3.125 \(\int \frac{x^9 (A+B x^2)}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=254 \[ -\frac{x^2 \left (x^2 \left (16 a^2 B c^2+6 a A b c^2-15 a b^2 B c+2 b^4 B\right )+2 a \left (6 a A c^2-7 a b B c+b^3 B\right )\right )}{4 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (-12 a^2 A c^3+30 a^2 b B c^2-10 a b^3 B c+b^5 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \left (b^2-4 a c\right )^{5/2}}-\frac{x^6 \left (x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{B \log \left (a+b x^2+c x^4\right )}{4 c^3} \]

[Out]

-(x^6*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x^2))/(4*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (x^2*(2*a
*(b^3*B - 7*a*b*B*c + 6*a*A*c^2) + (2*b^4*B - 15*a*b^2*B*c + 6*a*A*b*c^2 + 16*a^2*B*c^2)*x^2))/(4*c^2*(b^2 - 4
*a*c)^2*(a + b*x^2 + c*x^4)) + ((b^5*B - 10*a*b^3*B*c + 30*a^2*b*B*c^2 - 12*a^2*A*c^3)*ArcTanh[(b + 2*c*x^2)/S
qrt[b^2 - 4*a*c]])/(2*c^3*(b^2 - 4*a*c)^(5/2)) + (B*Log[a + b*x^2 + c*x^4])/(4*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.404194, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1251, 818, 634, 618, 206, 628} \[ -\frac{x^2 \left (x^2 \left (16 a^2 B c^2+6 a A b c^2-15 a b^2 B c+2 b^4 B\right )+2 a \left (6 a A c^2-7 a b B c+b^3 B\right )\right )}{4 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (-12 a^2 A c^3+30 a^2 b B c^2-10 a b^3 B c+b^5 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \left (b^2-4 a c\right )^{5/2}}-\frac{x^6 \left (x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{B \log \left (a+b x^2+c x^4\right )}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(a + b*x^2 + c*x^4)^3,x]

[Out]

-(x^6*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x^2))/(4*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (x^2*(2*a
*(b^3*B - 7*a*b*B*c + 6*a*A*c^2) + (2*b^4*B - 15*a*b^2*B*c + 6*a*A*b*c^2 + 16*a^2*B*c^2)*x^2))/(4*c^2*(b^2 - 4
*a*c)^2*(a + b*x^2 + c*x^4)) + ((b^5*B - 10*a*b^3*B*c + 30*a^2*b*B*c^2 - 12*a^2*A*c^3)*ArcTanh[(b + 2*c*x^2)/S
qrt[b^2 - 4*a*c]])/(2*c^3*(b^2 - 4*a*c)^(5/2)) + (B*Log[a + b*x^2 + c*x^4])/(4*c^3)

Rule 1251

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

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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+B x^2\right )}{\left (a+b x^2+c x^4\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4 (A+B x)}{\left (a+b x+c x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac{x^6 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a (b B-2 A c)+2 B \left (b^2-4 a c\right ) x\right )}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )}{4 c \left (b^2-4 a c\right )}\\ &=-\frac{x^6 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{x^2 \left (2 a \left (b^3 B-7 a b B c+6 a A c^2\right )+\left (2 b^4 B-15 a b^2 B c+6 a A b c^2+16 a^2 B c^2\right ) x^2\right )}{4 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 a \left (b^3 B-7 a b B c+6 a A c^2\right )+2 B \left (b^2-4 a c\right )^2 x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2 \left (b^2-4 a c\right )^2}\\ &=-\frac{x^6 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{x^2 \left (2 a \left (b^3 B-7 a b B c+6 a A c^2\right )+\left (2 b^4 B-15 a b^2 B c+6 a A b c^2+16 a^2 B c^2\right ) x^2\right )}{4 c^2 \left (b^2-4 a c\right )^2 \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 )}{4 c^3}-\frac{\left (b^5 B-10 a b^3 B c+30 a^2 b B c^2-12 a^2 A c^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3 \left (b^2-4 a c\right )^2}\\ &=-\frac{x^6 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{x^2 \left (2 a \left (b^3 B-7 a b B c+6 a A c^2\right )+\left (2 b^4 B-15 a b^2 B c+6 a A b c^2+16 a^2 B c^2\right ) x^2\right )}{4 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{B \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac{\left (b^5 B-10 a b^3 B c+30 a^2 b B c^2-12 a^2 A c^3\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3 \left (b^2-4 a c\right )^2}\\ &=-\frac{x^6 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{x^2 \left (2 a \left (b^3 B-7 a b B c+6 a A c^2\right )+\left (2 b^4 B-15 a b^2 B c+6 a A b c^2+16 a^2 B c^2\right ) x^2\right )}{4 c^2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{\left (b^5 B-10 a b^3 B c+30 a^2 b B c^2-12 a^2 A c^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \left (b^2-4 a c\right )^{5/2}}+\frac{B \log \left (a+b x^2+c x^4\right )}{4 c^3}\\ \end{align*}

