3.352 \(\int \frac {x (c+d x+e x^2)}{(a+b x^3)^3} \, dx\)

Optimal. Leaf size=239 \[ \frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{4/3}}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{4/3}}-\frac {\left (a^{2/3} e+2 b^{2/3} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{4/3}}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2} \]

[Out]

-1/6*x*(-b*d*x^2-b*c*x+a*e)/a/b/(b*x^3+a)^2+1/18*(-3*a*d+x*(4*b*c*x+a*e))/a^2/b/(b*x^3+a)-1/27*(2*b^(2/3)*c-a^
(2/3)*e)*ln(a^(1/3)+b^(1/3)*x)/a^(7/3)/b^(4/3)+1/54*(2*b^(2/3)*c-a^(2/3)*e)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/
3)*x^2)/a^(7/3)/b^(4/3)-1/27*(2*b^(2/3)*c+a^(2/3)*e)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(7/3)
/b^(4/3)*3^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1828, 1854, 1860, 31, 634, 617, 204, 628} \[ \frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{4/3}}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{4/3}}-\frac {\left (a^{2/3} e+2 b^{2/3} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{4/3}}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x + e*x^2))/(a + b*x^3)^3,x]

[Out]

-(x*(a*e - b*c*x - b*d*x^2))/(6*a*b*(a + b*x^3)^2) - (3*a*d - x*(a*e + 4*b*c*x))/(18*a^2*b*(a + b*x^3)) - ((2*
b^(2/3)*c + a^(2/3)*e)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(7/3)*b^(4/3)) - ((2*b^
(2/3)*c - a^(2/3)*e)*Log[a^(1/3) + b^(1/3)*x])/(27*a^(7/3)*b^(4/3)) + ((2*b^(2/3)*c - a^(2/3)*e)*Log[a^(2/3) -
a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(7/3)*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 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 1828

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 1854

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

Rule 1860

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]

Rubi steps

\begin {align*} \int \frac {x \left (c+d x+e x^2\right )}{\left (a+b x^3\right )^3} \, dx &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {\int \frac {-a e-4 b c x-3 b d x^2}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}+\frac {\int \frac {2 a e+4 b c x}{a+b x^3} \, dx}{18 a^2 b}\\ &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}+\frac {\int \frac {\sqrt [3]{a} \left (4 \sqrt [3]{a} b c+4 a \sqrt [3]{b} e\right )+\sqrt [3]{b} \left (4 \sqrt [3]{a} b c-2 a \sqrt [3]{b} e\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{54 a^{8/3} b^{4/3}}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{7/3} b}\\ &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{4/3}}+\frac {\left (2 b^{2/3} c-a^{2/3} e\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}{54 a^{7/3} b^{4/3}}+\frac {\left (2 b^{2/3} c+a^{2/3} e\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{18 a^2 b}\\ &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{4/3}}+\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{4/3}}+\frac {\left (2 b^{2/3} c+a^{2/3} e\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{9 a^{7/3} b^{4/3}}\\ &=-\frac {x \left (a e-b c x-b d x^2\right )}{6 a b \left (a+b x^3\right )^2}-\frac {3 a d-x (a e+4 b c x)}{18 a^2 b \left (a+b x^3\right )}-\frac {\left (2 b^{2/3} c+a^{2/3} e\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{4/3}}-\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{4/3}}+\frac {\left (2 b^{2/3} c-a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{4/3}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 214, normalized size = 0.90 \[ \frac {\left (2 a^{2/3} b c-a^{4/3} \sqrt [3]{b} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-2 \sqrt {3} a^{2/3} \sqrt [3]{b} \left (a^{2/3} e+2 b^{2/3} c\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \left (a^{4/3} \sqrt [3]{b} e-2 a^{2/3} b c\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+\frac {3 a b^{2/3} \left (-a^2 (3 d+2 e x)+a b x^2 \left (7 c+e x^2\right )+4 b^2 c x^5\right )}{\left (a+b x^3\right )^2}}{54 a^3 b^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x + e*x^2))/(a + b*x^3)^3,x]

[Out]

((3*a*b^(2/3)*(4*b^2*c*x^5 - a^2*(3*d + 2*e*x) + a*b*x^2*(7*c + e*x^2)))/(a + b*x^3)^2 - 2*Sqrt[3]*a^(2/3)*b^(
1/3)*(2*b^(2/3)*c + a^(2/3)*e)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 2*(-2*a^(2/3)*b*c + a^(4/3)*b^(1/
3)*e)*Log[a^(1/3) + b^(1/3)*x] + (2*a^(2/3)*b*c - a^(4/3)*b^(1/3)*e)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)
*x^2])/(54*a^3*b^(5/3))

