3.147 \(\int \frac {A+C x^2}{(a+b x+c x^2)^4} \, dx\)

Optimal. Leaf size=206 \[ -\frac {2 (b+2 c x) \left (C \left (a c+b^2\right )+5 A c^2\right )}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}-\frac {x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac {a C}{c}+A\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {8 c \left (C \left (a c+b^2\right )+5 A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}+\frac {(b+2 c x) \left (C \left (a+\frac {b^2}{c}\right )+5 A c\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2} \]

[Out]

1/3*(-b*c*(A+a*C/c)-(2*A*c^2+(-2*a*c+b^2)*C)*x)/c/(-4*a*c+b^2)/(c*x^2+b*x+a)^3+1/3*(5*A*c+(a+1/c*b^2)*C)*(2*c*
x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^2-2*(5*A*c^2+(a*c+b^2)*C)*(2*c*x+b)/(-4*a*c+b^2)^3/(c*x^2+b*x+a)+8*c*(5*A*c^
2+(a*c+b^2)*C)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(7/2)

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Rubi [A]  time = 0.19, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1660, 12, 614, 618, 206} \[ -\frac {2 (b+2 c x) \left (C \left (a c+b^2\right )+5 A c^2\right )}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}-\frac {x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac {a C}{c}+A\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {8 c \left (C \left (a c+b^2\right )+5 A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}+\frac {(b+2 c x) \left (C \left (a+\frac {b^2}{c}\right )+5 A c\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*(A + (a*C)/c) + (2*A*c^2 + (b^2 - 2*a*c)*C)*x)/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^3) + ((5*A*c + (a +
b^2/c)*C)*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^2) - (2*(5*A*c^2 + (b^2 + a*c)*C)*(b + 2*c*x))/((b
^2 - 4*a*c)^3*(a + b*x + c*x^2)) + (8*c*(5*A*c^2 + (b^2 + a*c)*C)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2
- 4*a*c)^(7/2)

Rule 12

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

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 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 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
]

Rubi steps

\begin {align*} \int \frac {A+C x^2}{\left (a+b x+c x^2\right )^4} \, 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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}-\frac {\int \frac {2 \left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right )}{\left (a+b x+c x^2\right )^3} \, dx}{3 \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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}-\frac {\left (2 \left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right )\right ) \int \frac {1}{\left (a+b x+c x^2\right )^3} \, dx}{3 \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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {\left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}+\frac {\left (2 \left (5 A c^2+\left (b^2+a c\right ) C\right )\right ) \int \frac {1}{\left (a+b x+c x^2\right )^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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {\left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {2 \left (5 A c^2+\left (b^2+a c\right ) C\right ) (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}-\frac {\left (4 c \left (5 A c^2+\left (b^2+a c\right ) C\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3}\\ &=-\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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {\left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {2 \left (5 A c^2+\left (b^2+a c\right ) C\right ) (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {\left (8 c \left (5 A c^2+\left (b^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 )^3}\\ &=-\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}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {\left (5 A c+\left (a+\frac {b^2}{c}\right ) C\right ) (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {2 \left (5 A c^2+\left (b^2+a c\right ) C\right ) (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {8 c \left (5 A c^2+\left (b^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 )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 204, normalized size = 0.99 \[ \frac {1}{3} \left (-\frac {6 (b+2 c x) \left (C \left (a c+b^2\right )+5 A c^2\right )}{\left (b^2-4 a c\right )^3 (a+x (b+c x))}+\frac {(b+2 c x) \left (C \left (a c+b^2\right )+5 A c^2\right )}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))^2}+\frac {24 c \left (C \left (a c+b^2\right )+5 A c^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{7/2}}+\frac {a C (b-2 c x)+A c (b+2 c x)+b^2 C x}{c \left (4 a c-b^2\right ) (a+x (b+c x))^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*A*c^2 + (b^2 + a*c)*C)*(b + 2*c*x))/(c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^2) - (6*(5*A*c^2 + (b^2 + a*c)*C
)*(b + 2*c*x))/((b^2 - 4*a*c)^3*(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))^3) + (24*c*(5*A*c^2 + (b^2 + a*c)*C)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 +
4*a*c)^(7/2))/3

