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

Optimal. Leaf size=199 \[ -\frac {\left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{2/3}}+\frac {\left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{2/3}}-\frac {\left (\sqrt [3]{a} d+2 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{2/3}}-\frac {a e-b x (c+d x)}{3 a b \left (a+b x^3\right )} \]

[Out]

1/3*(-a*e+b*x*(d*x+c))/a/b/(b*x^3+a)+1/9*(2*b^(1/3)*c-a^(1/3)*d)*ln(a^(1/3)+b^(1/3)*x)/a^(5/3)/b^(2/3)-1/18*(2
*b^(1/3)*c-a^(1/3)*d)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(5/3)/b^(2/3)-1/9*(2*b^(1/3)*c+a^(1/3)*d)*ar
ctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(5/3)/b^(2/3)*3^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1854, 1860, 31, 634, 617, 204, 628} \[ -\frac {\left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{2/3}}+\frac {\left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{2/3}}-\frac {\left (\sqrt [3]{a} d+2 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} b^{2/3}}-\frac {a e-b x (c+d x)}{3 a b \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.28, size = 189, normalized size = 0.95 \[ \frac {\sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a} d-2 \sqrt [3]{b} c\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+\left (4 \sqrt [3]{a} b^{2/3} c-2 a^{2/3} \sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+\frac {6 a (b x (c+d x)-a e)}{a+b x^3}-2 \sqrt {3} \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a} d+2 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{18 a^2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 2.60, size = 2118, normalized size = 10.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(12*b*d*x^2 + 12*b*c*x - 2*(a*b^2*x^3 + a^2*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2)
+ (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) +
(8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))*log(1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b
*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*
c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^4*b*d - 2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b
*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*
c^3 - a*d^3)/(a^5*b^2))^(1/3)))*a^2*b*c^2 + 4*a*c*d^2 + (8*b*c^3 + a*d^3)*x) - 12*a*e + ((a*b^2*x^3 + a^2*b)*(
(1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*
c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3))) + 3*sqrt(1/3)*(
a*b^2*x^3 + a^2*b)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b
^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^
2))^(1/3)))^2*a^3*b + 32*c*d)/(a^3*b)))*log(-1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (
8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8
*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^4*b*d + 2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (
8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8
*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))*a^2*b*c^2 - 4*a*c*d^2 + 2*(8*b*c^3 + a*d^3)*x + 3/4*sqrt(1/3)*(((1/2)^(1/3)
*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt
(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))*a^4*b*d + 8*a^2*b*c^2)*sqr
t(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(
2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^3*b +
32*c*d)/(a^3*b))) + ((a*b^2*x^3 + a^2*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3
- a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 -
a*d^3)/(a^5*b^2))^(1/3))) - 3*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a
*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*
d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^3*b + 32*c*d)/(a^3*b)))*log(-1/4*((1/2)^(1/3)*(I*sqr
t(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1
)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^4*b*d + 2*((1/2)^(1/3)*(I*sqr
t(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1
)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^5*b^2))^(1/3)))*a^2*b*c^2 - 4*a*c*d^2 + 2*(8*b*c^
3 + a*d^3)*x - 3/4*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a
^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*c^3 - a*d^3)/(a^
5*b^2))^(1/3)))*a^4*b*d + 8*a^2*b*c^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b
*c^3 - a*d^3)/(a^5*b^2))^(1/3) + 4*(1/2)^(2/3)*c*d*(I*sqrt(3) - 1)/(a^3*b*((8*b*c^3 + a*d^3)/(a^5*b^2) + (8*b*
c^3 - a*d^3)/(a^5*b^2))^(1/3)))^2*a^3*b + 32*c*d)/(a^3*b))))/(a*b^2*x^3 + a^2*b)

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giac [A]  time = 0.18, size = 184, normalized size = 0.92 \[ -\frac {\sqrt {3} {\left (2 \, b c - \left (-a b^{2}\right )^{\frac {1}{3}} d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} a} - \frac {{\left (2 \, b c + \left (-a b^{2}\right )^{\frac {1}{3}} d\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} a} - \frac {{\left (d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, c\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2}} + \frac {b d x^{2} + b c x - a e}{3 \, {\left (b x^{3} + a\right )} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 253, normalized size = 1.27 \[ \frac {d \,x^{2}}{3 \left (b \,x^{3}+a \right ) a}+\frac {c x}{3 \left (b \,x^{3}+a \right ) a}+\frac {2 \sqrt {3}\, c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a b}+\frac {2 c \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a b}-\frac {c \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} a b}+\frac {\sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} a b}-\frac {d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} a b}+\frac {d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \left (\frac {a}{b}\right )^{\frac {1}{3}} a b}-\frac {e}{3 \left (b \,x^{3}+a \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.02, size = 179, normalized size = 0.90 \[ \frac {b d x^{2} + b c x - a e}{3 \, {\left (a b^{2} x^{3} + a^{2} b\right )}} + \frac {\sqrt {3} {\left (d \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, c\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, c\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, c\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.25, size = 175, normalized size = 0.88 \[ \left (\sum _{k=1}^3\ln \left (\frac {b\,\left (2\,c\,d+d^2\,x+{\mathrm {root}\left (729\,a^5\,b^2\,z^3+54\,a^2\,b\,c\,d\,z-8\,b\,c^3+a\,d^3,z,k\right )}^2\,a^3\,b\,81+\mathrm {root}\left (729\,a^5\,b^2\,z^3+54\,a^2\,b\,c\,d\,z-8\,b\,c^3+a\,d^3,z,k\right )\,a\,b\,c\,x\,18\right )}{a^2\,9}\right )\,\mathrm {root}\left (729\,a^5\,b^2\,z^3+54\,a^2\,b\,c\,d\,z-8\,b\,c^3+a\,d^3,z,k\right )\right )+\frac {\frac {d\,x^2}{3\,a}-\frac {e}{3\,b}+\frac {c\,x}{3\,a}}{b\,x^3+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((b*(2*c*d + d^2*x + 81*root(729*a^5*b^2*z^3 + 54*a^2*b*c*d*z - 8*b*c^3 + a*d^3, z, k)^2*a^3*b + 18*
root(729*a^5*b^2*z^3 + 54*a^2*b*c*d*z - 8*b*c^3 + a*d^3, z, k)*a*b*c*x))/(9*a^2))*root(729*a^5*b^2*z^3 + 54*a^
2*b*c*d*z - 8*b*c^3 + a*d^3, z, k), k, 1, 3) + ((d*x^2)/(3*a) - e/(3*b) + (c*x)/(3*a))/(a + b*x^3)

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sympy [A]  time = 1.38, size = 116, normalized size = 0.58 \[ \operatorname {RootSum} {\left (729 t^{3} a^{5} b^{2} + 54 t a^{2} b c d + a d^{3} - 8 b c^{3}, \left (t \mapsto t \log {\left (x + \frac {81 t^{2} a^{4} b d + 36 t a^{2} b c^{2} + 4 a c d^{2}}{a d^{3} + 8 b c^{3}} \right )} \right )\right )} + \frac {- a e + b c x + b d x^{2}}{3 a^{2} b + 3 a b^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(729*_t**3*a**5*b**2 + 54*_t*a**2*b*c*d + a*d**3 - 8*b*c**3, Lambda(_t, _t*log(x + (81*_t**2*a**4*b*d +
36*_t*a**2*b*c**2 + 4*a*c*d**2)/(a*d**3 + 8*b*c**3)))) + (-a*e + b*c*x + b*d*x**2)/(3*a**2*b + 3*a*b**2*x**3)

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