3.91 \(\int \frac{x^9}{(a x+b x^3+c x^5)^2} \, dx\)

Optimal. Leaf size=132 \[ \frac{b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

[Out]

-(b*x^2)/(2*c*(b^2 - 4*a*c)) + (x^4*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (b*(b^2 - 6*a*c)*Ar
cTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^2*(b^2 - 4*a*c)^(3/2)) + Log[a + b*x^2 + c*x^4]/(4*c^2)

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Rubi [A]  time = 0.151706, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1585, 1114, 738, 773, 634, 618, 206, 628} \[ \frac{b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

-(b*x^2)/(2*c*(b^2 - 4*a*c)) + (x^4*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (b*(b^2 - 6*a*c)*Ar
cTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^2*(b^2 - 4*a*c)^(3/2)) + Log[a + b*x^2 + c*x^4]/(4*c^2)

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 738

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

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

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{x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx &=\int \frac{x^7}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{x (4 a+b x)}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a b+\left (-b^2+4 a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c \left (b^2-4 a c\right )}\\ &=-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}-\frac{\left (b \left (b^2-6 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2 \left (b^2-4 a c\right )}\\ &=-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{\left (b \left (b^2-6 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^2 \left (b^2-4 a c\right )}\\ &=-\frac{b x^2}{2 c \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac{\log \left (a+b x^2+c x^4\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.185909, size = 121, normalized size = 0.92 \[ \frac{\frac{2 \left (-2 a^2 c+a b \left (b-3 c x^2\right )+b^3 x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{2 b \left (b^2-6 a c\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

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

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Maple [A]  time = 0.011, size = 222, normalized size = 1.7 \begin{align*}{\frac{1}{2\,c{x}^{4}+2\,b{x}^{2}+2\,a} \left ({\frac{b \left ( 3\,ac-{b}^{2} \right ){x}^{2}}{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{a \left ( 2\,ac-{b}^{2} \right ) }{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }} \right ) }+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) a}{c \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}}{4\,{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-3\,{\frac{ab}{c \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}}{2\,{c}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(c*x^5+b*x^3+a*x)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a b^{2} - 2 \, a^{2} c +{\left (b^{3} - 3 \, a b c\right )} x^{2}}{2 \,{\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} +{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{4} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2}\right )}} - \frac{-\int \frac{{\left (b^{2} - 4 \, a c\right )} x^{3} + a b x}{c x^{4} + b x^{2} + a}\,{d x}}{b^{2} c - 4 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

1/2*(a*b^2 - 2*a^2*c + (b^3 - 3*a*b*c)*x^2)/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*
b*c^3)*x^2) - integrate(-((b^2 - 4*a*c)*x^3 + a*b*x)/(c*x^4 + b*x^2 + a), x)/(b^2*c - 4*a*c^2)

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Fricas [B]  time = 1.38481, size = 1412, normalized size = 10.7 \begin{align*} \left [\frac{2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \,{\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} x^{2} +{\left ({\left (b^{3} c - 6 \, a b c^{2}\right )} x^{4} + a b^{3} - 6 \, a^{2} b c +{\left (b^{4} - 6 \, a b^{2} c\right )} x^{2}\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c +{\left (2 \, c x^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) +{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} +{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} +{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}, \frac{2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \,{\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} x^{2} + 2 \,{\left ({\left (b^{3} c - 6 \, a b c^{2}\right )} x^{4} + a b^{3} - 6 \, a^{2} b c +{\left (b^{4} - 6 \, a b^{2} c\right )} x^{2}\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{{\left (2 \, c x^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) +{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} +{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} +{\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

[1/4*(2*a*b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + 2*(b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*x^2 + ((b^3*c - 6*a*b*c^2)*x^4
+ a*b^3 - 6*a^2*b*c + (b^4 - 6*a*b^2*c)*x^2)*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)) + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 +
16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 +
16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2), 1/4*(2*a*
b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + 2*(b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*x^2 + 2*((b^3*c - 6*a*b*c^2)*x^4 + a*b^3
- 6*a^2*b*c + (b^4 - 6*a*b^2*c)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)
) + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c
^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5
)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2)]

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Sympy [B]  time = 3.50461, size = 745, normalized size = 5.64 \begin{align*} \left (- \frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) \log{\left (x^{2} + \frac{- 32 a^{2} c^{3} \left (- \frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) + 8 a^{2} c + 16 a b^{2} c^{2} \left (- \frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) - a b^{2} - 2 b^{4} c \left (- \frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right )}{6 a b c - b^{3}} \right )} + \left (\frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) \log{\left (x^{2} + \frac{- 32 a^{2} c^{3} \left (\frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) + 8 a^{2} c + 16 a b^{2} c^{2} \left (\frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right ) - a b^{2} - 2 b^{4} c \left (\frac{b \sqrt{- \left (4 a c - b^{2}\right )^{3}} \left (6 a c - b^{2}\right )}{4 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )} + \frac{1}{4 c^{2}}\right )}{6 a b c - b^{3}} \right )} + \frac{2 a^{2} c - a b^{2} + x^{2} \left (3 a b c - b^{3}\right )}{8 a^{2} c^{3} - 2 a b^{2} c^{2} + x^{4} \left (8 a c^{4} - 2 b^{2} c^{3}\right ) + x^{2} \left (8 a b c^{3} - 2 b^{3} c^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**5+b*x**3+a*x)**2,x)

[Out]

(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) +
1/(4*c**2))*log(x**2 + (-32*a**2*c**3*(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*
a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + 1/(4*c**2)) + 8*a**2*c + 16*a*b**2*c**2*(-b*sqrt(-(4*a*c - b**2)**3)*(
6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + 1/(4*c**2)) - a*b**2 - 2*b**4
*c*(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)
) + 1/(4*c**2)))/(6*a*b*c - b**3)) + (b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**
2*b**2*c**2 + 12*a*b**4*c - b**6)) + 1/(4*c**2))*log(x**2 + (-32*a**2*c**3*(b*sqrt(-(4*a*c - b**2)**3)*(6*a*c
- b**2)/(4*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) + 1/(4*c**2)) + 8*a**2*c + 16*a*b**2*
c**2*(b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6
)) + 1/(4*c**2)) - a*b**2 - 2*b**4*c*(b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*c**2*(64*a**3*c**3 - 48*a**
2*b**2*c**2 + 12*a*b**4*c - b**6)) + 1/(4*c**2)))/(6*a*b*c - b**3)) + (2*a**2*c - a*b**2 + x**2*(3*a*b*c - b**
3))/(8*a**2*c**3 - 2*a*b**2*c**2 + x**4*(8*a*c**4 - 2*b**2*c**3) + x**2*(8*a*b*c**3 - 2*b**3*c**2))

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Giac [A]  time = 25.6046, size = 205, normalized size = 1.55 \begin{align*} -\frac{{\left (b^{3} - 6 \, a b c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{b^{2} c x^{4} - 4 \, a c^{2} x^{4} - b^{3} x^{2} + 2 \, a b c x^{2} - a b^{2}}{4 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} + \frac{\log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")

[Out]

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