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

Optimal. Leaf size=170 \[ \frac{\left (b+2 c x^2\right ) \left (2 a B c-3 A b c+b^2 B\right )}{4 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\left (2 a B c-3 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

-(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) + ((b^2*B - 3*A*b
*c + 2*a*B*c)*(b + 2*c*x^2))/(4*c*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - ((b^2*B - 3*A*b*c + 2*a*B*c)*ArcTanh[
(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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Rubi [A]  time = 0.163101, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1251, 777, 614, 618, 206} \[ \frac{\left (b+2 c x^2\right ) \left (2 a B c-3 A b c+b^2 B\right )}{4 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac{x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{4 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac{\left (2 a B c-3 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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) + ((b^2*B - 3*A*b
*c + 2*a*B*c)*(b + 2*c*x^2))/(4*c*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - ((b^2*B - 3*A*b*c + 2*a*B*c)*ArcTanh[
(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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 777

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

Rule 614

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

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])

Rubi steps

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

Mathematica [A]  time = 0.237169, size = 172, normalized size = 1.01 \[ \frac{1}{4} \left (\frac{\left (b+2 c x^2\right ) \left (2 a B c-3 A b c+b^2 B\right )}{c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac{-2 a c \left (A+B x^2\right )+a b B+b x^2 (b B-A c)}{c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )^2}+\frac{4 \left (2 a B c-3 A b c+b^2 B\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^2*B - 3*A*b*c + 2*a*B*c)*(b + 2*c*x^2))/(c*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (a*b*B + b*(b*B - A*c)*
x^2 - 2*a*c*(A + B*x^2))/(c*(-b^2 + 4*a*c)*(a + b*x^2 + c*x^4)^2) + (4*(b^2*B - 3*A*b*c + 2*a*B*c)*ArcTan[(b +
2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/4

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Maple [B]  time = 0.013, size = 379, normalized size = 2.2 \begin{align*}{\frac{1}{2\, \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{c \left ( 3\,Abc-2\,aBc-{b}^{2}B \right ){x}^{6}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}-{\frac{3\,b \left ( 3\,Abc-2\,aBc-{b}^{2}B \right ){x}^{4}}{32\,{a}^{2}{c}^{2}-16\,a{b}^{2}c+2\,{b}^{4}}}-{\frac{ \left ( 5\,Aabc+A{b}^{3}+2\,{a}^{2}Bc-5\,Ba{b}^{2} \right ){x}^{2}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}-{\frac{a \left ( 8\,aAc+A{b}^{2}-6\,abB \right ) }{32\,{a}^{2}{c}^{2}-16\,a{b}^{2}c+2\,{b}^{4}}} \right ) }-3\,{\frac{Abc}{ \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 ) }+2\,{\frac{aBc}{ \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}^{2}B}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}\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^3*(B*x^2+A)/(c*x^4+b*x^2+a)^3,x)

[Out]

1/2*(-c*(3*A*b*c-2*B*a*c-B*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^6-3/2*b*(3*A*b*c-2*B*a*c-B*b^2)/(16*a^2*c^2-8*a*b
^2*c+b^4)*x^4-(5*A*a*b*c+A*b^3+2*B*a^2*c-5*B*a*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-1/2*a*(8*A*a*c+A*b^2-6*B*a*
b)/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^4+b*x^2+a)^2-3/(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))*A*b*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))*a*B*c+1/(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^2*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^3*(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 = 1.57635, size = 2595, normalized size = 15.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(B*b^4*c - 4*(2*B*a^2 - 3*A*a*b)*c^3 - (2*B*a*b^2 + 3*A*b^3)*c^2)*x^6 + 6*B*a^2*b^3 - A*a*b^4 + 32*A*a
^3*c^2 + 3*(B*b^5 - 4*(2*B*a^2*b - 3*A*a*b^2)*c^2 - (2*B*a*b^3 + 3*A*b^4)*c)*x^4 + 2*(5*B*a*b^4 - A*b^5 + 4*(2
*B*a^3 + 5*A*a^2*b)*c^2 - (22*B*a^2*b^2 + A*a*b^3)*c)*x^2 - 2*((B*b^2*c^2 + (2*B*a - 3*A*b)*c^3)*x^8 + 2*(B*b^
3*c + (2*B*a*b - 3*A*b^2)*c^2)*x^6 + B*a^2*b^2 + (B*b^4 + 2*(2*B*a^2 - 3*A*a*b)*c^2 + (4*B*a*b^2 - 3*A*b^3)*c)
*x^4 + 2*(B*a*b^3 + (2*B*a^2*b - 3*A*a*b^2)*c)*x^2 + (2*B*a^3 - 3*A*a^2*b)*c)*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)) - 4*(6*B*a^3*b + A*a^2*b^2)
*c)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64
*a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 +
32*a^3*b^2*c^3 - 128*a^4*c^4)*x^4 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x^2), 1/4*(2*(B*
b^4*c - 4*(2*B*a^2 - 3*A*a*b)*c^3 - (2*B*a*b^2 + 3*A*b^3)*c^2)*x^6 + 6*B*a^2*b^3 - A*a*b^4 + 32*A*a^3*c^2 + 3*
(B*b^5 - 4*(2*B*a^2*b - 3*A*a*b^2)*c^2 - (2*B*a*b^3 + 3*A*b^4)*c)*x^4 + 2*(5*B*a*b^4 - A*b^5 + 4*(2*B*a^3 + 5*
A*a^2*b)*c^2 - (22*B*a^2*b^2 + A*a*b^3)*c)*x^2 - 4*((B*b^2*c^2 + (2*B*a - 3*A*b)*c^3)*x^8 + 2*(B*b^3*c + (2*B*
a*b - 3*A*b^2)*c^2)*x^6 + B*a^2*b^2 + (B*b^4 + 2*(2*B*a^2 - 3*A*a*b)*c^2 + (4*B*a*b^2 - 3*A*b^3)*c)*x^4 + 2*(B
*a*b^3 + (2*B*a^2*b - 3*A*a*b^2)*c)*x^2 + (2*B*a^3 - 3*A*a^2*b)*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sq
rt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - 4*(6*B*a^3*b + A*a^2*b^2)*c)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*
a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3
- 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^4 + 2*(a*b^7 - 12*
a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x^2)]

