### 3.146 $$\int \frac{A+C x^2}{(a+b x+c x^2)^3} \, dx$$

Optimal. Leaf size=161 $-\frac{x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{2 \left (C \left (2 a c+b^2\right )+6 A c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac{(b+2 c x) \left (2 a C+6 A c+\frac{b^2 C}{c}\right )}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}$

[Out]

-(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x)/(2*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + ((6*A*c + 2*a*C
+ (b^2*C)/c)*(b + 2*c*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) - (2*(6*A*c^2 + (b^2 + 2*a*c)*C)*ArcTanh[(b +
2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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Rubi [A]  time = 0.113923, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {1660, 12, 614, 618, 206} $-\frac{x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{2 \left (C \left (2 a c+b^2\right )+6 A c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac{(b+2 c x) \left (2 a C+6 A c+\frac{b^2 C}{c}\right )}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x)/(2*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + ((6*A*c + 2*a*C
+ (b^2*C)/c)*(b + 2*c*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) - (2*(6*A*c^2 + (b^2 + 2*a*c)*C)*ArcTanh[(b +
2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*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[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.220095, size = 160, normalized size = 0.99 $\frac{1}{2} \left (\frac{(b+2 c x) \left (C \left (2 a c+b^2\right )+6 A c^2\right )}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac{4 \left (C \left (2 a c+b^2\right )+6 A c^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}+\frac{a C (b-2 c x)+A c (b+2 c x)+b^2 C x}{c \left (4 a c-b^2\right ) (a+x (b+c x))^2}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(((6*A*c^2 + (b^2 + 2*a*c)*C)*(b + 2*c*x))/(c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))) + (b^2*C*x + a*C*(b - 2*c*x)
+ A*c*(b + 2*c*x))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))^2) + (4*(6*A*c^2 + (b^2 + 2*a*c)*C)*ArcTan[(b + 2*c*x)/
Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/2

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Maple [B]  time = 0.183, size = 373, normalized size = 2.3 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ({\frac{c \left ( 6\,A{c}^{2}+2\,Cac+C{b}^{2} \right ){x}^{3}}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}}+{\frac{3\,b \left ( 6\,A{c}^{2}+2\,Cac+C{b}^{2} \right ){x}^{2}}{32\,{a}^{2}{c}^{2}-16\,ac{b}^{2}+2\,{b}^{4}}}+{\frac{ \left ( 10\,aA{c}^{2}+2\,A{b}^{2}c-2\,C{a}^{2}c+5\,Ca{b}^{2} \right ) x}{16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4}}}+{\frac{b \left ( 10\,aAc-A{b}^{2}+6\,{a}^{2}C \right ) }{32\,{a}^{2}{c}^{2}-16\,ac{b}^{2}+2\,{b}^{4}}} \right ) }+12\,{\frac{A{c}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+4\,{\frac{Cac}{ \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{C{b}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,ac{b}^{2}+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+A)/(c*x^2+b*x+a)^3,x)

[Out]

(c*(6*A*c^2+2*C*a*c+C*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+3/2*b*(6*A*c^2+2*C*a*c+C*b^2)/(16*a^2*c^2-8*a*b^2*c+
b^4)*x^2+(10*A*a*c^2+2*A*b^2*c-2*C*a^2*c+5*C*a*b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x+1/2*b*(10*A*a*c-A*b^2+6*C*a^2
)/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2+12/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)
/(4*a*c-b^2)^(1/2))*A*c^2+4/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*C
*a*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+b)/(4*a*c-b^2)^(1/2))*C*b^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.73505, size = 2592, normalized size = 16.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(6*C*a^2*b^3 - A*b^5 - 40*A*a^2*b*c^2 + 2*(C*b^4*c - 2*C*a*b^2*c^2 - 24*A*a*c^4 - 2*(4*C*a^2 - 3*A*b^2)*c
^3)*x^3 + 3*(C*b^5 - 2*C*a*b^3*c - 24*A*a*b*c^3 - 2*(4*C*a^2*b - 3*A*b^3)*c^2)*x^2 + 2*(C*a^2*b^2 + 2*C*a^3*c
+ 6*A*a^2*c^2 + (C*b^2*c^2 + 2*C*a*c^3 + 6*A*c^4)*x^4 + 2*(C*b^3*c + 2*C*a*b*c^2 + 6*A*b*c^3)*x^3 + (C*b^4 + 4
*C*a*b^2*c + 12*A*a*c^3 + 2*(2*C*a^2 + 3*A*b^2)*c^2)*x^2 + 2*(C*a*b^3 + 2*C*a^2*b*c + 6*A*a*b*c^2)*x)*sqrt(b^2
- 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - 2*(12*C
*a^3*b - 7*A*a*b^3)*c + 2*(5*C*a*b^4 - 40*A*a^2*c^3 + 2*(4*C*a^3 + A*a*b^2)*c^2 - 2*(11*C*a^2*b^2 - A*b^4)*c)*
x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c
^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (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^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x), 1/2*(6*C*a^2*
b^3 - A*b^5 - 40*A*a^2*b*c^2 + 2*(C*b^4*c - 2*C*a*b^2*c^2 - 24*A*a*c^4 - 2*(4*C*a^2 - 3*A*b^2)*c^3)*x^3 + 3*(C
*b^5 - 2*C*a*b^3*c - 24*A*a*b*c^3 - 2*(4*C*a^2*b - 3*A*b^3)*c^2)*x^2 - 4*(C*a^2*b^2 + 2*C*a^3*c + 6*A*a^2*c^2
+ (C*b^2*c^2 + 2*C*a*c^3 + 6*A*c^4)*x^4 + 2*(C*b^3*c + 2*C*a*b*c^2 + 6*A*b*c^3)*x^3 + (C*b^4 + 4*C*a*b^2*c + 1
2*A*a*c^3 + 2*(2*C*a^2 + 3*A*b^2)*c^2)*x^2 + 2*(C*a*b^3 + 2*C*a^2*b*c + 6*A*a*b*c^2)*x)*sqrt(-b^2 + 4*a*c)*arc
tan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 2*(12*C*a^3*b - 7*A*a*b^3)*c + 2*(5*C*a*b^4 - 40*A*a^2*c^
3 + 2*(4*C*a^3 + A*a*b^2)*c^2 - 2*(11*C*a^2*b^2 - A*b^4)*c)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a
^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3
- 64*a^3*b*c^4)*x^3 + (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^2 + 2*(a*b^7 - 12*
a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]

