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

Optimal. Leaf size=203 \[ \frac{b^{2/3} \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3}}-\frac{\sqrt [3]{b} \left (\sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3}}+\frac{\sqrt [3]{b} \left (\sqrt [3]{a} d+\sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3}}-\frac{e \log \left (a+b x^3\right )}{3 a}-\frac{c}{2 a x^2}-\frac{d}{a x}+\frac{e \log (x)}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.193687, antiderivative size = 203, normalized size of antiderivative = 1., 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 = {1834, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ \frac{b^{2/3} \left (c-\frac{\sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3}}-\frac{\sqrt [3]{b} \left (\sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3}}+\frac{\sqrt [3]{b} \left (\sqrt [3]{a} d+\sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3}}-\frac{e \log \left (a+b x^3\right )}{3 a}-\frac{c}{2 a x^2}-\frac{d}{a x}+\frac{e \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

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 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.177999, size = 192, normalized size = 0.95 \[ \frac{\sqrt [3]{b} \left (\sqrt [3]{a} \sqrt [3]{b} c-a^{2/3} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \sqrt [3]{b} \left (a^{2/3} d-\sqrt [3]{a} \sqrt [3]{b} c\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt{3} \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a} d+\sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )-2 a e \log \left (a+b x^3\right )-\frac{3 a c}{x^2}-\frac{6 a d}{x}+6 a e \log (x)}{6 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 225, normalized size = 1.1 \begin{align*} -{\frac{c}{3\,a}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{c}{6\,a}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{c\sqrt{3}}{3\,a}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{d}{3\,a}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{d}{6\,a}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{d\sqrt{3}}{3\,a}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{e\ln \left ( b{x}^{3}+a \right ) }{3\,a}}-{\frac{d}{ax}}+{\frac{e\ln \left ( x \right ) }{a}}-{\frac{c}{2\,a{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [C]  time = 8.21964, size = 9917, normalized size = 48.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(2*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*
(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e
^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a
*b)/a^5)^(1/3) + 6*e/a)*a*x^2*log(1/36*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18
*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3
) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3
+ a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)^2*a^4*d + 2*a*b*c*d^2 - a*b*c^2*e + a^2*d*e^2 + 1/6*(a^2
*b*c^2 - 2*a^3*d*e)*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/
a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) +
1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 -
3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a) + (b^2*c^3 + a*b*d^3)*x) - 36*e*x^2*log(x) + 36*d*x - (((-I*sqrt(3) + 1)*(e
^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*
(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)
*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a*x^2
+ 3*sqrt(1/3)*a*x^2*sqrt(-(((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a
*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sq
rt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3
- (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)^2*a^3 - 12*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27
*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)
*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5
- 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^2*e + 144*b*c*d + 36*a*e^2)/a^3) - 18*e
*x^2)*log(-1/36*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4
+ 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(
-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c
*d*e)*a*b)/a^5)^(1/3) + 6*e/a)^2*a^4*d - 2*a*b*c*d^2 + a*b*c^2*e - a^2*d*e^2 - 1/6*(a^2*b*c^2 - 2*a^3*d*e)*((-
I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d
^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/1
8*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/
3) + 6*e/a) + 2*(b^2*c^3 + a*b*d^3)*x + 1/12*sqrt(1/3)*(((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/
27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*
e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^
5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^4*d - 6*a^2*b*c^2 - 6*a^3*d*e)*sqrt(-
(((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 +
a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 +
1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)
^(1/3) + 6*e/a)^2*a^3 - 12*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*
e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqr
t(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 -
(d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^2*e + 144*b*c*d + 36*a*e^2)/a^3)) - (((-I*sqrt(3) + 1)*(e^2/a^2 -
(b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3
+ a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 +
1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a*x^2 - 3*sqrt
(1/3)*a*x^2*sqrt(-(((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a
^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1
)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 -
3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)^2*a^3 - 12*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3
+ 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5
)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b
^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^2*e + 144*b*c*d + 36*a*e^2)/a^3) - 18*e*x^2)*log
(-1/36*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b
*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3
/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b
)/a^5)^(1/3) + 6*e/a)^2*a^4*d - 2*a*b*c*d^2 + a*b*c^2*e - a^2*d*e^2 - 1/6*(a^2*b*c^2 - 2*a^3*d*e)*((-I*sqrt(3)
+ 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5
- 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d
+ a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/
a) + 2*(b^2*c^3 + a*b*d^3)*x - 1/12*sqrt(1/3)*(((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^
3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a
^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*
(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^4*d - 6*a^2*b*c^2 - 6*a^3*d*e)*sqrt(-(((-I*sqr
t(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b
/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27*e^3/a^3 + 1/18*(b*
c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) +
6*e/a)^2*a^3 - 12*((-I*sqrt(3) + 1)*(e^2/a^2 - (b*c*d + a*e^2)/a^3)/(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^
4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3*c*d*e)*a*b)/a^5)^(1/3) + 9*(I*sqrt(3) + 1)
*(-1/27*e^3/a^3 + 1/18*(b*c*d + a*e^2)*e/a^4 + 1/54*(b*c^3 + a*d^3)*b/a^5 - 1/54*(b^2*c^3 + a^2*e^3 - (d^3 - 3
*c*d*e)*a*b)/a^5)^(1/3) + 6*e/a)*a^2*e + 144*b*c*d + 36*a*e^2)/a^3)) + 18*c)/(a*x^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.07828, size = 285, normalized size = 1.4 \begin{align*} -\frac{e \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a} + \frac{e \log \left ({\left | x \right |}\right )}{a} - \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b c + \left (-a b^{2}\right )^{\frac{2}{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 )}{6 \, a^{2} b} + \frac{{\left (a b^{2} d \left (-\frac{a}{b}\right )^{\frac{1}{3}} + a b^{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 )}{3 \, a^{3} b} - \frac{2 \, d x + c}{2 \, a x^{2}} - \frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} a b^{3} c - \left (-a b^{2}\right )^{\frac{2}{3}} a b^{2} 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 )}{3 \, a^{3} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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