\left (b c - a d\right )} x^{3} - 2 \, a c}{10 \, a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*x/b + 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
^2*b^2*(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^2*b^2*
(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^2*b^2*(a/b)^(2/3)) + 1/10*(5*(b
*c - a*d)*x^3 - 2*a*c)/(a^2*x^5)

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mupad [B]  time = 5.09, size = 207, normalized size = 0.92 \[ \frac {f\,x}{b}-\frac {\frac {b\,c}{5\,a}+\frac {b\,x^3\,\left (a\,d-b\,c\right )}{2\,a^2}}{b\,x^5}+\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{8/3}\,b^{4/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{8/3}\,b^{4/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{3\,a^{8/3}\,b^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 19.68, size = 328, normalized size = 1.46 \[ \operatorname {RootSum} {\left (27 t^{3} a^{8} b^{4} + a^{9} f^{3} - 3 a^{8} b e f^{2} + 3 a^{7} b^{2} d f^{2} + 3 a^{7} b^{2} e^{2} f - 3 a^{6} b^{3} c f^{2} - 6 a^{6} b^{3} d e f - a^{6} b^{3} e^{3} + 6 a^{5} b^{4} c e f + 3 a^{5} b^{4} d^{2} f + 3 a^{5} b^{4} d e^{2} - 6 a^{4} b^{5} c d f - 3 a^{4} b^{5} c e^{2} - 3 a^{4} b^{5} d^{2} e + 3 a^{3} b^{6} c^{2} f + 6 a^{3} b^{6} c d e + a^{3} b^{6} d^{3} - 3 a^{2} b^{7} c^{2} e - 3 a^{2} b^{7} c d^{2} + 3 a b^{8} c^{2} d - b^{9} c^{3}, \left (t \mapsto t \log {\left (- \frac {3 t a^{3} b}{a^{3} f - a^{2} b e + a b^{2} d - b^{3} c} + x \right )} \right )\right )} + \frac {f x}{b} + \frac {- 2 a c + x^{3} \left (- 5 a d + 5 b c\right )}{10 a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(27*_t**3*a**8*b**4 + 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(-3*_t*a**3
*b/(a**3*f - a**2*b*e + a*b**2*d - b**3*c) + x))) + f*x/b + (-2*a*c + x**3*(-5*a*d + 5*b*c))/(10*a**2*x**5)

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