3.103 \(\int \frac{1}{x^3 (a x+b x^3+c x^5)^2} \, dx\)

Optimal. Leaf size=219 \[ \frac{b \left (30 a^2 c^2-20 a b^2 c+3 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^4 \left (b^2-4 a c\right )^{3/2}}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 x^2 \left (b^2-4 a c\right )}-\frac{3 b^2-8 a c}{4 a^2 x^4 \left (b^2-4 a c\right )}-\frac{\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^4}+\frac{\log (x) \left (3 b^2-2 a c\right )}{a^4}+\frac{-2 a c+b^2+b c x^2}{2 a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

-(3*b^2 - 8*a*c)/(4*a^2*(b^2 - 4*a*c)*x^4) + (b*(3*b^2 - 11*a*c))/(2*a^3*(b^2 - 4*a*c)*x^2) + (b^2 - 2*a*c + b
*c*x^2)/(2*a*(b^2 - 4*a*c)*x^4*(a + b*x^2 + c*x^4)) + (b*(3*b^4 - 20*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2*c*x^
2)/Sqrt[b^2 - 4*a*c]])/(2*a^4*(b^2 - 4*a*c)^(3/2)) + ((3*b^2 - 2*a*c)*Log[x])/a^4 - ((3*b^2 - 2*a*c)*Log[a + b
*x^2 + c*x^4])/(4*a^4)

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Rubi [A]  time = 0.312461, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1585, 1114, 740, 800, 634, 618, 206, 628} \[ \frac{b \left (30 a^2 c^2-20 a b^2 c+3 b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^4 \left (b^2-4 a c\right )^{3/2}}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 x^2 \left (b^2-4 a c\right )}-\frac{3 b^2-8 a c}{4 a^2 x^4 \left (b^2-4 a c\right )}-\frac{\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^4}+\frac{\log (x) \left (3 b^2-2 a c\right )}{a^4}+\frac{-2 a c+b^2+b c x^2}{2 a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a*x + b*x^3 + c*x^5)^2),x]

[Out]

-(3*b^2 - 8*a*c)/(4*a^2*(b^2 - 4*a*c)*x^4) + (b*(3*b^2 - 11*a*c))/(2*a^3*(b^2 - 4*a*c)*x^2) + (b^2 - 2*a*c + b
*c*x^2)/(2*a*(b^2 - 4*a*c)*x^4*(a + b*x^2 + c*x^4)) + (b*(3*b^4 - 20*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2*c*x^
2)/Sqrt[b^2 - 4*a*c]])/(2*a^4*(b^2 - 4*a*c)^(3/2)) + ((3*b^2 - 2*a*c)*Log[x])/a^4 - ((3*b^2 - 2*a*c)*Log[a + b
*x^2 + c*x^4])/(4*a^4)

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx &=\int \frac{1}{x^5 \left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 b^2+8 a c-3 b c x}{x^3 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{-3 b^2+8 a c}{a x^3}+\frac{3 b^3-11 a b c}{a^2 x^2}+\frac{\left (b^2-4 a c\right ) \left (-3 b^2+2 a c\right )}{a^3 x}+\frac{b \left (3 b^4-17 a b^2 c+19 a^2 c^2\right )+c \left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}\\ &=-\frac{3 b^2-8 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}+\frac{\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac{\operatorname{Subst}\left (\int \frac{b \left (3 b^4-17 a b^2 c+19 a^2 c^2\right )+c \left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^4 \left (b^2-4 a c\right )}\\ &=-\frac{3 b^2-8 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}+\frac{\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac{\left (3 b^2-2 a c\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^4}-\frac{\left (b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^4 \left (b^2-4 a c\right )}\\ &=-\frac{3 b^2-8 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}+\frac{\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac{\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^4}+\frac{\left (b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 a^4 \left (b^2-4 a c\right )}\\ &=-\frac{3 b^2-8 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac{b \left (3 b^2-11 a c\right )}{2 a^3 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}+\frac{b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^4 \left (b^2-4 a c\right )^{3/2}}+\frac{\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac{\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^4}\\ \end{align*}

