3.350 \(\int \frac{c+d x+e x^2}{x^4 (a+b x^3)^2} \, dx\)

Optimal. Leaf size=262 \[ -\frac{x \left (-\frac{b^2 c x^2}{a}+b d+b e x\right )}{3 a^2 \left (a+b x^3\right )}+\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3}}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}-\frac{2 b c \log (x)}{a^3}-\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{\sqrt [3]{b} \left (4 \sqrt [3]{a} e+5 \sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3}}-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x} \]

[Out]

-c/(3*a^2*x^3) - d/(2*a^2*x^2) - e/(a^2*x) - (x*(b*d + b*e*x - (b^2*c*x^2)/a))/(3*a^2*(a + b*x^3)) + (b^(1/3)*
(5*b^(1/3)*d + 4*a^(1/3)*e)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)) - (2*b*c*Lo
g[x])/a^3 - (b^(1/3)*(5*b^(1/3)*d - 4*a^(1/3)*e)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(8/3)) + (b^(1/3)*(5*b^(1/3)*d
- 4*a^(1/3)*e)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(8/3)) + (2*b*c*Log[a + b*x^3])/(3*a^3)

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Rubi [A]  time = 0.403572, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {1829, 1834, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ -\frac{x \left (-\frac{b^2 c x^2}{a}+b d+b e x\right )}{3 a^2 \left (a+b x^3\right )}+\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3}}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}-\frac{2 b c \log (x)}{a^3}-\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{\sqrt [3]{b} \left (4 \sqrt [3]{a} e+5 \sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3}}-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(3*a^2*x^3) - d/(2*a^2*x^2) - e/(a^2*x) - (x*(b*d + b*e*x - (b^2*c*x^2)/a))/(3*a^2*(a + b*x^3)) + (b^(1/3)*
(5*b^(1/3)*d + 4*a^(1/3)*e)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)) - (2*b*c*Lo
g[x])/a^3 - (b^(1/3)*(5*b^(1/3)*d - 4*a^(1/3)*e)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(8/3)) + (b^(1/3)*(5*b^(1/3)*d
- 4*a^(1/3)*e)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(8/3)) + (2*b*c*Log[a + b*x^3])/(3*a^3)

Rule 1829

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)*Coeff[R, x, i]*x^(i - m))/a, {i, 0, n - 1}], x], x], x] - S
imp[(x*R*(a + b*x^n)^(p + 1))/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

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^4 \left (a+b x^3\right )^2} \, dx &=-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{\int \frac{-3 b c-3 b d x-3 b e x^2+\frac{3 b^2 c x^3}{a}+\frac{2 b^2 d x^4}{a}+\frac{b^2 e x^5}{a}}{x^4 \left (a+b x^3\right )} \, dx}{3 a b}\\ &=-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{\int \left (-\frac{3 b c}{a x^4}-\frac{3 b d}{a x^3}-\frac{3 b e}{a x^2}+\frac{6 b^2 c}{a^2 x}+\frac{b^2 \left (5 a d+4 a e x-6 b c x^2\right )}{a^2 \left (a+b x^3\right )}\right ) \, dx}{3 a b}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{2 b c \log (x)}{a^3}-\frac{b \int \frac{5 a d+4 a e x-6 b c x^2}{a+b x^3} \, dx}{3 a^3}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{2 b c \log (x)}{a^3}-\frac{b \int \frac{5 a d+4 a e x}{a+b x^3} \, dx}{3 a^3}+\frac{\left (2 b^2 c\right ) \int \frac{x^2}{a+b x^3} \, dx}{a^3}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{2 b c \log (x)}{a^3}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}-\frac{b^{2/3} \int \frac{\sqrt [3]{a} \left (10 a \sqrt [3]{b} d+4 a^{4/3} e\right )+\sqrt [3]{b} \left (-5 a \sqrt [3]{b} d+4 a^{4/3} e\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{9 a^{11/3}}-\frac{\left (b \left (5 d-\frac{4 \sqrt [3]{a} e}{\sqrt [3]{b}}\right )\right ) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 a^{8/3}}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{2 b c \log (x)}{a^3}-\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}+\frac{\left (\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\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}{18 a^{8/3}}-\frac{\left (b^{2/3} \left (5 \sqrt [3]{b} d+4 \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}{6 a^{7/3}}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}-\frac{2 b c \log (x)}{a^3}-\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3}}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}-\frac{\left (\sqrt [3]{b} \left (5 \sqrt [3]{b} d+4 \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 )}{3 a^{8/3}}\\ &=-\frac{c}{3 a^2 x^3}-\frac{d}{2 a^2 x^2}-\frac{e}{a^2 x}-\frac{x \left (b d+b e x-\frac{b^2 c x^2}{a}\right )}{3 a^2 \left (a+b x^3\right )}+\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d+4 \sqrt [3]{a} e\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3}}-\frac{2 b c \log (x)}{a^3}-\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{\sqrt [3]{b} \left (5 \sqrt [3]{b} d-4 \sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3}}+\frac{2 b c \log \left (a+b x^3\right )}{3 a^3}\\ \end{align*}