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fricas [B]  time = 0.97, size = 2103, normalized size = 10.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(C*a^2*b^5 + A*b^7 - 264*A*a^3*b*c^3 + 12*(C*b^4*c^3 - 3*C*a*b^2*c^4 - 20*A*a*c^6 - (4*C*a^2 - 5*A*b^2)*
c^5)*x^5 + 30*(C*b^5*c^2 - 3*C*a*b^3*c^3 - 20*A*a*b*c^5 - (4*C*a^2*b - 5*A*b^3)*c^4)*x^4 + 2*(11*C*b^6*c - 17*
C*a*b^4*c^2 - 320*A*a^2*c^5 - 4*(16*C*a^3 + 35*A*a*b^2)*c^4 - (92*C*a^2*b^2 - 55*A*b^4)*c^3)*x^3 - 2*(52*C*a^4
*b - 59*A*a^2*b^3)*c^2 + 3*(C*b^7 + 13*C*a*b^5*c - 320*A*a^2*b*c^4 - 4*(16*C*a^3*b - 15*A*a*b^3)*c^3 - (52*C*a
^2*b^3 - 5*A*b^5)*c^2)*x^2 + 12*(C*a^3*b^2*c + C*a^4*c^2 + 5*A*a^3*c^3 + (C*b^2*c^4 + C*a*c^5 + 5*A*c^6)*x^6 +
3*(C*b^3*c^3 + C*a*b*c^4 + 5*A*b*c^5)*x^5 + 3*(C*b^4*c^2 + 2*C*a*b^2*c^3 + 5*A*a*c^5 + (C*a^2 + 5*A*b^2)*c^4)
*x^4 + (C*b^5*c + 7*C*a*b^3*c^2 + 30*A*a*b*c^4 + (6*C*a^2*b + 5*A*b^3)*c^3)*x^3 + 3*(C*a*b^4*c + 2*C*a^2*b^2*c
^2 + 5*A*a^2*c^4 + (C*a^3 + 5*A*a*b^2)*c^3)*x^2 + 3*(C*a^2*b^3*c + C*a^3*b*c^2 + 5*A*a^2*b*c^3)*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)) + (22*C*a^3*
b^3 - 17*A*a*b^5)*c + 3*(C*a*b^6 - 176*A*a^3*c^4 + 4*(4*C*a^4 - 7*A*a^2*b^2)*c^3 - 2*(46*C*a^3*b^2 - 11*A*a*b^
4)*c^2 + (18*C*a^2*b^4 - A*b^6)*c)*x)/(a^3*b^8 - 16*a^4*b^6*c + 96*a^5*b^4*c^2 - 256*a^6*b^2*c^3 + 256*a^7*c^4
+ (b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6 + 256*a^4*c^7)*x^6 + 3*(b^9*c^2 - 16*a*b^7*c^3 +
96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*x^5 + 3*(b^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c^3 - 160*a^3*b
^4*c^4 + 256*a^5*c^6)*x^4 + (b^11 - 10*a*b^9*c + 320*a^3*b^5*c^3 - 1280*a^4*b^3*c^4 + 1536*a^5*b*c^5)*x^3 + 3*
(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4*b^4*c^3 + 256*a^6*c^5)*x^2 + 3*(a^2*b^9 - 16*a^3*b^7*c + 96*
a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*x), -1/3*(C*a^2*b^5 + A*b^7 - 264*A*a^3*b*c^3 + 12*(C*b^4*c^3 -
3*C*a*b^2*c^4 - 20*A*a*c^6 - (4*C*a^2 - 5*A*b^2)*c^5)*x^5 + 30*(C*b^5*c^2 - 3*C*a*b^3*c^3 - 20*A*a*b*c^5 - (4
*C*a^2*b - 5*A*b^3)*c^4)*x^4 + 2*(11*C*b^6*c - 17*C*a*b^4*c^2 - 320*A*a^2*c^5 - 4*(16*C*a^3 + 35*A*a*b^2)*c^4
- (92*C*a^2*b^2 - 55*A*b^4)*c^3)*x^3 - 2*(52*C*a^4*b - 59*A*a^2*b^3)*c^2 + 3*(C*b^7 + 13*C*a*b^5*c - 320*A*a^2
*b*c^4 - 4*(16*C*a^3*b - 15*A*a*b^3)*c^3 - (52*C*a^2*b^3 - 5*A*b^5)*c^2)*x^2 - 24*(C*a^3*b^2*c + C*a^4*c^2 + 5
*A*a^3*c^3 + (C*b^2*c^4 + C*a*c^5 + 5*A*c^6)*x^6 + 3*(C*b^3*c^3 + C*a*b*c^4 + 5*A*b*c^5)*x^5 + 3*(C*b^4*c^2 +
2*C*a*b^2*c^3 + 5*A*a*c^5 + (C*a^2 + 5*A*b^2)*c^4)*x^4 + (C*b^5*c + 7*C*a*b^3*c^2 + 30*A*a*b*c^4 + (6*C*a^2*b
+ 5*A*b^3)*c^3)*x^3 + 3*(C*a*b^4*c + 2*C*a^2*b^2*c^2 + 5*A*a^2*c^4 + (C*a^3 + 5*A*a*b^2)*c^3)*x^2 + 3*(C*a^2*b
^3*c + C*a^3*b*c^2 + 5*A*a^2*b*c^3)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)
) + (22*C*a^3*b^3 - 17*A*a*b^5)*c + 3*(C*a*b^6 - 176*A*a^3*c^4 + 4*(4*C*a^4 - 7*A*a^2*b^2)*c^3 - 2*(46*C*a^3*b
^2 - 11*A*a*b^4)*c^2 + (18*C*a^2*b^4 - A*b^6)*c)*x)/(a^3*b^8 - 16*a^4*b^6*c + 96*a^5*b^4*c^2 - 256*a^6*b^2*c^3
+ 256*a^7*c^4 + (b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6 + 256*a^4*c^7)*x^6 + 3*(b^9*c^2 -
16*a*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*x^5 + 3*(b^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c
^3 - 160*a^3*b^4*c^4 + 256*a^5*c^6)*x^4 + (b^11 - 10*a*b^9*c + 320*a^3*b^5*c^3 - 1280*a^4*b^3*c^4 + 1536*a^5*b
*c^5)*x^3 + 3*(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4*b^4*c^3 + 256*a^6*c^5)*x^2 + 3*(a^2*b^9 - 16*a
^3*b^7*c + 96*a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*x)]