Mathematica [A]  time = 0.365738, size = 328, normalized size = 1.5 \[ \frac{\frac{2 a \left (2 a^2 c^2-4 a b^2 c-3 a b c^2 x^2+b^3 c x^2+b^4\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (8 a^2 c^2 \sqrt{b^2-4 a c}+30 a^2 b c^2+3 b^4 \sqrt{b^2-4 a c}-20 a b^3 c-14 a b^2 c \sqrt{b^2-4 a c}+3 b^5\right ) \log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\left (-8 a^2 c^2 \sqrt{b^2-4 a c}+30 a^2 b c^2-3 b^4 \sqrt{b^2-4 a c}-20 a b^3 c+14 a b^2 c \sqrt{b^2-4 a c}+3 b^5\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{a^2}{x^4}+4 \log (x) \left (3 b^2-2 a c\right )+\frac{4 a b}{x^2}}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a*x + b*x^3 + c*x^5)^2),x]

[Out]

(-(a^2/x^4) + (4*a*b)/x^2 + (2*a*(b^4 - 4*a*b^2*c + 2*a^2*c^2 + b^3*c*x^2 - 3*a*b*c^2*x^2))/((b^2 - 4*a*c)*(a
+ b*x^2 + c*x^4)) + 4*(3*b^2 - 2*a*c)*Log[x] - ((3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2 + 3*b^4*Sqrt[b^2 - 4*a*c] -
14*a*b^2*c*Sqrt[b^2 - 4*a*c] + 8*a^2*c^2*Sqrt[b^2 - 4*a*c])*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*
c)^(3/2) + ((3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2 - 3*b^4*Sqrt[b^2 - 4*a*c] + 14*a*b^2*c*Sqrt[b^2 - 4*a*c] - 8*a^
2*c^2*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/(4*a^4)

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Maple [B]  time = 0.019, size = 443, normalized size = 2. \begin{align*}{\frac{3\,{c}^{2}b{x}^{2}}{2\,{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{b}^{3}c{x}^{2}}{2\,{a}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{c}^{2}}{a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{{b}^{2}c}{{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{b}^{4}}{2\,{a}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{{c}^{2}\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) }{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{7\,c\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}}{2\,{a}^{3} \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{3\,\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{4}}{4\,{a}^{4} \left ( 4\,ac-{b}^{2} \right ) }}+15\,{\frac{{c}^{2}b}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-10\,{\frac{{b}^{3}c}{{a}^{3} \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{3\,{b}^{5}}{2\,{a}^{4}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{4\,{a}^{2}{x}^{4}}}-2\,{\frac{\ln \left ( x \right ) c}{{a}^{3}}}+3\,{\frac{\ln \left ( x \right ){b}^{2}}{{a}^{4}}}+{\frac{b}{{x}^{2}{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^5+b*x^3+a*x)^2,x)

[Out]

3/2/a^2/(c*x^4+b*x^2+a)*b*c^2/(4*a*c-b^2)*x^2-1/2/a^3/(c*x^4+b*x^2+a)*b^3*c/(4*a*c-b^2)*x^2-1/a/(c*x^4+b*x^2+a
)/(4*a*c-b^2)*c^2+2/a^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b^2*c-1/2/a^3/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b^4+2/a^2/(4*a*c
-b^2)*c^2*ln(c*x^4+b*x^2+a)-7/2/a^3/(4*a*c-b^2)*c*ln(c*x^4+b*x^2+a)*b^2+3/4/a^4/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*
b^4+15/a^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*c^2-10/a^3/(4*a*c-b^2)^(3/2)*arctan((2*c*
x^2+b)/(4*a*c-b^2)^(1/2))*b^3*c+3/2/a^4/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^5-1/4/a^2/x^
4-2/a^3*ln(x)*c+3/a^4*ln(x)*b^2+1/a^3*b/x^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (3 \, b^{3} c - 11 \, a b c^{2}\right )} x^{6} +{\left (6 \, b^{4} - 25 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x^{4} - a^{2} b^{2} + 4 \, a^{3} c + 3 \,{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}}{4 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{8} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{6} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{4}\right )}} - \frac{\frac{1}{4} \,{\left (3 \, b^{4} - 14 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right ) + \frac{{\left (3 \, b^{5} - 20 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c}}}{a^{4} b^{2} - 4 \, a^{5} c} + \frac{{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (x\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