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fricas [C]  time = 2.73, size = 2519, normalized size = 10.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*(24*b^2*c*x^5 + 6*a*b*e*x^4 + 42*a*b*c*x^2 - 12*a^2*e*x - 18*a^2*d - 2*(a^2*b^3*x^6 + 2*a^3*b^2*x^3 + a^
4*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) +
4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))
^(1/3)))*log(1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^
4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3
)/(a^7*b^4))^(1/3)))^2*a^5*b^3*c - 1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*
c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4)
- (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))*a^4*b*e^2 + 8*a*b*c^2*e + (8*b^2*c^3 + a^2*e^3)*x) + ((a^2*b^3*x^6
+ 2*a^3*b^2*x^3 + a^4*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)
/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 -
a^2*e^3)/(a^7*b^4))^(1/3))) + 3*sqrt(1/3)*(a^2*b^3*x^6 + 2*a^3*b^2*x^3 + a^4*b)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3
) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3
) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))^2*a^4*b^2 + 32*c*e
)/(a^4*b^2)))*log(-1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(
a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a
^2*e^3)/(a^7*b^4))^(1/3)))^2*a^5*b^3*c + 1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (
8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7
*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))*a^4*b*e^2 - 8*a*b*c^2*e + 2*(8*b^2*c^3 + a^2*e^3)*x + 3/2*sqr
t(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)
+ 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^
4))^(1/3)))*a^5*b^3*c + a^4*b*e^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b
^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^
4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))^2*a^4*b^2 + 32*c*e)/(a^4*b^2))) + ((a^2*b^3*x^6 + 2*a^3*b^2*x^3
+ a^4*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3
) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b
^4))^(1/3))) - 3*sqrt(1/3)*(a^2*b^3*x^6 + 2*a^3*b^2*x^3 + a^4*b)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c
^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*
((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))^2*a^4*b^2 + 32*c*e)/(a^4*b^2)))*lo
g(-1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)
+ 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4
))^(1/3)))^2*a^5*b^3*c + 1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*
e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c
^3 - a^2*e^3)/(a^7*b^4))^(1/3)))*a^4*b*e^2 - 8*a*b*c^2*e + 2*(8*b^2*c^3 + a^2*e^3)*x - 3/2*sqrt(1/3)*(((1/2)^(
1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)
*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3)/(a^7*b^4))^(1/3)))*a^5*
b^3*c + a^4*b*e^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3 - a^2*e^3
)/(a^7*b^4))^(1/3) + 4*(1/2)^(2/3)*c*e*(I*sqrt(3) - 1)/(a^4*b^2*((8*b^2*c^3 + a^2*e^3)/(a^7*b^4) - (8*b^2*c^3
- a^2*e^3)/(a^7*b^4))^(1/3)))^2*a^4*b^2 + 32*c*e)/(a^4*b^2))))/(a^2*b^3*x^6 + 2*a^3*b^2*x^3 + a^4*b)

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giac [A]  time = 0.20, size = 215, normalized size = 0.90 \[ -\frac {\sqrt {3} {\left (a e - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} c\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (a e + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} c\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (2 \, b c \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a e\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{3} b} + \frac {4 \, b^{2} c x^{5} + a b x^{4} e + 7 \, a b c x^{2} - 2 \, a^{2} x e - 3 \, a^{2} d}{18 \, {\left (b x^{3} + a\right )}^{2} a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*sqrt(3)*(a*e - 2*(-a*b^2)^(1/3)*c)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)
*a^2) - 1/54*(a*e + 2*(-a*b^2)^(1/3)*c)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a^2) - 1/27*(
2*b*c*(-a/b)^(1/3) + a*e)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b) + 1/18*(4*b^2*c*x^5 + a*b*x^4*e + 7*
a*b*c*x^2 - 2*a^2*x*e - 3*a^2*d)/((b*x^3 + a)^2*a^2*b)

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maple [A]  time = 0.05, size = 256, normalized size = 1.07 \[ \frac {\sqrt {3}\, e \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}+\frac {e \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}-\frac {e \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \left (\frac {a}{b}\right )^{\frac {2}{3}} a \,b^{2}}+\frac {2 \sqrt {3}\, c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b}-\frac {2 c \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b}+\frac {c \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b}+\frac {\frac {2 b c \,x^{5}}{9 a^{2}}+\frac {e \,x^{4}}{18 a}+\frac {7 c \,x^{2}}{18 a}-\frac {e x}{9 b}-\frac {d}{6 b}}{\left (b \,x^{3}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d*x+c)/(b*x^3+a)^3,x)

[Out]

(2/9/a^2*c*b*x^5+1/18/a*e*x^4+7/18/a*c*x^2-1/9/b*e*x-1/6/b*d)/(b*x^3+a)^2+1/27/(a/b)^(2/3)/a/b^2*e*ln(x+(a/b)^
(1/3))-1/54/(a/b)^(2/3)/a/b^2*e*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/27/(a/b)^(2/3)*3^(1/2)/a/b^2*e*arctan(1/3*
3^(1/2)*(2/(a/b)^(1/3)*x-1))-2/27/(a/b)^(1/3)/a^2/b*c*ln(x+(a/b)^(1/3))+1/27/(a/b)^(1/3)/a^2/b*c*ln(x^2-(a/b)^
(1/3)*x+(a/b)^(2/3))+2/27*3^(1/2)/(a/b)^(1/3)/a^2/b*c*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))