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Sympy [B]  time = 2.6655, size = 774, normalized size = 4.81 \begin{align*} - \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) \log{\left (x + \frac{6 A b c^{2} + 2 C a b c + C b^{3} - 64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) + 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) - 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) + b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right )}{12 A c^{3} + 4 C a c^{2} + 2 C b^{2} c} \right )} + \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) \log{\left (x + \frac{6 A b c^{2} + 2 C a b c + C b^{3} + 64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) - 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) + 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right ) - b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (6 A c^{2} + 2 C a c + C b^{2}\right )}{12 A c^{3} + 4 C a c^{2} + 2 C b^{2} c} \right )} + \frac{10 A a b c - A b^{3} + 6 C a^{2} b + x^{3} \left (12 A c^{3} + 4 C a c^{2} + 2 C b^{2} c\right ) + x^{2} \left (18 A b c^{2} + 6 C a b c + 3 C b^{3}\right ) + x \left (20 A a c^{2} + 4 A b^{2} c - 4 C a^{2} c + 10 C a b^{2}\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2)*log(x + (6*A*b*c**2 + 2*C*a*b*c + C*b**3 - 64*a**3*c
**3*sqrt(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2) + 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(6
*A*c**2 + 2*C*a*c + C*b**2) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2) + b**6*sqrt
(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2))/(12*A*c**3 + 4*C*a*c**2 + 2*C*b**2*c)) + sqrt(-1/(4*a*c
- b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2)*log(x + (6*A*b*c**2 + 2*C*a*b*c + C*b**3 + 64*a**3*c**3*sqrt(-1/(4*a
*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2) - 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*
c + C*b**2) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2) - b**6*sqrt(-1/(4*a*c - b**
2)**5)*(6*A*c**2 + 2*C*a*c + C*b**2))/(12*A*c**3 + 4*C*a*c**2 + 2*C*b**2*c)) + (10*A*a*b*c - A*b**3 + 6*C*a**2
*b + x**3*(12*A*c**3 + 4*C*a*c**2 + 2*C*b**2*c) + x**2*(18*A*b*c**2 + 6*C*a*b*c + 3*C*b**3) + x*(20*A*a*c**2 +
4*A*b**2*c - 4*C*a**2*c + 10*C*a*b**2))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a**2*b**4 + x**4*(32*a**2*c**4 - 1
6*a*b**2*c**3 + 2*b**4*c**2) + x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) + x**2*(64*a**3*c**3 - 12*a*b
**4*c + 2*b**6) + x*(64*a**3*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))

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Giac [A]  time = 1.27177, size = 293, normalized size = 1.82 \begin{align*} \frac{2 \,{\left (C b^{2} + 2 \, C a c + 6 \, A c^{2}\right )} \arctan \left (\frac{2 \, c x + 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 \, C b^{2} c x^{3} + 4 \, C a c^{2} x^{3} + 12 \, A c^{3} x^{3} + 3 \, C b^{3} x^{2} + 6 \, C a b c x^{2} + 18 \, A b c^{2} x^{2} + 10 \, C a b^{2} x - 4 \, C a^{2} c x + 4 \, A b^{2} c x + 20 \, A a c^{2} x + 6 \, C a^{2} b - A b^{3} + 10 \, A a b c}{2 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}{\left (c x^{2} + b x + a\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(C*b^2 + 2*C*a*c + 6*A*c^2)*arctan((2*c*x + 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)) + 1/2*(2*C*b^2*c*x^3 + 4*C*a*c^2*x^3 + 12*A*c^3*x^3 + 3*C*b^3*x^2 + 6*C*a*b*c*x^2 + 18*A*b*c^2*x^2
+ 10*C*a*b^2*x - 4*C*a^2*c*x + 4*A*b^2*c*x + 20*A*a*c^2*x + 6*C*a^2*b - A*b^3 + 10*A*a*b*c)/((b^4 - 8*a*b^2*c
+ 16*a^2*c^2)*(c*x^2 + b*x + a)^2)