\(\int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{(a+b x^3)^2} \, dx\) [416]

Optimal result

Integrand size = 35, antiderivative size = 276 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=\frac {x \left (b c-a f+(b d-a g) x+(b e-a h) x^2\right )}{3 a b \left (a+b x^3\right )}-\frac {\left (2 b^{4/3} c+\sqrt [3]{a} b d+a \sqrt [3]{b} f+2 a^{4/3} g\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{5/3}}+\frac {\left (\sqrt [3]{b} (2 b c+a f)-\sqrt [3]{a} (b d+2 a g)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{5/3}}-\frac {\left (\sqrt [3]{b} (2 b c+a f)-\sqrt [3]{a} (b d+2 a g)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{5/3}}+\frac {h \log \left (a+b x^3\right )}{3 b^2} \]

[Out]

1/3*x*(b*c-a*f+(-a*g+b*d)*x+(-a*h+b*e)*x^2)/a/b/(b*x^3+a)+1/9*(b^(1/3)*(a*f+2*b*c)-a^(1/3)*(2*a*g+b*d))*ln(a^(
1/3)+b^(1/3)*x)/a^(5/3)/b^(5/3)-1/18*(b^(1/3)*(a*f+2*b*c)-a^(1/3)*(2*a*g+b*d))*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^
(2/3)*x^2)/a^(5/3)/b^(5/3)+1/3*h*ln(b*x^3+a)/b^2-1/9*(2*b^(4/3)*c+a^(1/3)*b*d+a*b^(1/3)*f+2*a^(4/3)*g)*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 = 0.25 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {1872, 1885, 1874, 31, 648, 631, 210, 642, 266} \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (2 a^{4/3} g+\sqrt [3]{a} b d+a \sqrt [3]{b} f+2 b^{4/3} c\right )}{3 \sqrt {3} a^{5/3} b^{5/3}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (\sqrt [3]{b} (a f+2 b c)-\sqrt [3]{a} (2 a g+b d)\right )}{18 a^{5/3} b^{5/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (a f+2 b c)-\sqrt [3]{a} (2 a g+b d)\right )}{9 a^{5/3} b^{5/3}}+\frac {h \log \left (a+b x^3\right )}{3 b^2}+\frac {x \left (x (b d-a g)+x^2 (b e-a h)-a f+b c\right )}{3 a b \left (a+b x^3\right )} \]

[In]

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

[Out]

(x*(b*c - a*f + (b*d - a*g)*x + (b*e - a*h)*x^2))/(3*a*b*(a + b*x^3)) - ((2*b^(4/3)*c + a^(1/3)*b*d + a*b^(1/3
)*f + 2*a^(4/3)*g)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(5/3)) + ((b^(1/3)*
(2*b*c + a*f) - a^(1/3)*(b*d + 2*a*g))*Log[a^(1/3) + b^(1/3)*x])/(9*a^(5/3)*b^(5/3)) - ((b^(1/3)*(2*b*c + a*f)
- a^(1/3)*(b*d + 2*a*g))*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(5/3)*b^(5/3)) + (h*Log[a + b*
x^3])/(3*b^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 1872

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] + Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x]] /
; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=\frac {\frac {6 \left (a^2 h+b^2 x (c+d x)-a b (e+x (f+g x))\right )}{a \left (a+b x^3\right )}-\frac {2 \sqrt {3} \sqrt [3]{b} \left (2 b^{4/3} c+\sqrt [3]{a} b d+a \sqrt [3]{b} f+2 a^{4/3} g\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3}}+\frac {2 \sqrt [3]{b} \left (2 b^{4/3} c-\sqrt [3]{a} b d+a \sqrt [3]{b} f-2 a^{4/3} g\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{5/3}}+\frac {\sqrt [3]{b} \left (-2 b^{4/3} c+\sqrt [3]{a} b d-a \sqrt [3]{b} f+2 a^{4/3} g\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{5/3}}+6 h \log \left (a+b x^3\right )}{18 b^2} \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 1.52 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {-\frac {\left (a g -b d \right ) x^{2}}{3 a b}-\frac {\left (a f -b c \right ) x}{3 a b}+\frac {a h -b e}{3 b^{2}}}{b \,x^{3}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (3 h \,\textit {\_R}^{2}+\frac {\left (2 a g +b d \right ) \textit {\_R}}{a}+\frac {a f +2 b c}{a}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{2}}\)	\(120\)
default	\(\frac {-\frac {\left (a g -b d \right ) x^{2}}{3 a b}-\frac {\left (a f -b c \right ) x}{3 a b}+\frac {a h -b e}{3 b^{2}}}{b \,x^{3}+a}+\frac {\left (a f +2 b c \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (2 a g +b d \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {a h \ln \left (b \,x^{3}+a \right )}{b}}{3 b a}\)	\(283\)

[In]

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

[Out]