Mathematica [A]  time = 0.525902, size = 354, normalized size = 1.39 \[ \frac{\frac{2 a^2 b c^3 \left (11 A+25 B x^2\right )+4 a^2 c^3 \left (8 a B-5 A c x^2\right )-2 a b^3 c^2 \left (4 A+15 B x^2\right )+a b^2 c^2 \left (16 A c x^2-39 a B\right )+b^4 c \left (11 a B-2 A c x^2\right )+b^5 c \left (A+4 B x^2\right )+b^6 (-B)}{\left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{a^2 c \left (b c \left (3 A+5 B x^2\right )-2 A c^2 x^2-4 b^2 B\right )+2 a^3 B c^2+a b^2 \left (-b c \left (A+5 B x^2\right )+4 A c^2 x^2+b^2 B\right )+b^4 x^2 (b B-A c)}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{2 c \left (-12 a^2 A c^3+30 a^2 b B c^2-10 a b^3 B c+b^5 B\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}+B c \log \left (a+b x^2+c x^4\right )}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(a + b*x^2 + c*x^4)^3,x]

[Out]

((-(b^6*B) + b^5*c*(A + 4*B*x^2) - 2*a*b^3*c^2*(4*A + 15*B*x^2) + 2*a^2*b*c^3*(11*A + 25*B*x^2) + 4*a^2*c^3*(8
*a*B - 5*A*c*x^2) + b^4*c*(11*a*B - 2*A*c*x^2) + a*b^2*c^2*(-39*a*B + 16*A*c*x^2))/((b^2 - 4*a*c)^2*(a + b*x^2
+ c*x^4)) + (2*a^3*B*c^2 + b^4*(b*B - A*c)*x^2 + a*b^2*(b^2*B + 4*A*c^2*x^2 - b*c*(A + 5*B*x^2)) + a^2*c*(-4*
b^2*B - 2*A*c^2*x^2 + b*c*(3*A + 5*B*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (2*c*(b^5*B - 10*a*b^3*B*c
+ 30*a^2*b*B*c^2 - 12*a^2*A*c^3)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2) + B*c*Log[a +
b*x^2 + c*x^4])/(4*c^4)

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Maple [B]  time = 0.02, size = 723, normalized size = 2.9 \begin{align*}{\frac{1}{2\, \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{ \left ( 10\,A{a}^{2}{c}^{3}-8\,Aa{b}^{2}{c}^{2}+A{b}^{4}c-25\,B{a}^{2}b{c}^{2}+15\,Ba{b}^{3}c-2\,B{b}^{5} \right ){x}^{6}}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{ \left ( 2\,A{a}^{2}b{c}^{3}+8\,Aa{b}^{3}{c}^{2}-A{b}^{5}c+32\,B{a}^{3}{c}^{3}+11\,B{a}^{2}{b}^{2}{c}^{2}-19\,Ba{b}^{4}c+3\,B{b}^{6} \right ){x}^{4}}{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-{\frac{a \left ( 6\,A{a}^{2}{c}^{3}-10\,Aa{b}^{2}{c}^{2}+A{b}^{4}c-31\,B{a}^{2}b{c}^{2}+22\,Ba{b}^{3}c-3\,B{b}^{5} \right ){x}^{2}}{{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{{a}^{2} \left ( 10\,aAb{c}^{2}-A{b}^{3}c+24\,{a}^{2}B{c}^{2}-21\,a{b}^{2}Bc+3\,{b}^{4}B \right ) }{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }} \right ) }+4\,{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){a}^{2}B}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-2\,{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) a{b}^{2}B}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{4}B}{4\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+6\,{\frac{A{a}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-15\,{\frac{B{a}^{2}b}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+5\,{\frac{Ba{b}^{3}}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{B{b}^{5}}{2\,{c}^{3} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(B*x^2+A)/(c*x^4+b*x^2+a)^3,x)

