\(\int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x (a+b x^3)^2} \, dx\) [417]

Optimal result

Integrand size = 38, antiderivative size = 289 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\frac {x \left (a (b d-a g)+a (b e-a h) x-b (b c-a f) x^2\right )}{3 a^2 b \left (a+b x^3\right )}-\frac {\left (2 b^{4/3} d+\sqrt [3]{a} b e+a \sqrt [3]{b} g+2 a^{4/3} h\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{5/3}}+\frac {c \log (x)}{a^2}+\frac {\left (\sqrt [3]{b} (2 b d+a g)-\sqrt [3]{a} (b e+2 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{5/3}}-\frac {\left (\sqrt [3]{b} (2 b d+a g)-\sqrt [3]{a} (b e+2 a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{5/3}}-\frac {c \log \left (a+b x^3\right )}{3 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1843, 1848, 1885, 1874, 31, 648, 631, 210, 642, 266} \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (2 a^{4/3} h+\sqrt [3]{a} b e+a \sqrt [3]{b} g+2 b^{4/3} d\right )}{3 \sqrt {3} a^{5/3} b^{5/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {\sqrt [3]{a} (2 a h+b e)}{\sqrt [3]{b}}+a g+2 b d\right )}{18 a^{5/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (a g+2 b d)-\sqrt [3]{a} (2 a h+b e)\right )}{9 a^{5/3} b^{5/3}}+\frac {x \left (-b x^2 (b c-a f)+a (b d-a g)+a x (b e-a h)\right )}{3 a^2 b \left (a+b x^3\right )}-\frac {c \log \left (a+b x^3\right )}{3 a^2}+\frac {c \log (x)}{a^2} \]

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x*(a + b*x^3)^2),x]

[Out]

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

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 266

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

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 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]

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]

Rule 1874

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, Dist[(-r)*((B*r - A*s)/(3*a*s)), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) +
s*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[
a/b]

Rule 1885

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\frac {-\frac {6 a (-b (c+x (d+e x))+a (f+x (g+h x)))}{b \left (a+b x^3\right )}-\frac {2 \sqrt {3} \sqrt [3]{a} \left (2 b^{4/3} d+\sqrt [3]{a} b e+a \sqrt [3]{b} g+2 a^{4/3} h\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{5/3}}+18 c \log (x)+\frac {2 \sqrt [3]{a} \left (2 b^{4/3} d-\sqrt [3]{a} b e+a \sqrt [3]{b} g-2 a^{4/3} h\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{5/3}}+\frac {\sqrt [3]{a} \left (-2 b^{4/3} d+\sqrt [3]{a} b e-a \sqrt [3]{b} g+2 a^{4/3} h\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{5/3}}-6 c \log \left (a+b x^3\right )}{18 a^2} \]

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x*(a + b*x^3)^2),x]

[Out]

((-6*a*(-(b*(c + x*(d + e*x))) + a*(f + x*(g + h*x))))/(b*(a + b*x^3)) - (2*Sqrt[3]*a^(1/3)*(2*b^(4/3)*d + a^(
1/3)*b*e + a*b^(1/3)*g + 2*a^(4/3)*h)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(5/3) + 18*c*Log[x] + (2*
a^(1/3)*(2*b^(4/3)*d - a^(1/3)*b*e + a*b^(1/3)*g - 2*a^(4/3)*h)*Log[a^(1/3) + b^(1/3)*x])/b^(5/3) + (a^(1/3)*(
-2*b^(4/3)*d + a^(1/3)*b*e - a*b^(1/3)*g + 2*a^(4/3)*h)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(5/3
) - 6*c*Log[a + b*x^3])/(18*a^2)