(-1/3*(a*g-b*d)/a/b*x^2-1/3*(a*f-b*c)/a/b*x+1/3*(a*h-b*e)/b^2)/(b*x^3+a)+1/9/b^2*sum((3*h*_R^2+1/a*(2*a*g+b*d)
*_R+(a*f+2*b*c)/a)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.52 (sec) , antiderivative size = 12636, normalized size of antiderivative = 45.78 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=-\frac {a b e - a^{2} h - {\left (b^{2} d - a b g\right )} x^{2} - {\left (b^{2} c - a b f\right )} x}{3 \, {\left (a b^{3} x^{3} + a^{2} b^{2}\right )}} + \frac {\sqrt {3} {\left (b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b f \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b^{2}} + \frac {{\left (6 \, a h \left (\frac {a}{b}\right )^{\frac {2}{3}} + b d \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, b c - a f\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (3 \, a h \left (\frac {a}{b}\right )^{\frac {2}{3}} - b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a g \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, b c + a f\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=\frac {h \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{2}} - \frac {\sqrt {3} {\left (2 \, b^{2} c + a b f - \left (-a b^{2}\right )^{\frac {1}{3}} b d - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b} - \frac {{\left (2 \, b^{2} c + a b f + \left (-a b^{2}\right )^{\frac {1}{3}} b d + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a g\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b} + \frac {{\left (b d - a g\right )} x^{2} + {\left (b c - a f\right )} x - \frac {a b e - a^{2} h}{b}}{3 \, {\left (b x^{3} + a\right )} a b} - \frac {{\left (a b^{3} d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a^{2} b^{2} g \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b^{3} c + a^{2} b^{2} f\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3} b^{3}} \]

[In]

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

[Out]

1/3*h*log(abs(b*x^3 + a))/b^2 - 1/9*sqrt(3)*(2*b^2*c + a*b*f - (-a*b^2)^(1/3)*b*d - 2*(-a*b^2)^(1/3)*a*g)*arct
an(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*c + a*b*f + (-a*b^2)^(1/3
)*b*d + 2*(-a*b^2)^(1/3)*a*g)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a*b) + 1/3*((b*d - a*g)
*x^2 + (b*c - a*f)*x - (a*b*e - a^2*h)/b)/((b*x^3 + a)*a*b) - 1/9*(a*b^3*d*(-a/b)^(1/3) + 2*a^2*b^2*g*(-a/b)^(
1/3) + 2*a*b^3*c + a^2*b^2*f)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b^3)

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 835, normalized size of antiderivative = 3.03 \[ \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a+b x^3\right )^2} \, dx=\left (\sum _{k=1}^3\ln \left (\frac {\mathrm {root}\left (729\,a^5\,b^6\,z^3-729\,a^5\,b^4\,h\,z^2+54\,a^4\,b^3\,f\,g\,z+108\,a^3\,b^4\,c\,g\,z+27\,a^3\,b^4\,d\,f\,z+54\,a^2\,b^5\,c\,d\,z+243\,a^5\,b^2\,h^2\,z-18\,a^4\,b\,f\,g\,h-36\,a^3\,b^2\,c\,g\,h-9\,a^3\,b^2\,d\,f\,h-18\,a^2\,b^3\,c\,d\,h-12\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,d\,g^2+6\,a^2\,b^3\,d^2\,g-6\,a^2\,b^3\,c\,f^2+8\,a^4\,b\,g^3+a\,b^4\,d^3-27\,a^5\,h^3-8\,b^5\,c^3-a^3\,b^2\,f^3,z,k\right )\,\left (-6\,a^2\,h+\mathrm {root}\left (729\,a^5\,b^6\,z^3-729\,a^5\,b^4\,h\,z^2+54\,a^4\,b^3\,f\,g\,z+108\,a^3\,b^4\,c\,g\,z+27\,a^3\,b^4\,d\,f\,z+54\,a^2\,b^5\,c\,d\,z+243\,a^5\,b^2\,h^2\,z-18\,a^4\,b\,f\,g\,h-36\,a^3\,b^2\,c\,g\,h-9\,a^3\,b^2\,d\,f\,h-18\,a^2\,b^3\,c\,d\,h-12\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,d\,g^2+6\,a^2\,b^3\,d^2\,g-6\,a^2\,b^3\,c\,f^2+8\,a^4\,b\,g^3+a\,b^4\,d^3-27\,a^5\,h^3-8\,b^5\,c^3-a^3\,b^2\,f^3,z,k\right )\,a^2\,b^2\,9+2\,b^2\,c\,x+a\,b\,f\,x\right )}{a}+\frac {9\,a^3\,h^2+2\,b^3\,c\,d+4\,a\,b^2\,c\,g+a\,b^2\,d\,f+2\,a^2\,b\,f\,g}{9\,a^2\,b^2}+\frac {x\,\left (4\,a^2\,g^2-3\,f\,h\,a^2+4\,a\,b\,d\,g-6\,c\,h\,a\,b+b^2\,d^2\right )}{9\,a^2\,b}\right )\,\mathrm {root}\left (729\,a^5\,b^6\,z^3-729\,a^5\,b^4\,h\,z^2+54\,a^4\,b^3\,f\,g\,z+108\,a^3\,b^4\,c\,g\,z+27\,a^3\,b^4\,d\,f\,z+54\,a^2\,b^5\,c\,d\,z+243\,a^5\,b^2\,h^2\,z-18\,a^4\,b\,f\,g\,h-36\,a^3\,b^2\,c\,g\,h-9\,a^3\,b^2\,d\,f\,h-18\,a^2\,b^3\,c\,d\,h-12\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,d\,g^2+6\,a^2\,b^3\,d^2\,g-6\,a^2\,b^3\,c\,f^2+8\,a^4\,b\,g^3+a\,b^4\,d^3-27\,a^5\,h^3-8\,b^5\,c^3-a^3\,b^2\,f^3,z,k\right )\right )+\frac {\frac {x\,\left (b\,c-a\,f\right )}{3\,a\,b}-\frac {b\,e-a\,h}{3\,b^2}+\frac {x^2\,\left (b\,d-a\,g\right )}{3\,a\,b}}{b\,x^3+a} \]

[In]

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

[Out]

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