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Sympy [B]  time = 27.4522, size = 789, normalized size = 4.64 \begin{align*} - \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) \log{\left (x^{2} + \frac{- 3 A b^{2} c + 2 B a b c + B b^{3} - 64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) + 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) - 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) + b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right )}{- 6 A b c^{2} + 4 B a c^{2} + 2 B b^{2} c} \right )}}{2} + \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) \log{\left (x^{2} + \frac{- 3 A b^{2} c + 2 B a b c + B b^{3} + 64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) - 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) + 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right ) - b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 3 A b c + 2 B a c + B b^{2}\right )}{- 6 A b c^{2} + 4 B a c^{2} + 2 B b^{2} c} \right )}}{2} + \frac{- 8 A a^{2} c - A a b^{2} + 6 B a^{2} b + x^{6} \left (- 6 A b c^{2} + 4 B a c^{2} + 2 B b^{2} c\right ) + x^{4} \left (- 9 A b^{2} c + 6 B a b c + 3 B b^{3}\right ) + x^{2} \left (- 10 A a b c - 2 A b^{3} - 4 B a^{2} c + 10 B a b^{2}\right )}{64 a^{4} c^{2} - 32 a^{3} b^{2} c + 4 a^{2} b^{4} + x^{8} \left (64 a^{2} c^{4} - 32 a b^{2} c^{3} + 4 b^{4} c^{2}\right ) + x^{6} \left (128 a^{2} b c^{3} - 64 a b^{3} c^{2} + 8 b^{5} c\right ) + x^{4} \left (128 a^{3} c^{3} - 24 a b^{4} c + 4 b^{6}\right ) + x^{2} \left (128 a^{3} b c^{2} - 64 a^{2} b^{3} c + 8 a b^{5}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2)*log(x**2 + (-3*A*b**2*c + 2*B*a*b*c + B*b**3 - 64*a*
*3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2) + 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5
)*(-3*A*b*c + 2*B*a*c + B*b**2) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2) + b**6*
sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2))/(-6*A*b*c**2 + 4*B*a*c**2 + 2*B*b**2*c))/2 + sqrt(-1
/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2)*log(x**2 + (-3*A*b**2*c + 2*B*a*b*c + B*b**3 + 64*a**3*c**3*
sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2) - 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(-3*A*
b*c + 2*B*a*c + B*b**2) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2) - b**6*sqrt(-1/
(4*a*c - b**2)**5)*(-3*A*b*c + 2*B*a*c + B*b**2))/(-6*A*b*c**2 + 4*B*a*c**2 + 2*B*b**2*c))/2 + (-8*A*a**2*c -
A*a*b**2 + 6*B*a**2*b + x**6*(-6*A*b*c**2 + 4*B*a*c**2 + 2*B*b**2*c) + x**4*(-9*A*b**2*c + 6*B*a*b*c + 3*B*b**
3) + x**2*(-10*A*a*b*c - 2*A*b**3 - 4*B*a**2*c + 10*B*a*b**2))/(64*a**4*c**2 - 32*a**3*b**2*c + 4*a**2*b**4 +
x**8*(64*a**2*c**4 - 32*a*b**2*c**3 + 4*b**4*c**2) + x**6*(128*a**2*b*c**3 - 64*a*b**3*c**2 + 8*b**5*c) + x**4
*(128*a**3*c**3 - 24*a*b**4*c + 4*b**6) + x**2*(128*a**3*b*c**2 - 64*a**2*b**3*c + 8*a*b**5))

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Giac [A]  time = 27.6781, size = 308, normalized size = 1.81 \begin{align*} \frac{{\left (B b^{2} + 2 \, B a c - 3 \, A b c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{2 \, B b^{2} c x^{6} + 4 \, B a c^{2} x^{6} - 6 \, A b c^{2} x^{6} + 3 \, B b^{3} x^{4} + 6 \, B a b c x^{4} - 9 \, A b^{2} c x^{4} + 10 \, B a b^{2} x^{2} - 2 \, A b^{3} x^{2} - 4 \, B a^{2} c x^{2} - 10 \, A a b c x^{2} + 6 \, B a^{2} b - A a b^{2} - 8 \, A a^{2} c}{4 \,{\left (c x^{4} + b x^{2} + a\right )}^{2}{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b^2 + 2*B*a*c - 3*A*b*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2
+ 4*a*c)) + 1/4*(2*B*b^2*c*x^6 + 4*B*a*c^2*x^6 - 6*A*b*c^2*x^6 + 3*B*b^3*x^4 + 6*B*a*b*c*x^4 - 9*A*b^2*c*x^4
+ 10*B*a*b^2*x^2 - 2*A*b^3*x^2 - 4*B*a^2*c*x^2 - 10*A*a*b*c*x^2 + 6*B*a^2*b - A*a*b^2 - 8*A*a^2*c)/((c*x^4 + b
*x^2 + a)^2*(b^4 - 8*a*b^2*c + 16*a^2*c^2))