1/4*(2*(3*b^3*c - 11*a*b*c^2)*x^6 + (6*b^4 - 25*a*b^2*c + 8*a^2*c^2)*x^4 - a^2*b^2 + 4*a^3*c + 3*(a*b^3 - 4*a^
2*b*c)*x^2)/((a^3*b^2*c - 4*a^4*c^2)*x^8 + (a^3*b^3 - 4*a^4*b*c)*x^6 + (a^4*b^2 - 4*a^5*c)*x^4) - integrate(((
3*b^4*c - 14*a*b^2*c^2 + 8*a^2*c^3)*x^3 + (3*b^5 - 17*a*b^3*c + 19*a^2*b*c^2)*x)/(c*x^4 + b*x^2 + a), x)/(a^4*
b^2 - 4*a^5*c) + (3*b^2 - 2*a*c)*log(x)/a^4

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Fricas [B]  time = 2.7237, size = 2627, normalized size = 12. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

[-1/4*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2*(3*a*b^5*c - 23*a^2*b^3*c^2 + 44*a^3*b*c^3)*x^6 - (6*a*b^6 - 49*
a^2*b^4*c + 108*a^3*b^2*c^2 - 32*a^4*c^3)*x^4 - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^2 + ((3*b^5*c - 20*
a*b^3*c^2 + 30*a^2*b*c^3)*x^8 + (3*b^6 - 20*a*b^4*c + 30*a^2*b^2*c^2)*x^6 + (3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b
*c^2)*x^4)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^
4 + b*x^2 + a)) + ((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^7 - 26*a*b^5*c + 64*a^2*b
^3*c^2 - 32*a^3*b*c^3)*x^6 + (3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(c*x^4 + b*x^2 + a
) - 4*((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*
a^3*b*c^3)*x^6 + (3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(x))/((a^4*b^4*c - 8*a^5*b^2*c
^2 + 16*a^6*c^3)*x^8 + (a^4*b^5 - 8*a^5*b^3*c + 16*a^6*b*c^2)*x^6 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*x^4),
-1/4*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2*(3*a*b^5*c - 23*a^2*b^3*c^2 + 44*a^3*b*c^3)*x^6 - (6*a*b^6 - 49*
a^2*b^4*c + 108*a^3*b^2*c^2 - 32*a^4*c^3)*x^4 - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^2 - 2*((3*b^5*c - 2
0*a*b^3*c^2 + 30*a^2*b*c^3)*x^8 + (3*b^6 - 20*a*b^4*c + 30*a^2*b^2*c^2)*x^6 + (3*a*b^5 - 20*a^2*b^3*c + 30*a^3
*b*c^2)*x^4)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + ((3*b^6*c - 26*a*b^4
*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^6 + (3*a*b^6
- 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(c*x^4 + b*x^2 + a) - 4*((3*b^6*c - 26*a*b^4*c^2 + 64*a^
2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^6 + (3*a*b^6 - 26*a^2*b^4
*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(x))/((a^4*b^4*c - 8*a^5*b^2*c^2 + 16*a^6*c^3)*x^8 + (a^4*b^5 - 8*a^
5*b^3*c + 16*a^6*b*c^2)*x^6 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*x^4)]

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Sympy [B]  time = 131.412, size = 1074, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(c*x**5+b*x**3+a*x)**2,x)

