3.5.15 \(\int \frac {x (c+d x+e x^2+f x^3+g x^4+h x^5)}{(a+b x^3)^2} \, dx\) [415]

Optimal. Leaf size=289 \[ \frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\left (b^{5/3} c+a^{2/3} b e+2 a b^{2/3} f-4 a^{5/3} h\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{7/3}}-\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{7/3}}+\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{7/3}}+\frac {g \log \left (a+b x^3\right )}{3 b^2} \]

[Out]

h*x/b^2-1/3*x*(a*(-a*h+b*e)-b*(-a*f+b*c)*x-b*(-a*g+b*d)*x^2)/a/b^2/(b*x^3+a)-1/9*(b^(2/3)*(2*a*f+b*c)-a^(2/3)*
(-4*a*h+b*e))*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)/b^(7/3)+1/18*(b^(2/3)*(2*a*f+b*c)-a^(2/3)*(-4*a*h+b*e))*ln(a^(2/3)
-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(4/3)/b^(7/3)+1/3*g*ln(b*x^3+a)/b^2-1/9*(b^(5/3)*c+a^(2/3)*b*e+2*a*b^(2/3)*f
-4*a^(5/3)*h)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(4/3)/b^(7/3)*3^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1842, 1901, 1885, 1874, 31, 648, 631, 210, 642, 266} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{2/3} b e-4 a^{5/3} h+2 a b^{2/3} f+b^{5/3} c\right )}{3 \sqrt {3} a^{4/3} b^{7/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (b^{2/3} (2 a f+b c)-a^{2/3} (b e-4 a h)\right )}{18 a^{4/3} b^{7/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (b^{2/3} (2 a f+b c)-a^{2/3} (b e-4 a h)\right )}{9 a^{4/3} b^{7/3}}-\frac {x \left (-b x (b c-a f)-b x^2 (b d-a g)+a (b e-a h)\right )}{3 a b^2 \left (a+b x^3\right )}+\frac {g \log \left (a+b x^3\right )}{3 b^2}+\frac {h x}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(h*x)/b^2 - (x*(a*(b*e - a*h) - b*(b*c - a*f)*x - b*(b*d - a*g)*x^2))/(3*a*b^2*(a + b*x^3)) - ((b^(5/3)*c + a^
(2/3)*b*e + 2*a*b^(2/3)*f - 4*a^(5/3)*h)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(4/3)
*b^(7/3)) - ((b^(2/3)*(b*c + 2*a*f) - a^(2/3)*(b*e - 4*a*h))*Log[a^(1/3) + b^(1/3)*x])/(9*a^(4/3)*b^(7/3)) + (
(b^(2/3)*(b*c + 2*a*f) - a^(2/3)*(b*e - 4*a*h))*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(4/3)*b^
(7/3)) + (g*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 1842

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = Pol
ynomialQuotient[b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] +
1)*x^m*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] && 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]

