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

Optimal. Leaf size=193 \[ \frac {\sqrt [3]{a} \left (d-\frac {\sqrt [3]{a} e}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{4/3}}-\frac {\sqrt [3]{a} \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {\sqrt [3]{a} \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3}}+\frac {c \log \left (a+b x^3\right )}{3 b}+\frac {d x}{b}+\frac {e x^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1887, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ \frac {\sqrt [3]{a} \left (d-\frac {\sqrt [3]{a} e}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{4/3}}-\frac {\sqrt [3]{a} \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {\sqrt [3]{a} \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3}}+\frac {c \log \left (a+b x^3\right )}{3 b}+\frac {d x}{b}+\frac {e x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 260

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 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 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]

Rule 1871

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 1887

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

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Mathematica [A]  time = 0.10, size = 184, normalized size = 0.95 \[ \frac {-\left (a^{2/3} e-\sqrt [3]{a} \sqrt [3]{b} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \left (a^{2/3} e-\sqrt [3]{a} \sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 b^{2/3} c \log \left (a+b x^3\right )+2 \sqrt {3} \sqrt [3]{a} \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+6 b^{2/3} d x+3 b^{2/3} e x^2}{6 b^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 2.76, size = 4261, normalized size = 22.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*e*x^2 - 2*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*
e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(
3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*
b)/b^5)^(1/3) - 2*c/b)*b*log(1/4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 -
3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)
^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^
3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3*e + b*c*d^2 + b*c^2*e + 2*a*d*e^2 + 1/2*(b^2*d^2 + 2*b^2*c*e)*(2*(
1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*
e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3
*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)
+ (b*d^3 + a*e^3)*x) + 12*d*x + ((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 -
3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2
)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d
^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b + 3*sqrt(1/3)*b*sqrt(-((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*
c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3
*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a
/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)
*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2
*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b
*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b^2*c + 4*b*c^2 + 16*a*d*e
)/b^3) + 6*c)*log(-1/4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 +
a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*
sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*
e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3*e - b*c*d^2 - b*c^2*e - 2*a*d*e^2 - 1/2*(b^2*d^2 + 2*b^2*c*e)*(2*(1/2)^(2/3)
*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5
+ (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 +
a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b) + 2*(b*d^
3 + a*e^3)*x + 3/4*sqrt(1/3)*((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(
b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1
/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 -
3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b^3*e - 2*b^2*d^2 + 2*b^2*c*e)*sqrt(-((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/
b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 -
(d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 +
a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(-I*sqr
t(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*
c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c
/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b^2*c + 4*b*c^2 +
16*a*d*e)/b^3)) + ((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*
d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqr
t(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*
a*b)/b^5)^(1/3) - 2*c/b)*b - 3*sqrt(1/3)*b*sqrt(-((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b
^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b
^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^
3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b
*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 -
3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*
a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b^2*c + 4*b*c^2 + 16*a*d*e)/b^3) + 6*c)*
log(-1/4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4
+ (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(
2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(
1/3) - 2*c/b)^2*b^3*e - b*c*d^2 - b*c^2*e - 2*a*d*e^2 - 1/2*(b^2*d^2 + 2*b^2*c*e)*(2*(1/2)^(2/3)*(-I*sqrt(3) +
1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 +
a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 +
(b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b) + 2*(b*d^3 + a*e^3)*x -
3/4*sqrt(1/3)*((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)
*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3)
+ 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)
/b^5)^(1/3) - 2*c/b)*b^3*e - 2*b^2*d^2 + 2*b^2*c*e)*sqrt(-((2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2/b^2 - (b*c^2 +
a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*
e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5
+ (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(c^2
/b^2 - (b*c^2 + a*d*e)/b^3)/(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3 + a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3
- (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(2*c^3/b^3 - 3*(b*c^2 + a*d*e)*c/b^4 + (b*d^3
+ a*e^3)*a/b^5 + (b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/b^5)^(1/3) - 2*c/b)*b^2*c + 4*b*c^2 + 16*a*d*e)/b^3
)))/b