[Out]

1/2*(-1/c^2*(10*A*a^2*c^3-8*A*a*b^2*c^2+A*b^4*c-25*B*a^2*b*c^2+15*B*a*b^3*c-2*B*b^5)/(16*a^2*c^2-8*a*b^2*c+b^4
)*x^6+1/2*(2*A*a^2*b*c^3+8*A*a*b^3*c^2-A*b^5*c+32*B*a^3*c^3+11*B*a^2*b^2*c^2-19*B*a*b^4*c+3*B*b^6)/c^3/(16*a^2
*c^2-8*a*b^2*c+b^4)*x^4-a*(6*A*a^2*c^3-10*A*a*b^2*c^2+A*b^4*c-31*B*a^2*b*c^2+22*B*a*b^3*c-3*B*b^5)/c^3/(16*a^2
*c^2-8*a*b^2*c+b^4)*x^2+1/2*a^2*(10*A*a*b*c^2-A*b^3*c+24*B*a^2*c^2-21*B*a*b^2*c+3*B*b^4)/c^3/(16*a^2*c^2-8*a*b
^2*c+b^4))/(c*x^4+b*x^2+a)^2+4/c/(16*a^2*c^2-8*a*b^2*c+b^4)*ln(c*x^4+b*x^2+a)*a^2*B-2/c^2/(16*a^2*c^2-8*a*b^2*
c+b^4)*ln(c*x^4+b*x^2+a)*a*b^2*B+1/4/c^3/(16*a^2*c^2-8*a*b^2*c+b^4)*ln(c*x^4+b*x^2+a)*b^4*B+6/(16*a^2*c^2-8*a*
b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*A*a^2-15/c/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*
c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a^2*b*B+5/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*
arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*B*a*b^3-1/2/c^3/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c
*x^2+b)/(4*a*c-b^2)^(1/2))*b^5*B