Mathematica [A]  time = 0.174743, size = 225, normalized size = 0.86 \[ \frac{\sqrt [3]{b} \left (5 \sqrt [3]{a} \sqrt [3]{b} d-4 a^{2/3} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \sqrt [3]{b} \left (4 a^{2/3} e-5 \sqrt [3]{a} \sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-\frac{6 a b (c+x (d+e x))}{a+b x^3}+12 b c \log \left (a+b x^3\right )+2 \sqrt{3} \sqrt [3]{a} \sqrt [3]{b} \left (4 \sqrt [3]{a} e+5 \sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )-\frac{6 a c}{x^3}-\frac{9 a d}{x^2}-\frac{18 a e}{x}-36 b c \log (x)}{18 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a*c)/x^3 - (9*a*d)/x^2 - (18*a*e)/x - (6*a*b*(c + x*(d + e*x)))/(a + b*x^3) + 2*Sqrt[3]*a^(1/3)*b^(1/3)*(
5*b^(1/3)*d + 4*a^(1/3)*e)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 36*b*c*Log[x] + 2*b^(1/3)*(-5*a^(1/3)
*b^(1/3)*d + 4*a^(2/3)*e)*Log[a^(1/3) + b^(1/3)*x] + b^(1/3)*(5*a^(1/3)*b^(1/3)*d - 4*a^(2/3)*e)*Log[a^(2/3) -
a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 12*b*c*Log[a + b*x^3])/(18*a^3)

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Maple [A]  time = 0.013, size = 289, normalized size = 1.1 \begin{align*} -{\frac{be{x}^{2}}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{bdx}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{bc}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{5\,d}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{5\,d}{18\,{a}^{2}}\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{5\,d\sqrt{3}}{9\,{a}^{2}}\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{4\,e}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{2\,e}{9\,{a}^{2}}\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{4\,e\sqrt{3}}{9\,{a}^{2}}\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{2\,bc\ln \left ( b{x}^{3}+a \right ) }{3\,{a}^{3}}}-{\frac{d}{2\,{a}^{2}{x}^{2}}}-{\frac{e}{{a}^{2}x}}-{\frac{c}{3\,{x}^{3}{a}^{2}}}-2\,{\frac{bc\ln \left ( x \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^4/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 10.1793, size = 12670, normalized size = 48.36 \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^4/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

