3.89 \(\int \frac{x^{11}}{(a x+b x^3+c x^5)^2} \, dx\)

Optimal. Leaf size=166 \[ -\frac{\left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{3/2}}+\frac{x^2 \left (b^2-3 a c\right )}{c^2 \left (b^2-4 a c\right )}+\frac{x^6 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}-\frac{b \log \left (a+b x^2+c x^4\right )}{2 c^3} \]

[Out]

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

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Rubi [A]  time = 0.222667, antiderivative size = 166, 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, 738, 800, 634, 618, 206, 628} \[ -\frac{\left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{3/2}}+\frac{x^2 \left (b^2-3 a c\right )}{c^2 \left (b^2-4 a c\right )}+\frac{x^6 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}-\frac{b \log \left (a+b x^2+c x^4\right )}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

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

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 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{x^{11}}{\left (a x+b x^3+c x^5\right )^2} \, dx &=\int \frac{x^9}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{x^6 \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^2 (6 a+2 b x)}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=\frac{x^6 \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 \left (-\frac{2 \left (b^2-3 a c\right )}{c^2}+\frac{2 b x}{c}+\frac{2 \left (a \left (b^2-3 a c\right )+b \left (b^2-4 a c\right ) x\right )}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=\frac{\left (b^2-3 a c\right ) x^2}{c^2 \left (b^2-4 a c\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}+\frac{x^6 \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 \left (b^2-3 a c\right )+b \left (b^2-4 a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{c^2 \left (b^2-4 a c\right )}\\ &=\frac{\left (b^2-3 a c\right ) x^2}{c^2 \left (b^2-4 a c\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}+\frac{x^6 \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 \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^3}+\frac{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^3 \left (b^2-4 a c\right )}\\ &=\frac{\left (b^2-3 a c\right ) x^2}{c^2 \left (b^2-4 a c\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}+\frac{x^6 \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 \log \left (a+b x^2+c x^4\right )}{2 c^3}-\frac{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{c^3 \left (b^2-4 a c\right )}\\ &=\frac{\left (b^2-3 a c\right ) x^2}{c^2 \left (b^2-4 a c\right )}-\frac{b x^4}{2 c \left (b^2-4 a c\right )}+\frac{x^6 \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{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{3/2}}-\frac{b \log \left (a+b x^2+c x^4\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.20864, size = 151, normalized size = 0.91 \[ \frac{-\frac{2 \left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\frac{a^2 c \left (3 b-2 c x^2\right )-a b^2 \left (b-4 c x^2\right )+b^4 \left (-x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-b \log \left (a+b x^2+c x^4\right )+c x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

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

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Maple [B]  time = 0.013, size = 383, normalized size = 2.3 \begin{align*}{\frac{{x}^{2}}{2\,{c}^{2}}}+{\frac{{a}^{2}{x}^{2}}{c \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-2\,{\frac{a{x}^{2}{b}^{2}}{{c}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{{x}^{2}{b}^{4}}{2\,{c}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{3\,{a}^{2}b}{2\,{c}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{a{b}^{3}}{2\,{c}^{3} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-2\,{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) ab}{ \left ( 4\,ac-{b}^{2} \right ){c}^{2}}}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{3}}{2\,{c}^{3} \left ( 4\,ac-{b}^{2} \right ) }}-6\,{\frac{{a}^{2}}{c \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+6\,{\frac{a{b}^{2}}{{c}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{4}}{{c}^{3}}\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^11/(c*x^5+b*x^3+a*x)^2,x)

[Out]

1/2/c^2*x^2+1/c/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a^2-2/c^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a*b^2+1/2/c^3/(c*x^4
+b*x^2+a)/(4*a*c-b^2)*x^2*b^4-3/2/c^2/(c*x^4+b*x^2+a)*a^2*b/(4*a*c-b^2)+1/2/c^3/(c*x^4+b*x^2+a)*a*b^3/(4*a*c-b
^2)-2/c^2/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*a*b+1/2/c^3/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*b^3-6/c/(4*a*c-b^2)^(3/2)*ar
ctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a^2+6/c^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*b^2-1/
c^3/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a b^{3} - 3 \, a^{2} b c +{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} x^{2}}{2 \,{\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} +{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} +{\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{2}\right )}} + \frac{x^{2}}{2 \, c^{2}} + \frac{-2 \, \int \frac{{\left (b^{3} - 4 \, a b c\right )} x^{3} +{\left (a b^{2} - 3 \, a^{2} c\right )} x}{c x^{4} + b x^{2} + a}\,{d x}}{b^{2} c^{2} - 4 \, a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