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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(x^9*(B*x^2+A)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.04094, size = 4617, normalized size = 18.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/4*(3*B*a^2*b^6 + 2*(2*B*b^7*c + 40*A*a^3*c^5 - 2*(50*B*a^3*b + 21*A*a^2*b^2)*c^4 + (85*B*a^2*b^3 + 12*A*a*b
^4)*c^3 - (23*B*a*b^5 + A*b^6)*c^2)*x^6 + (3*B*b^8 - 8*(16*B*a^4 + A*a^3*b)*c^4 - 6*(2*B*a^3*b^2 + 5*A*a^2*b^3
)*c^3 + 3*(29*B*a^2*b^4 + 4*A*a*b^5)*c^2 - (31*B*a*b^6 + A*b^7)*c)*x^4 - 8*(12*B*a^5 + 5*A*a^4*b)*c^3 + 2*(54*
B*a^4*b^2 + 7*A*a^3*b^3)*c^2 + 2*(3*B*a*b^7 + 24*A*a^4*c^4 - 2*(62*B*a^4*b + 23*A*a^3*b^2)*c^3 + 7*(17*B*a^3*b
^3 + 2*A*a^2*b^4)*c^2 - (34*B*a^2*b^5 + A*a*b^6)*c)*x^2 - ((B*b^5*c^2 - 10*B*a*b^3*c^3 + 30*B*a^2*b*c^4 - 12*A
*a^2*c^5)*x^8 + B*a^2*b^5 - 10*B*a^3*b^3*c + 30*B*a^4*b*c^2 - 12*A*a^4*c^3 + 2*(B*b^6*c - 10*B*a*b^4*c^2 + 30*
B*a^2*b^2*c^3 - 12*A*a^2*b*c^4)*x^6 + (B*b^7 - 8*B*a*b^5*c + 10*B*a^2*b^3*c^2 - 24*A*a^3*c^4 + 12*(5*B*a^3*b -
A*a^2*b^2)*c^3)*x^4 + 2*(B*a*b^6 - 10*B*a^2*b^4*c + 30*B*a^3*b^2*c^2 - 12*A*a^3*b*c^3)*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)) - (33*B*a^3*
b^4 + A*a^2*b^5)*c + (B*a^2*b^6 - 12*B*a^3*b^4*c + 48*B*a^4*b^2*c^2 - 64*B*a^5*c^3 + (B*b^6*c^2 - 12*B*a*b^4*c
^3 + 48*B*a^2*b^2*c^4 - 64*B*a^3*c^5)*x^8 + 2*(B*b^7*c - 12*B*a*b^5*c^2 + 48*B*a^2*b^3*c^3 - 64*B*a^3*b*c^4)*x
^6 + (B*b^8 - 10*B*a*b^6*c + 24*B*a^2*b^4*c^2 + 32*B*a^3*b^2*c^3 - 128*B*a^4*c^4)*x^4 + 2*(B*a*b^7 - 12*B*a^2*
b^5*c + 48*B*a^3*b^3*c^2 - 64*B*a^4*b*c^3)*x^2)*log(c*x^4 + b*x^2 + a))/(a^2*b^6*c^3 - 12*a^3*b^4*c^4 + 48*a^4
*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*x^8 + 2*(b^7*c^4 - 12*a*b^5*c^5
+ 48*a^2*b^3*c^6 - 64*a^3*b*c^7)*x^6 + (b^8*c^3 - 10*a*b^6*c^4 + 24*a^2*b^4*c^5 + 32*a^3*b^2*c^6 - 128*a^4*c^
7)*x^4 + 2*(a*b^7*c^3 - 12*a^2*b^5*c^4 + 48*a^3*b^3*c^5 - 64*a^4*b*c^6)*x^2), 1/4*(3*B*a^2*b^6 + 2*(2*B*b^7*c
+ 40*A*a^3*c^5 - 2*(50*B*a^3*b + 21*A*a^2*b^2)*c^4 + (85*B*a^2*b^3 + 12*A*a*b^4)*c^3 - (23*B*a*b^5 + A*b^6)*c^
2)*x^6 + (3*B*b^8 - 8*(16*B*a^4 + A*a^3*b)*c^4 - 6*(2*B*a^3*b^2 + 5*A*a^2*b^3)*c^3 + 3*(29*B*a^2*b^4 + 4*A*a*b
^5)*c^2 - (31*B*a*b^6 + A*b^7)*c)*x^4 - 8*(12*B*a^5 + 5*A*a^4*b)*c^3 + 2*(54*B*a^4*b^2 + 7*A*a^3*b^3)*c^2 + 2*
(3*B*a*b^7 + 24*A*a^4*c^4 - 2*(62*B*a^4*b + 23*A*a^3*b^2)*c^3 + 7*(17*B*a^3*b^3 + 2*A*a^2*b^4)*c^2 - (34*B*a^2
*b^5 + A*a*b^6)*c)*x^2 + 2*((B*b^5*c^2 - 10*B*a*b^3*c^3 + 30*B*a^2*b*c^4 - 12*A*a^2*c^5)*x^8 + B*a^2*b^5 - 10*
B*a^3*b^3*c + 30*B*a^4*b*c^2 - 12*A*a^4*c^3 + 2*(B*b^6*c - 10*B*a*b^4*c^2 + 30*B*a^2*b^2*c^3 - 12*A*a^2*b*c^4)
*x^6 + (B*b^7 - 8*B*a*b^5*c + 10*B*a^2*b^3*c^2 - 24*A*a^3*c^4 + 12*(5*B*a^3*b - A*a^2*b^2)*c^3)*x^4 + 2*(B*a*b
^6 - 10*B*a^2*b^4*c + 30*B*a^3*b^2*c^2 - 12*A*a^3*b*c^3)*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)) - (33*B*a^3*b^4 + A*a^2*b^5)*c + (B*a^2*b^6 - 12*B*a^3*b^4*c + 48*B*a^4*b^2*c^2 - 6
4*B*a^5*c^3 + (B*b^6*c^2 - 12*B*a*b^4*c^3 + 48*B*a^2*b^2*c^4 - 64*B*a^3*c^5)*x^8 + 2*(B*b^7*c - 12*B*a*b^5*c^2
+ 48*B*a^2*b^3*c^3 - 64*B*a^3*b*c^4)*x^6 + (B*b^8 - 10*B*a*b^6*c + 24*B*a^2*b^4*c^2 + 32*B*a^3*b^2*c^3 - 128*
B*a^4*c^4)*x^4 + 2*(B*a*b^7 - 12*B*a^2*b^5*c + 48*B*a^3*b^3*c^2 - 64*B*a^4*b*c^3)*x^2)*log(c*x^4 + b*x^2 + a))
/(a^2*b^6*c^3 - 12*a^3*b^4*c^4 + 48*a^4*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a
^3*c^8)*x^8 + 2*(b^7*c^4 - 12*a*b^5*c^5 + 48*a^2*b^3*c^6 - 64*a^3*b*c^7)*x^6 + (b^8*c^3 - 10*a*b^6*c^4 + 24*a^
2*b^4*c^5 + 32*a^3*b^2*c^6 - 128*a^4*c^7)*x^4 + 2*(a*b^7*c^3 - 12*a^2*b^5*c^4 + 48*a^3*b^3*c^5 - 64*a^4*b*c^6)
*x^2)]