-1/36*(48*a*b*e*x^5 + 30*a*b*d*x^4 + 24*a*b*c*x^3 + 36*a^2*e*x^2 + 18*a^2*d*x + 12*a^2*c + 2*(a^3*b*x^6 + a^4*
x^3)*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d
^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e
)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*
c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)
*log((8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d
^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e
)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*
c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)
^2*a^6*e + 150*b^2*c*d^2 + 144*b^2*c^2*e + 160*a*b*d*e^2 + 1/2*(25*a^3*b*d^2 + 48*a^3*b*c*e)*(8*(1/2)^(2/3)*(-
I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 -
72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) +
(1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a
^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3) + (125*b^2*d^3 + 64*a
*b*e^3)*x) - (36*b^2*c*x^6 + 36*a*b*c*x^3 + (a^3*b*x^6 + a^4*x^3)*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a
^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)
*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)
*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*
b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3) + 3*sqrt(1/3)*(a^3*b*x^6 + a^4*x^3)*sqrt(-((8*(1
/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*
e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a
^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*
b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)^2*a^6 + 2
4*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3
+ 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a
*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2
+ 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)*a^
3*b*c + 144*b^2*c^2 + 320*a*b*d*e)/a^6))*log(-(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*
a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*
c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 +
(125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 -
72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)^2*a^6*e - 150*b^2*c*d^2 - 144*b^2*c^2*e - 160*a*b*d*e^2 - 1/2*(25*a^
3*b*d^2 + 48*a^3*b*c*e)*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3
*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5
*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)
*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^
(1/3) - 12*b*c/a^3) + 2*(125*b^2*d^3 + 64*a*b*e^3)*x + 3/2*sqrt(1/3)*(2*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2
*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*
b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3
) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 6
4*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)*a^6*e - 25*a^3*b*d^2 + 24*a^3*b*c*e)*sqrt(
-((8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3
+ 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a
*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2
+ 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)^2*
a^6 + 24*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125
*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c
*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*
b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/
a^3)*a^3*b*c + 144*b^2*c^2 + 320*a*b*d*e)/a^6)) - (36*b^2*c*x^6 + 36*a*b*c*x^3 + (a^3*b*x^6 + a^4*x^3)*(8*(1/2
)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^
3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9
)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*
d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3) - 3*sqrt(1/
3)*(a^3*b*x^6 + a^4*x^3)*sqrt(-((8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/
(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b
*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 6
4*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^
2)/a^9)^(1/3) - 12*b*c/a^3)^2*a^6 + 24*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e
)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 6
4*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*
d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*
e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)*a^3*b*c + 144*b^2*c^2 + 320*a*b*d*e)/a^6))*log(-(8*(1/2)^(2/3)*(-I*sqrt(3)
+ 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2
*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3
)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*
b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)^2*a^6*e - 150*b^2*c*d^2 - 144*b
^2*c^2*e - 160*a*b*d*e^2 - 1/2*(25*a^3*b*d^2 + 48*a^3*b*c*e)*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 -
(9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/
a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432
*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3
- 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3) + 2*(125*b^2*d^3 + 64*a*b*e^3)*x - 3/2*sqrt(1/3)*(2*(
8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 6
4*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^
2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 +
5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)*a^6*e
- 25*a^3*b*d^2 + 24*a^3*b*c*e)*sqrt(-((8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*a*b*d*e
)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 6
4*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 + (125*b*
d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*
e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)^2*a^6 + 24*(8*(1/2)^(2/3)*(-I*sqrt(3) + 1)*(9*b^2*c^2/a^6 - (9*b^2*c^2 + 5*
a*b*d*e)/a^6)/(432*b^3*c^3/a^9 + (125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*
c^3 + 64*a^2*b*e^3 - 5*(25*d^3 - 72*c*d*e)*a*b^2)/a^9)^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(432*b^3*c^3/a^9 +
(125*b*d^3 + 64*a*e^3)*b/a^8 - 72*(9*b^2*c^2 + 5*a*b*d*e)*b*c/a^9 + (216*b^3*c^3 + 64*a^2*b*e^3 - 5*(25*d^3 -
72*c*d*e)*a*b^2)/a^9)^(1/3) - 12*b*c/a^3)*a^3*b*c + 144*b^2*c^2 + 320*a*b*d*e)/a^6)) + 72*(b^2*c*x^6 + a*b*c*x
^3)*log(x))/(a^3*b*x^6 + a^4*x^3)

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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**4/(b*x**3+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.07759, size = 373, normalized size = 1.42 \begin{align*} \frac{2 \, b c \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3}} - \frac{2 \, b c \log \left ({\left | x \right |}\right )}{a^{3}} - \frac{{\left (5 \, \left (-a b^{2}\right )^{\frac{1}{3}} b d + 4 \, \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 )}{18 \, a^{3} b} - \frac{\sqrt{3}{\left (5 \, \left (-a b^{2}\right )^{\frac{1}{3}} a b^{3} d - 4 \, \left (-a b^{2}\right )^{\frac{2}{3}} a b^{2} 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^{4} b^{3}} + \frac{{\left (4 \, a^{4} b^{2} \left (-\frac{a}{b}\right )^{\frac{1}{3}} e + 5 \, a^{4} b^{2} d\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^{7} b} - \frac{8 \, a b x^{5} e + 5 \, a b d x^{4} + 4 \, a b c x^{3} + 6 \, a^{2} x^{2} e + 3 \, a^{2} d x + 2 \, a^{2} c}{6 \,{\left (b x^{3} + a\right )} a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b*c*log(abs(b*x^3 + a))/a^3 - 2*b*c*log(abs(x))/a^3 - 1/18*(5*(-a*b^2)^(1/3)*b*d + 4*(-a*b^2)^(2/3)*e)*log
(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b) - 1/9*sqrt(3)*(5*(-a*b^2)^(1/3)*a*b^3*d - 4*(-a*b^2)^(2/3)*a*b^2
*e)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^4*b^3) + 1/9*(4*a^4*b^2*(-a/b)^(1/3)*e + 5*a^4*b^
2*d)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a^7*b) - 1/6*(8*a*b*x^5*e + 5*a*b*d*x^4 + 4*a*b*c*x^3 + 6*a^2*x^
2*e + 3*a^2*d*x + 2*a^2*c)/((b*x^3 + a)*a^3*x^3)