Rule 1901

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

Rubi steps

\begin {align*} \int \frac {x \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{\left (a+b x^3\right )^2} \, dx &=-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\int \frac {-a (b e-a h)-b (b c+2 a f) x-3 a b g x^2-3 a b h x^3}{a+b x^3} \, dx}{3 a b^2}\\ &=-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\int \left (-3 a h-\frac {a (b e-4 a h)+b (b c+2 a f) x+3 a b g x^2}{a+b x^3}\right ) \, dx}{3 a b^2}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}+\frac {\int \frac {a (b e-4 a h)+b (b c+2 a f) x+3 a b g x^2}{a+b x^3} \, dx}{3 a b^2}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}+\frac {\int \frac {a (b e-4 a h)+b (b c+2 a f) x}{a+b x^3} \, dx}{3 a b^2}+\frac {g \int \frac {x^2}{a+b x^3} \, dx}{b}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}+\frac {g \log \left (a+b x^3\right )}{3 b^2}+\frac {\int \frac {\sqrt [3]{a} \left (\sqrt [3]{a} b (b c+2 a f)+2 a \sqrt [3]{b} (b e-4 a h)\right )+\sqrt [3]{b} \left (\sqrt [3]{a} b (b c+2 a f)-a \sqrt [3]{b} (b e-4 a h)\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 (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 a^{4/3} b^2}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{7/3}}+\frac {g \log \left (a+b x^3\right )}{3 b^2}+\frac {\left (b^{5/3} c+a^{2/3} b e+2 a b^{2/3} f-4 a^{5/3} h\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 a b^2}+\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\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^{4/3} b^{7/3}}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{7/3}}+\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{7/3}}+\frac {g \log \left (a+b x^3\right )}{3 b^2}+\frac {\left (b^{5/3} c+a^{2/3} b e+2 a b^{2/3} f-4 a^{5/3} 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^{4/3} b^{7/3}}\\ &=\frac {h x}{b^2}-\frac {x \left (a (b e-a h)-b (b c-a f) x-b (b d-a g) x^2\right )}{3 a b^2 \left (a+b x^3\right )}-\frac {\left (b^{5/3} c+a^{2/3} b e+2 a b^{2/3} f-4 a^{5/3} h\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{7/3}}-\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{7/3}}+\frac {\left (b^{2/3} (b c+2 a f)-a^{2/3} (b e-4 a h)\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{7/3}}+\frac {g \log \left (a+b x^3\right )}{3 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 285, normalized size = 0.99 \begin {gather*} \frac {18 b^{2/3} h x+\frac {6 b^{2/3} \left (b^2 c x^2+a^2 (g+h x)-a b (d+x (e+f x))\right )}{a \left (a+b x^3\right )}-\frac {2 \sqrt {3} \left (b^2 c+a^{2/3} b^{4/3} e+2 a b f-4 a^{5/3} \sqrt [3]{b} h\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{4/3}}-\frac {2 \left (b^2 c-a^{2/3} b^{4/3} e+2 a b f+4 a^{5/3} \sqrt [3]{b} h\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{4/3}}+\frac {\left (b^2 c-a^{2/3} b^{4/3} e+2 a b f+4 a^{5/3} \sqrt [3]{b} h\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{4/3}}+6 b^{2/3} g \log \left (a+b x^3\right )}{18 b^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(18*b^(2/3)*h*x + (6*b^(2/3)*(b^2*c*x^2 + a^2*(g + h*x) - a*b*(d + x*(e + f*x))))/(a*(a + b*x^3)) - (2*Sqrt[3]
*(b^2*c + a^(2/3)*b^(4/3)*e + 2*a*b*f - 4*a^(5/3)*b^(1/3)*h)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(4
/3) - (2*(b^2*c - a^(2/3)*b^(4/3)*e + 2*a*b*f + 4*a^(5/3)*b^(1/3)*h)*Log[a^(1/3) + b^(1/3)*x])/a^(4/3) + ((b^2
*c - a^(2/3)*b^(4/3)*e + 2*a*b*f + 4*a^(5/3)*b^(1/3)*h)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(4/3
) + 6*b^(2/3)*g*Log[a + b*x^3])/(18*b^(8/3))

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Maple [A]
time = 0.40, size = 285, normalized size = 0.99