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {c \ln \left (x \right )}{a^{2}}+\frac {\frac {-\frac {a \left (a h -b e \right ) x^{2}}{3 b}-\frac {a \left (a g -b d \right ) x}{3 b}-\frac {a \left (a f -b c \right )}{3 b}}{b \,x^{3}+a}+\frac {\left (a^{2} g +2 a b d \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (2 a^{2} h +a e b \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )-b c \ln \left (b \,x^{3}+a \right )}{3 b}}{a^{2}}\)	\(293\)
risch	\(\frac {-\frac {\left (a h -b e \right ) x^{2}}{3 a b}-\frac {\left (a g -b d \right ) x}{3 a b}-\frac {a f -b c}{3 a b}}{b \,x^{3}+a}+\frac {c \ln \left (-x \right )}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{6} b^{5} \textit {\_Z}^{3}+9 a^{4} b^{5} c \,\textit {\_Z}^{2}+\left (6 a^{5} b^{2} g h +12 a^{4} b^{3} d h +3 a^{4} b^{3} e g +6 a^{3} b^{4} d e +27 a^{2} b^{5} c^{2}\right ) \textit {\_Z} +8 a^{5} h^{3}+12 a^{4} b e \,h^{2}-a^{4} b \,g^{3}+18 a^{3} b^{2} c g h -6 a^{3} b^{2} d \,g^{2}+6 a^{3} b^{2} e^{2} h +36 a^{2} b^{3} c d h +9 a^{2} b^{3} c e g -12 a^{2} b^{3} d^{2} g +a^{2} b^{3} e^{3}+18 a \,b^{4} c d e -8 a \,b^{4} d^{3}+27 b^{5} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{5} b^{5}-24 \textit {\_R}^{2} a^{3} b^{5} c +\left (-20 a^{4} b^{2} g h -40 a^{3} b^{3} d h -10 a^{3} b^{3} e g -20 a^{2} b^{4} d e -36 a \,b^{5} c^{2}\right ) \textit {\_R} -24 a^{4} h^{3}-36 a^{3} b e \,h^{2}+3 a^{3} b \,g^{3}-36 a^{2} b^{2} c g h +18 a^{2} b^{2} d \,g^{2}-18 a^{2} b^{2} e^{2} h -72 a \,b^{3} c d h -18 a \,b^{3} c e g +36 a \,b^{3} d^{2} g -3 a \,b^{3} e^{3}-36 b^{4} c d e +24 b^{4} d^{3}\right ) x +\left (2 a^{5} b^{3} h +a^{4} b^{4} e \right ) \textit {\_R}^{2}+\left (-a^{4} g^{2} b^{2}-12 b^{3} c h \,a^{3}-4 b^{3} d g \,a^{3}-6 a^{2} b^{4} c e -4 a^{2} b^{4} d^{2}\right ) \textit {\_R} +9 a^{2} b^{2} c \,g^{2}-54 a \,b^{3} c^{2} h +36 a \,b^{3} c d g -27 b^{4} c^{2} e +36 b^{4} c \,d^{2}\right )\right )}{9}\)	\(612\)

[In]

int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 21.33 (sec) , antiderivative size = 12541, normalized size of antiderivative = 43.39 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x/(b*x**3+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.04 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\frac {{\left (b e - a h\right )} x^{2} + b c - a f + {\left (b d - a g\right )} x}{3 \, {\left (a b^{2} x^{3} + a^{2} b\right )}} + \frac {c \log \left (x\right )}{a^{2}} + \frac {\sqrt {3} {\left (a b e \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a^{2} h \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} g \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b} - \frac {{\left (6 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b d + a^{2} g\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (3 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a b d - a^{2} g\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/3*((b*e - a*h)*x^2 + b*c - a*f + (b*d - a*g)*x)/(a*b^2*x^3 + a^2*b) + c*log(x)/a^2 + 1/9*sqrt(3)*(a*b*e*(a/b
)^(2/3) + 2*a^2*h*(a/b)^(2/3) + 2*a*b*d*(a/b)^(1/3) + a^2*g*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3)
)/(a/b)^(1/3))/(a^3*b) - 1/18*(6*b^2*c*(a/b)^(2/3) - a*b*e*(a/b)^(1/3) - 2*a^2*h*(a/b)^(1/3) + 2*a*b*d + a^2*g
)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b^2*(a/b)^(2/3)) - 1/9*(3*b^2*c*(a/b)^(2/3) + a*b*e*(a/b)^(1/3)
+ 2*a^2*h*(a/b)^(1/3) - 2*a*b*d - a^2*g)*log(x + (a/b)^(1/3))/(a^2*b^2*(a/b)^(2/3))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=-\frac {c \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2}} + \frac {c \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {\sqrt {3} {\left (2 \, b^{2} d + a b g - \left (-a b^{2}\right )^{\frac {1}{3}} b e - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a h\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b} - \frac {{\left (2 \, b^{2} d + a b g + \left (-a b^{2}\right )^{\frac {1}{3}} b e + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a h\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b} + \frac {a b c - a^{2} f + {\left (a b e - a^{2} h\right )} x^{2} + {\left (a b d - a^{2} g\right )} x}{3 \, {\left (b x^{3} + a\right )} a^{2} b} - \frac {{\left (a^{3} b^{3} e \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a^{4} b^{2} h \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a^{3} b^{3} d + a^{4} b^{2} g\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{5} b^{3}} \]

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.78 (sec) , antiderivative size = 1660, normalized size of antiderivative = 5.74 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )^2} \, dx=\text {Too large to display} \]

[In]

int((c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(x*(a + b*x^3)^2),x)

[Out]