[Out]

(-b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3 - 48*a**2*b**2*c**2 +
12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(4*a**4))*log(x**2 + (32*a**6*c**2*(-b*sqrt(-(4*a*c - b**2)**3)*(30*a
**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + (2*a*c - 3
*b**2)/(4*a**4)) - 16*a**5*b**2*c*(-b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(
64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(4*a**4)) + 2*a**4*b**4*(-b*sqrt(-(
4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*
c - b**6)) + (2*a*c - 3*b**2)/(4*a**4)) - 16*a**3*c**3 + 47*a**2*b**2*c**2 - 23*a*b**4*c + 3*b**6)/(30*a**2*b*
c**3 - 20*a*b**3*c**2 + 3*b**5*c)) + (b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4
*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(4*a**4))*log(x**2 + (32*a**6*c**
2*(b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3 - 48*a**2*b**2*c**2
+ 12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(4*a**4)) - 16*a**5*b**2*c*(b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2
- 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(
4*a**4)) + 2*a**4*b**4*(b*sqrt(-(4*a*c - b**2)**3)*(30*a**2*c**2 - 20*a*b**2*c + 3*b**4)/(4*a**4*(64*a**3*c**3
- 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + (2*a*c - 3*b**2)/(4*a**4)) - 16*a**3*c**3 + 47*a**2*b**2*c**2 -
23*a*b**4*c + 3*b**6)/(30*a**2*b*c**3 - 20*a*b**3*c**2 + 3*b**5*c)) + (-4*a**3*c + a**2*b**2 + x**6*(22*a*b*c*
*2 - 6*b**3*c) + x**4*(-8*a**2*c**2 + 25*a*b**2*c - 6*b**4) + x**2*(12*a**2*b*c - 3*a*b**3))/(x**8*(16*a**4*c*
*2 - 4*a**3*b**2*c) + x**6*(16*a**4*b*c - 4*a**3*b**3) + x**4*(16*a**5*c - 4*a**4*b**2)) - (2*a*c - 3*b**2)*lo
g(x)/a**4

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Giac [A]  time = 23.7638, size = 370, normalized size = 1.69 \begin{align*} -\frac{{\left (3 \, b^{5} - 20 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{3 \, b^{4} c x^{4} - 14 \, a b^{2} c^{2} x^{4} + 8 \, a^{2} c^{3} x^{4} + 3 \, b^{5} x^{2} - 12 \, a b^{3} c x^{2} + 2 \, a^{2} b c^{2} x^{2} + 5 \, a b^{4} - 22 \, a^{2} b^{2} c + 12 \, a^{3} c^{2}}{4 \,{\left (a^{4} b^{2} - 4 \, a^{5} c\right )}{\left (c x^{4} + b x^{2} + a\right )}} - \frac{{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{4}} + \frac{{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (x^{2}\right )}{2 \, a^{4}} - \frac{9 \, b^{2} x^{4} - 6 \, a c x^{4} - 4 \, a b x^{2} + a^{2}}{4 \, a^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")

[Out]

-1/2*(3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((a^4*b^2 - 4*a^5*c)*sqrt(-b
^2 + 4*a*c)) + 1/4*(3*b^4*c*x^4 - 14*a*b^2*c^2*x^4 + 8*a^2*c^3*x^4 + 3*b^5*x^2 - 12*a*b^3*c*x^2 + 2*a^2*b*c^2*
x^2 + 5*a*b^4 - 22*a^2*b^2*c + 12*a^3*c^2)/((a^4*b^2 - 4*a^5*c)*(c*x^4 + b*x^2 + a)) - 1/4*(3*b^2 - 2*a*c)*log
(c*x^4 + b*x^2 + a)/a^4 + 1/2*(3*b^2 - 2*a*c)*log(x^2)/a^4 - 1/4*(9*b^2*x^4 - 6*a*c*x^4 - 4*a*b*x^2 + a^2)/(a^
4*x^4)