-1/2*(a*b^3 - 3*a^2*b*c + (b^4 - 4*a*b^2*c + 2*a^2*c^2)*x^2)/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^4
+ (b^3*c^3 - 4*a*b*c^4)*x^2) + 1/2*x^2/c^2 + 2*integrate(-((b^3 - 4*a*b*c)*x^3 + (a*b^2 - 3*a^2*c)*x)/(c*x^4 +
b*x^2 + a), x)/(b^2*c^2 - 4*a*c^3)

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Fricas [B]  time = 1.67139, size = 1806, normalized size = 10.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

[1/2*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^6 - a*b^5 + 7*a^2*b^3*c - 12*a^3*b*c^2 + (b^5*c - 8*a*b^3*c^2 + 1
6*a^2*b*c^3)*x^4 - (b^6 - 9*a*b^4*c + 26*a^2*b^2*c^2 - 24*a^3*c^3)*x^2 - (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2 + (b
^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*x^4 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*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^5 - 8*a^2*b^3*c + 16*a^
3*b*c^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^4 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2)*log(c*x^4 + b*x^2
+ a))/(a*b^4*c^3 - 8*a^2*b^2*c^4 + 16*a^3*c^5 + (b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^4 + (b^5*c^3 - 8*a*b^3
*c^4 + 16*a^2*b*c^5)*x^2), 1/2*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^6 - a*b^5 + 7*a^2*b^3*c - 12*a^3*b*c^2
+ (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^4 - (b^6 - 9*a*b^4*c + 26*a^2*b^2*c^2 - 24*a^3*c^3)*x^2 - 2*(a*b^4 -
6*a^2*b^2*c + 6*a^3*c^2 + (b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*x^4 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*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^5 - 8*a^2*b^3*c + 16*a^3*b*c^2 + (b
^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^4 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^
4*c^3 - 8*a^2*b^2*c^4 + 16*a^3*c^5 + (b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^4 + (b^5*c^3 - 8*a*b^3*c^4 + 16*a^
2*b*c^5)*x^2)]

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Sympy [B]  time = 4.75116, size = 877, normalized size = 5.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(c*x**5+b*x**3+a*x)**2,x)

[Out]

(-b/(2*c**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3 - 48*a**2*b**2
*c**2 + 12*a*b**4*c - b**6)))*log(x**2 + (-5*a**2*b*c - 16*a**2*c**4*(-b/(2*c**3) - sqrt(-(4*a*c - b**2)**3)*(
6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + a*b**3 +
8*a*b**2*c**3*(-b/(2*c**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3
- 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) - b**4*c**2*(-b/(2*c**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 -
6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))))/(6*a**2*c**2 - 6*a*b**2
*c + b**4)) + (-b/(2*c**3) + sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3
- 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)))*log(x**2 + (-5*a**2*b*c - 16*a**2*c**4*(-b/(2*c**3) + sqrt(-(4*a*c
- b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)
)) + a*b**3 + 8*a*b**2*c**3*(-b/(2*c**3) + sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*
(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) - b**4*c**2*(-b/(2*c**3) + sqrt(-(4*a*c - b**2)**3)*
(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*c**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))))/(6*a**2*c
**2 - 6*a*b**2*c + b**4)) + (-3*a**2*b*c + a*b**3 + x**2*(2*a**2*c**2 - 4*a*b**2*c + b**4))/(8*a**2*c**4 - 2*a
*b**2*c**3 + x**4*(8*a*c**5 - 2*b**2*c**4) + x**2*(8*a*b*c**4 - 2*b**3*c**3)) + x**2/(2*c**2)

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Giac [A]  time = 21.8737, size = 217, normalized size = 1.31 \begin{align*} \frac{{\left (b^{4} - 6 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{x^{2}}{2 \, c^{2}} + \frac{b^{3} x^{4} - 4 \, a b c x^{4} - 2 \, a^{2} c x^{2} - a^{2} b}{2 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} - \frac{b \log \left (c x^{4} + b x^{2} + a\right )}{2 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")

[Out]

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