((b*c - a*f)/(3*a*b) + (x*(b*d - a*g))/(3*a*b) + (x^2*(b*e - a*h))/(3*a*b))/(a + b*x^3) + symsum(log((c*(4*b^2
*d^2 + a^2*g^2 - 3*b^2*c*e - 6*a*b*c*h + 4*a*b*d*g))/(9*a^3) - (root(729*a^6*b^5*z^3 + 729*a^4*b^5*c*z^2 + 54*
a^5*b^2*g*h*z + 108*a^4*b^3*d*h*z + 27*a^4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b^5*c^2*z + 18*a*b^4*c*d*e +
18*a^3*b^2*c*g*h + 36*a^2*b^3*c*d*h + 9*a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2*e^2*h - 12*a^2*b^3*d^2*g -
6*a^3*b^2*d*g^2 - a^4*b*g^3 - 8*a*b^4*d^3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3, z, k)*(a^3*g^2 + 4*a*b^2*d^
2 + 36*b^3*c^2*x + 324*root(729*a^6*b^5*z^3 + 729*a^4*b^5*c*z^2 + 54*a^5*b^2*g*h*z + 108*a^4*b^3*d*h*z + 27*a^
4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b^5*c^2*z + 18*a*b^4*c*d*e + 18*a^3*b^2*c*g*h + 36*a^2*b^3*c*d*h + 9*
a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2*e^2*h - 12*a^2*b^3*d^2*g - 6*a^3*b^2*d*g^2 - a^4*b*g^3 - 8*a*b^4*d^
3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3, z, k)^2*a^4*b^3*x - 18*root(729*a^6*b^5*z^3 + 729*a^4*b^5*c*z^2 + 54
*a^5*b^2*g*h*z + 108*a^4*b^3*d*h*z + 27*a^4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b^5*c^2*z + 18*a*b^4*c*d*e
+ 18*a^3*b^2*c*g*h + 36*a^2*b^3*c*d*h + 9*a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2*e^2*h - 12*a^2*b^3*d^2*g
- 6*a^3*b^2*d*g^2 - a^4*b*g^3 - 8*a*b^4*d^3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3, z, k)*a^4*b*h + 6*a*b^2*c*
e + 12*a^2*b*c*h + 4*a^2*b*d*g + 20*a^3*g*h*x - 9*root(729*a^6*b^5*z^3 + 729*a^4*b^5*c*z^2 + 54*a^5*b^2*g*h*z
+ 108*a^4*b^3*d*h*z + 27*a^4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b^5*c^2*z + 18*a*b^4*c*d*e + 18*a^3*b^2*c*
g*h + 36*a^2*b^3*c*d*h + 9*a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2*e^2*h - 12*a^2*b^3*d^2*g - 6*a^3*b^2*d*g
^2 - a^4*b*g^3 - 8*a*b^4*d^3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3, z, k)*a^3*b^2*e + 216*root(729*a^6*b^5*z^
3 + 729*a^4*b^5*c*z^2 + 54*a^5*b^2*g*h*z + 108*a^4*b^3*d*h*z + 27*a^4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b
^5*c^2*z + 18*a*b^4*c*d*e + 18*a^3*b^2*c*g*h + 36*a^2*b^3*c*d*h + 9*a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2
*e^2*h - 12*a^2*b^3*d^2*g - 6*a^3*b^2*d*g^2 - a^4*b*g^3 - 8*a*b^4*d^3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3,
z, k)*a^2*b^3*c*x + 20*a*b^2*d*e*x + 40*a^2*b*d*h*x + 10*a^2*b*e*g*x))/(9*a^2) - (x*(8*a^4*h^3 - 8*b^4*d^3 + a
*b^3*e^3 - a^3*b*g^3 - 6*a^2*b^2*d*g^2 + 6*a^2*b^2*e^2*h + 12*b^4*c*d*e - 12*a*b^3*d^2*g + 12*a^3*b*e*h^2 + 12
*a^2*b^2*c*g*h + 24*a*b^3*c*d*h + 6*a*b^3*c*e*g))/(27*a^3*b^2))*root(729*a^6*b^5*z^3 + 729*a^4*b^5*c*z^2 + 54*
a^5*b^2*g*h*z + 108*a^4*b^3*d*h*z + 27*a^4*b^3*e*g*z + 54*a^3*b^4*d*e*z + 243*a^2*b^5*c^2*z + 18*a*b^4*c*d*e +
18*a^3*b^2*c*g*h + 36*a^2*b^3*c*d*h + 9*a^2*b^3*c*e*g + 12*a^4*b*e*h^2 + 6*a^3*b^2*e^2*h - 12*a^2*b^3*d^2*g -
6*a^3*b^2*d*g^2 - a^4*b*g^3 - 8*a*b^4*d^3 + 8*a^5*h^3 + 27*b^5*c^3 + a^2*b^3*e^3, z, k), k, 1, 3) + (c*log(x)
)/a^2