method	result	size
risch	\(\frac {h x}{b^{2}}+\frac {-\frac {b \left (a f -b c \right ) x^{2}}{3 a}+\left (\frac {a h}{3}-\frac {b e}{3}\right ) x +\frac {a g}{3}-\frac {b d}{3}}{b^{2} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (3 b g \,\textit {\_R}^{2}+\frac {b \left (2 a f +b c \right ) \textit {\_R}}{a}-4 a h +b e \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{3}}\)	\(113\)
default	\(\frac {h x}{b^{2}}-\frac {\frac {\frac {b \left (a f -b c \right ) x^{2}}{3 a}+\left (-\frac {a h}{3}+\frac {b e}{3}\right ) x -\frac {a g}{3}+\frac {b d}{3}}{b \,x^{3}+a}+\frac {\left (4 a^{2} h -a b e \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )+\left (-2 a b f -b^{2} c \right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )-a g \ln \left (b \,x^{3}+a \right )}{3 a}}{b^{2}}\)	\(285\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 316, normalized size = 1.09 \begin {gather*} -\frac {a b d - a^{2} g - {\left (b^{2} c - a b f\right )} x^{2} - {\left (a^{2} h - a b e\right )} x}{3 \, {\left (a b^{3} x^{3} + a^{2} b^{2}\right )}} + \frac {h x}{b^{2}} + \frac {\sqrt {3} {\left (b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4 \, a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b \left (\frac {a}{b}\right )^{\frac {1}{3}} 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 )}{9 \, a^{2} b^{2}} + \frac {{\left (6 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} + b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} + 4 \, a^{2} h - a 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 )}{18 \, a b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (3 \, a b g \left (\frac {a}{b}\right )^{\frac {2}{3}} - b^{2} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a b f \left (\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a^{2} h + a b e\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [C] Result contains complex when optimal does not.
time = 1.91, size = 12617, normalized size = 43.66 \begin {gather*} \text {too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(36*a*b*h*x^4 - 12*a*b*d + 12*a^2*g + 12*(b^2*c - a*b*f)*x^2 - 2*(a*b^3*x^3 + a^2*b^2)*(2*(1/2)^(2/3)*(-I
*sqrt(3) + 1)*(9*g^2/b^4 - (b^2*c*e + (9*g^2 - 8*f*h)*a^2 + 2*(e*f - 2*c*h)*a*b)/(a^2*b^4))/(54*g^3/b^6 - 9*(b
^2*c*e + (9*g^2 - 8*f*h)*a^2 + 2*(e*f - 2*c*h)*a*b)*g/(a^2*b^6) - (b^5*c^3 + a^2*b^3*e^3 + 6*a*b^4*c^2*f + 12*
a^2*b^3*c*f^2 + 8*a^3*b^2*f^3 - 12*a^3*b^2*e^2*h + 48*a^4*b*e*h^2 - 64*a^5*h^3)/(a^4*b^7) - (b^5*c^3 + 6*a*b^4
*c^2*f + 64*a^5*h^3 - 3*(9*g^3 - 24*f*g*h + 16*e*h^2)*a^4*b + 2*(4*f^3 - 9*e*f*g + 6*e^2*h + 18*c*g*h)*a^3*b^2
- (e^3 - 3*(4*f^2 - 3*e*g)*c)*a^2*b^3)/(a^4*b^7))^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(54*g^3/b^6 - 9*(b^2*c*
e + (9*g^2 - 8*f*h)*a^2 + 2*(e*f - 2*c*h)*a*b)*g/(a^2*b^6) - (b^5*c^3 + a^2*b^3*e^3 + 6*a*b^4*c^2*f + 12*a^2*b
^3*c*f^2 + 8*a^3*b^2*f^3 - 12*a^3*b^2*e^2*h + 48*a^4*b*e*h^2 - 64*a^5*h^3)/(a^4*b^7) - (b^5*c^3 + 6*a*b^4*c^2*
f + 64*a^5*h^3 - 3*(9*g^3 - 24*f*g*h + 16*e*h^2)*a^4*b + 2*(4*f^3 - 9*e*f*g + 6*e^2*h + 18*c*g*h)*a^3*b^2 - (e
^3 - 3*(4* ...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.46, size = 318, normalized size = 1.10 \begin {gather*} \frac {h x}{b^{2}} + \frac {g \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{2}} + \frac {\sqrt {3} {\left (4 \, a^{2} h - a b e + \left (-a b^{2}\right )^{\frac {1}{3}} b c + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} a 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 b} + \frac {{\left (4 \, a^{2} h - a b e - \left (-a b^{2}\right )^{\frac {1}{3}} b c - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} 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 \, \left (-a b^{2}\right )^{\frac {2}{3}} a b} - \frac {a b d - a^{2} g - {\left (b^{2} c - a b f\right )} x^{2} - {\left (a^{2} h - a b e\right )} x}{3 \, {\left (b x^{3} + a\right )} a b^{2}} - \frac {{\left (a b^{5} c \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a^{2} b^{4} f \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a^{3} b^{3} h + a^{2} b^{4} e\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^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.39, size = 827, normalized size = 2.86 \begin {gather*} \left (\sum _{k=1}^3\ln \left (\frac {9\,a^2\,g^2+b^2\,c\,e-8\,a^2\,f\,h-4\,a\,b\,c\,h+2\,a\,b\,e\,f}{9\,a\,b^2}-\mathrm {root}\left (729\,a^4\,b^7\,z^3-729\,a^4\,b^5\,g\,z^2-216\,a^4\,b^3\,f\,h\,z-108\,a^3\,b^4\,c\,h\,z+54\,a^3\,b^4\,e\,f\,z+27\,a^2\,b^5\,c\,e\,z+243\,a^4\,b^3\,g^2\,z+72\,a^4\,b\,f\,g\,h+36\,a^3\,b^2\,c\,g\,h-18\,a^3\,b^2\,e\,f\,g-9\,a^2\,b^3\,c\,e\,g-48\,a^4\,b\,e\,h^2+6\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,e^2\,h+12\,a^2\,b^3\,c\,f^2+8\,a^3\,b^2\,f^3-27\,a^4\,b\,g^3+64\,a^5\,h^3+b^5\,c^3-a^2\,b^3\,e^3,z,k\right )\,\left (6\,a\,g-b\,e\,x+4\,a\,h\,x-\mathrm {root}\left (729\,a^4\,b^7\,z^3-729\,a^4\,b^5\,g\,z^2-216\,a^4\,b^3\,f\,h\,z-108\,a^3\,b^4\,c\,h\,z+54\,a^3\,b^4\,e\,f\,z+27\,a^2\,b^5\,c\,e\,z+243\,a^4\,b^3\,g^2\,z+72\,a^4\,b\,f\,g\,h+36\,a^3\,b^2\,c\,g\,h-18\,a^3\,b^2\,e\,f\,g-9\,a^2\,b^3\,c\,e\,g-48\,a^4\,b\,e\,h^2+6\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,e^2\,h+12\,a^2\,b^3\,c\,f^2+8\,a^3\,b^2\,f^3-27\,a^4\,b\,g^3+64\,a^5\,h^3+b^5\,c^3-a^2\,b^3\,e^3,z,k\right )\,a\,b^2\,9\right )+\frac {x\,\left (12\,g\,h\,a^3+4\,a^2\,b\,f^2-3\,e\,g\,a^2\,b+4\,a\,b^2\,c\,f+b^3\,c^2\right )}{9\,a^2\,b^2}\right )\,\mathrm {root}\left (729\,a^4\,b^7\,z^3-729\,a^4\,b^5\,g\,z^2-216\,a^4\,b^3\,f\,h\,z-108\,a^3\,b^4\,c\,h\,z+54\,a^3\,b^4\,e\,f\,z+27\,a^2\,b^5\,c\,e\,z+243\,a^4\,b^3\,g^2\,z+72\,a^4\,b\,f\,g\,h+36\,a^3\,b^2\,c\,g\,h-18\,a^3\,b^2\,e\,f\,g-9\,a^2\,b^3\,c\,e\,g-48\,a^4\,b\,e\,h^2+6\,a\,b^4\,c^2\,f+12\,a^3\,b^2\,e^2\,h+12\,a^2\,b^3\,c\,f^2+8\,a^3\,b^2\,f^3-27\,a^4\,b\,g^3+64\,a^5\,h^3+b^5\,c^3-a^2\,b^3\,e^3,z,k\right )\right )-\frac {\frac {b\,d}{3}-\frac {a\,g}{3}+x\,\left (\frac {b\,e}{3}-\frac {a\,h}{3}\right )-\frac {b\,x^2\,\left (b\,c-a\,f\right )}{3\,a}}{b^3\,x^3+a\,b^2}+\frac {h\,x}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((9*a^2*g^2 + b^2*c*e - 8*a^2*f*h - 4*a*b*c*h + 2*a*b*e*f)/(9*a*b^2) - root(729*a^4*b^7*z^3 - 729*a^
4*b^5*g*z^2 - 216*a^4*b^3*f*h*z - 108*a^3*b^4*c*h*z + 54*a^3*b^4*e*f*z + 27*a^2*b^5*c*e*z + 243*a^4*b^3*g^2*z
+ 72*a^4*b*f*g*h + 36*a^3*b^2*c*g*h - 18*a^3*b^2*e*f*g - 9*a^2*b^3*c*e*g - 48*a^4*b*e*h^2 + 6*a*b^4*c^2*f + 12
*a^3*b^2*e^2*h + 12*a^2*b^3*c*f^2 + 8*a^3*b^2*f^3 - 27*a^4*b*g^3 + 64*a^5*h^3 + b^5*c^3 - a^2*b^3*e^3, z, k)*(
6*a*g - b*e*x + 4*a*h*x - 9*root(729*a^4*b^7*z^3 - 729*a^4*b^5*g*z^2 - 216*a^4*b^3*f*h*z - 108*a^3*b^4*c*h*z +
54*a^3*b^4*e*f*z + 27*a^2*b^5*c*e*z + 243*a^4*b^3*g^2*z + 72*a^4*b*f*g*h + 36*a^3*b^2*c*g*h - 18*a^3*b^2*e*f*
g - 9*a^2*b^3*c*e*g - 48*a^4*b*e*h^2 + 6*a*b^4*c^2*f + 12*a^3*b^2*e^2*h + 12*a^2*b^3*c*f^2 + 8*a^3*b^2*f^3 - 2
7*a^4*b*g^3 + 64*a^5*h^3 + b^5*c^3 - a^2*b^3*e^3, z, k)*a*b^2) + (x*(b^3*c^2 + 4*a^2*b*f^2 + 12*a^3*g*h + 4*a*
b^2*c*f - 3*a^2*b*e*g))/(9*a^2*b^2))*root(729*a^4*b^7*z^3 - 729*a^4*b^5*g*z^2 - 216*a^4*b^3*f*h*z - 108*a^3*b^
4*c*h*z + 54*a^3*b^4*e*f*z + 27*a^2*b^5*c*e*z + 243*a^4*b^3*g^2*z + 72*a^4*b*f*g*h + 36*a^3*b^2*c*g*h - 18*a^3
*b^2*e*f*g - 9*a^2*b^3*c*e*g - 48*a^4*b*e*h^2 + 6*a*b^4*c^2*f + 12*a^3*b^2*e^2*h + 12*a^2*b^3*c*f^2 + 8*a^3*b^
2*f^3 - 27*a^4*b*g^3 + 64*a^5*h^3 + b^5*c^3 - a^2*b^3*e^3, z, k), k, 1, 3) - ((b*d)/3 - (a*g)/3 + x*((b*e)/3 -
(a*h)/3) - (b*x^2*(b*c - a*f))/(3*a))/(a*b^2 + b^3*x^3) + (h*x)/b^2

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