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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(x**9*(B*x**2+A)/(c*x**4+b*x**2+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 39.023, size = 629, normalized size = 2.48 \begin{align*} -\frac{{\left (B b^{5} - 10 \, B a b^{3} c + 30 \, B a^{2} b c^{2} - 12 \, A a^{2} c^{3}\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{B \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac{3 \, B b^{4} c^{2} x^{8} - 24 \, B a b^{2} c^{3} x^{8} + 48 \, B a^{2} c^{4} x^{8} - 2 \, B b^{5} c x^{6} + 12 \, B a b^{3} c^{2} x^{6} + 4 \, A b^{4} c^{2} x^{6} - 4 \, B a^{2} b c^{3} x^{6} - 32 \, A a b^{2} c^{3} x^{6} + 40 \, A a^{2} c^{4} x^{6} - 3 \, B b^{6} x^{4} + 20 \, B a b^{4} c x^{4} + 2 \, A b^{5} c x^{4} - 22 \, B a^{2} b^{2} c^{2} x^{4} - 16 \, A a b^{3} c^{2} x^{4} + 32 \, B a^{3} c^{3} x^{4} - 4 \, A a^{2} b c^{3} x^{4} - 6 \, B a b^{5} x^{2} + 40 \, B a^{2} b^{3} c x^{2} + 4 \, A a b^{4} c x^{2} - 28 \, B a^{3} b c^{2} x^{2} - 40 \, A a^{2} b^{2} c^{2} x^{2} + 24 \, A a^{3} c^{3} x^{2} - 3 \, B a^{2} b^{4} + 18 \, B a^{3} b^{2} c + 2 \, A a^{2} b^{3} c - 20 \, A a^{3} b c^{2}}{8 \,{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )}{\left (c x^{4} + b x^{2} + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

-1/2*(B*b^5 - 10*B*a*b^3*c + 30*B*a^2*b*c^2 - 12*A*a^2*c^3)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^4*c^3
- 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(-b^2 + 4*a*c)) + 1/4*B*log(c*x^4 + b*x^2 + a)/c^3 - 1/8*(3*B*b^4*c^2*x^8 - 2
4*B*a*b^2*c^3*x^8 + 48*B*a^2*c^4*x^8 - 2*B*b^5*c*x^6 + 12*B*a*b^3*c^2*x^6 + 4*A*b^4*c^2*x^6 - 4*B*a^2*b*c^3*x^
6 - 32*A*a*b^2*c^3*x^6 + 40*A*a^2*c^4*x^6 - 3*B*b^6*x^4 + 20*B*a*b^4*c*x^4 + 2*A*b^5*c*x^4 - 22*B*a^2*b^2*c^2*
x^4 - 16*A*a*b^3*c^2*x^4 + 32*B*a^3*c^3*x^4 - 4*A*a^2*b*c^3*x^4 - 6*B*a*b^5*x^2 + 40*B*a^2*b^3*c*x^2 + 4*A*a*b
^4*c*x^2 - 28*B*a^3*b*c^2*x^2 - 40*A*a^2*b^2*c^2*x^2 + 24*A*a^3*c^3*x^2 - 3*B*a^2*b^4 + 18*B*a^3*b^2*c + 2*A*a
^2*b^3*c - 20*A*a^3*b*c^2)/((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*(c*x^4 + b*x^2 + a)^2)