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giac [A]  time = 0.21, size = 195, normalized size = 1.01 \[ \frac {c \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b} - \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} b d - \left (-a b^{2}\right )^{\frac {2}{3}} e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3}} + \frac {b x^{2} e + 2 \, b d x}{2 \, b^{2}} - \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b d + \left (-a b^{2}\right )^{\frac {2}{3}} e\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3}} + \frac {{\left (a b^{4} \left (-\frac {a}{b}\right )^{\frac {1}{3}} e + a b^{4} d\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 221, normalized size = 1.15 \[ \frac {e \,x^{2}}{2 b}-\frac {\sqrt {3}\, a d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {a d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}+\frac {a d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {\sqrt {3}\, a e \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}+\frac {a e \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}-\frac {a e \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}+\frac {c \ln \left (b \,x^{3}+a \right )}{3 b}+\frac {d x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 181, normalized size = 0.94 \[ -\frac {\sqrt {3} {\left (a e \left (\frac {a}{b}\right )^{\frac {2}{3}} + a d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b} + \frac {e x^{2} + 2 \, d x}{2 \, b} + \frac {{\left (2 \, b c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a e \left (\frac {a}{b}\right )^{\frac {1}{3}} + a d\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (b c \left (\frac {a}{b}\right )^{\frac {2}{3}} + a e \left (\frac {a}{b}\right )^{\frac {1}{3}} - a d\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.13, size = 340, normalized size = 1.76 \[ \left (\sum _{k=1}^3\ln \left (\frac {a\,\left (b\,c^2+{\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c\,z^2+9\,a\,b^2\,d\,e\,z+9\,b^3\,c^2\,z-3\,a\,b\,c\,d\,e+a\,b\,d^3-a^2\,e^3-b^2\,c^3,z,k\right )}^2\,b^3\,9+a\,d\,e-\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c\,z^2+9\,a\,b^2\,d\,e\,z+9\,b^3\,c^2\,z-3\,a\,b\,c\,d\,e+a\,b\,d^3-a^2\,e^3-b^2\,c^3,z,k\right )\,b^2\,c\,6+a\,e^2\,x+b\,c\,d\,x-\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c\,z^2+9\,a\,b^2\,d\,e\,z+9\,b^3\,c^2\,z-3\,a\,b\,c\,d\,e+a\,b\,d^3-a^2\,e^3-b^2\,c^3,z,k\right )\,b^2\,d\,x\,3\right )}{b}\right )\,\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c\,z^2+9\,a\,b^2\,d\,e\,z+9\,b^3\,c^2\,z-3\,a\,b\,c\,d\,e+a\,b\,d^3-a^2\,e^3-b^2\,c^3,z,k\right )\right )+\frac {e\,x^2}{2\,b}+\frac {d\,x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((a*(b*c^2 + 9*root(27*b^5*z^3 - 27*b^4*c*z^2 + 9*a*b^2*d*e*z + 9*b^3*c^2*z - 3*a*b*c*d*e + a*b*d^3
- a^2*e^3 - b^2*c^3, z, k)^2*b^3 + a*d*e - 6*root(27*b^5*z^3 - 27*b^4*c*z^2 + 9*a*b^2*d*e*z + 9*b^3*c^2*z - 3*
a*b*c*d*e + a*b*d^3 - a^2*e^3 - b^2*c^3, z, k)*b^2*c + a*e^2*x + b*c*d*x - 3*root(27*b^5*z^3 - 27*b^4*c*z^2 +
9*a*b^2*d*e*z + 9*b^3*c^2*z - 3*a*b*c*d*e + a*b*d^3 - a^2*e^3 - b^2*c^3, z, k)*b^2*d*x))/b)*root(27*b^5*z^3 -
27*b^4*c*z^2 + 9*a*b^2*d*e*z + 9*b^3*c^2*z - 3*a*b*c*d*e + a*b*d^3 - a^2*e^3 - b^2*c^3, z, k), k, 1, 3) + (e*x
^2)/(2*b) + (d*x)/b

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sympy [A]  time = 1.49, size = 150, normalized size = 0.78 \[ \operatorname {RootSum} {\left (27 t^{3} b^{5} - 27 t^{2} b^{4} c + t \left (9 a b^{2} d e + 9 b^{3} c^{2}\right ) - a^{2} e^{3} - 3 a b c d e + a b d^{3} - b^{2} c^{3}, \left (t \mapsto t \log {\left (x + \frac {9 t^{2} b^{3} e - 6 t b^{2} c e - 3 t b^{2} d^{2} + 2 a d e^{2} + b c^{2} e + b c d^{2}}{a e^{3} + b d^{3}} \right )} \right )\right )} + \frac {d x}{b} + \frac {e x^{2}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(27*_t**3*b**5 - 27*_t**2*b**4*c + _t*(9*a*b**2*d*e + 9*b**3*c**2) - a**2*e**3 - 3*a*b*c*d*e + a*b*d**3
- b**2*c**3, Lambda(_t, _t*log(x + (9*_t**2*b**3*e - 6*_t*b**2*c*e - 3*_t*b**2*d**2 + 2*a*d*e**2 + b*c**2*e +
b*c*d**2)/(a*e**3 + b*d**3)))) + d*x/b + e*x**2/(2*b)

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