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maxima [A]  time = 3.06, size = 223, normalized size = 0.93 \[ \frac {4 \, b^{2} c x^{5} + a b e x^{4} + 7 \, a b c x^{2} - 2 \, a^{2} e x - 3 \, a^{2} d}{18 \, {\left (a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{3} + a^{4} b\right )}} + \frac {\sqrt {3} {\left (2 \, b c \left (\frac {a}{b}\right )^{\frac {1}{3}} + a e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (2 \, b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a e\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (2 \, b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a e\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(4*b^2*c*x^5 + a*b*e*x^4 + 7*a*b*c*x^2 - 2*a^2*e*x - 3*a^2*d)/(a^2*b^3*x^6 + 2*a^3*b^2*x^3 + a^4*b) + 1/2
7*sqrt(3)*(2*b*c*(a/b)^(1/3) + a*e)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^2*b^2*(a/b)^(2/3))
+ 1/54*(2*b*c*(a/b)^(1/3) - a*e)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b^2*(a/b)^(2/3)) - 1/27*(2*b*c*(a
/b)^(1/3) - a*e)*log(x + (a/b)^(1/3))/(a^2*b^2*(a/b)^(2/3))

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mupad [B]  time = 0.23, size = 232, normalized size = 0.97 \[ \frac {\frac {7\,c\,x^2}{18\,a}-\frac {d}{6\,b}+\frac {e\,x^4}{18\,a}-\frac {e\,x}{9\,b}+\frac {2\,b\,c\,x^5}{9\,a^2}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\left (\sum _{k=1}^3\ln \left (\frac {2\,a\,c\,e+{\mathrm {root}\left (19683\,a^7\,b^4\,z^3+162\,a^3\,b^2\,c\,e\,z+8\,b^2\,c^3-a^2\,e^3,z,k\right )}^2\,a^5\,b^2\,729+4\,b\,c^2\,x+\mathrm {root}\left (19683\,a^7\,b^4\,z^3+162\,a^3\,b^2\,c\,e\,z+8\,b^2\,c^3-a^2\,e^3,z,k\right )\,a^3\,b\,e\,x\,27}{a^4\,81}\right )\,\mathrm {root}\left (19683\,a^7\,b^4\,z^3+162\,a^3\,b^2\,c\,e\,z+8\,b^2\,c^3-a^2\,e^3,z,k\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c + d*x + e*x^2))/(a + b*x^3)^3,x)

[Out]

((7*c*x^2)/(18*a) - d/(6*b) + (e*x^4)/(18*a) - (e*x)/(9*b) + (2*b*c*x^5)/(9*a^2))/(a^2 + b^2*x^6 + 2*a*b*x^3)
+ symsum(log((2*a*c*e + 729*root(19683*a^7*b^4*z^3 + 162*a^3*b^2*c*e*z + 8*b^2*c^3 - a^2*e^3, z, k)^2*a^5*b^2
+ 4*b*c^2*x + 27*root(19683*a^7*b^4*z^3 + 162*a^3*b^2*c*e*z + 8*b^2*c^3 - a^2*e^3, z, k)*a^3*b*e*x)/(81*a^4))*
root(19683*a^7*b^4*z^3 + 162*a^3*b^2*c*e*z + 8*b^2*c^3 - a^2*e^3, z, k), k, 1, 3)

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sympy [A]  time = 3.99, size = 170, normalized size = 0.71 \[ \operatorname {RootSum} {\left (19683 t^{3} a^{7} b^{4} + 162 t a^{3} b^{2} c e - a^{2} e^{3} + 8 b^{2} c^{3}, \left (t \mapsto t \log {\left (x + \frac {1458 t^{2} a^{5} b^{3} c + 27 t a^{4} b e^{2} + 8 a b c^{2} e}{a^{2} e^{3} + 8 b^{2} c^{3}} \right )} \right )\right )} + \frac {- 3 a^{2} d - 2 a^{2} e x + 7 a b c x^{2} + a b e x^{4} + 4 b^{2} c x^{5}}{18 a^{4} b + 36 a^{3} b^{2} x^{3} + 18 a^{2} b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x**2+d*x+c)/(b*x**3+a)**3,x)

[Out]

RootSum(19683*_t**3*a**7*b**4 + 162*_t*a**3*b**2*c*e - a**2*e**3 + 8*b**2*c**3, Lambda(_t, _t*log(x + (1458*_t
**2*a**5*b**3*c + 27*_t*a**4*b*e**2 + 8*a*b*c**2*e)/(a**2*e**3 + 8*b**2*c**3)))) + (-3*a**2*d - 2*a**2*e*x + 7
*a*b*c*x**2 + a*b*e*x**4 + 4*b**2*c*x**5)/(18*a**4*b + 36*a**3*b**2*x**3 + 18*a**2*b**3*x**6)

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