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giac [B]  time = 0.17, size = 407, normalized size = 1.98 \[ -\frac {8 \, {\left (C b^{2} c + C a c^{2} + 5 \, A c^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {12 \, C b^{2} c^{3} x^{5} + 12 \, C a c^{4} x^{5} + 60 \, A c^{5} x^{5} + 30 \, C b^{3} c^{2} x^{4} + 30 \, C a b c^{3} x^{4} + 150 \, A b c^{4} x^{4} + 22 \, C b^{4} c x^{3} + 54 \, C a b^{2} c^{2} x^{3} + 32 \, C a^{2} c^{3} x^{3} + 110 \, A b^{2} c^{3} x^{3} + 160 \, A a c^{4} x^{3} + 3 \, C b^{5} x^{2} + 51 \, C a b^{3} c x^{2} + 48 \, C a^{2} b c^{2} x^{2} + 15 \, A b^{3} c^{2} x^{2} + 240 \, A a b c^{3} x^{2} + 3 \, C a b^{4} x + 66 \, C a^{2} b^{2} c x - 3 \, A b^{4} c x - 12 \, C a^{3} c^{2} x + 54 \, A a b^{2} c^{2} x + 132 \, A a^{2} c^{3} x + C a^{2} b^{3} + A b^{5} + 26 \, C a^{3} b c - 13 \, A a b^{3} c + 66 \, A a^{2} b c^{2}}{3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} {\left (c x^{2} + b x + a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(C*b^2*c + C*a*c^2 + 5*A*c^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 -
64*a^3*c^3)*sqrt(-b^2 + 4*a*c)) - 1/3*(12*C*b^2*c^3*x^5 + 12*C*a*c^4*x^5 + 60*A*c^5*x^5 + 30*C*b^3*c^2*x^4 + 3
0*C*a*b*c^3*x^4 + 150*A*b*c^4*x^4 + 22*C*b^4*c*x^3 + 54*C*a*b^2*c^2*x^3 + 32*C*a^2*c^3*x^3 + 110*A*b^2*c^3*x^3
+ 160*A*a*c^4*x^3 + 3*C*b^5*x^2 + 51*C*a*b^3*c*x^2 + 48*C*a^2*b*c^2*x^2 + 15*A*b^3*c^2*x^2 + 240*A*a*b*c^3*x^
2 + 3*C*a*b^4*x + 66*C*a^2*b^2*c*x - 3*A*b^4*c*x - 12*C*a^3*c^2*x + 54*A*a*b^2*c^2*x + 132*A*a^2*c^3*x + C*a^2
*b^3 + A*b^5 + 26*C*a^3*b*c - 13*A*a*b^3*c + 66*A*a^2*b*c^2)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)
*(c*x^2 + b*x + a)^3)

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maple [B]  time = 0.02, size = 643, normalized size = 3.12 \[ \frac {40 A \,c^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \sqrt {4 a c -b^{2}}}+\frac {8 C a \,c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \sqrt {4 a c -b^{2}}}+\frac {8 C \,b^{2} c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \sqrt {4 a c -b^{2}}}+\frac {\frac {4 \left (5 A \,c^{2}+C a c +C \,b^{2}\right ) c^{3} x^{5}}{64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}+\frac {10 \left (5 A \,c^{2}+C a c +C \,b^{2}\right ) b \,c^{2} x^{4}}{64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}+\frac {2 \left (16 a c +11 b^{2}\right ) \left (5 A \,c^{2}+C a c +C \,b^{2}\right ) c \,x^{3}}{3 \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}+\frac {\left (16 a c +b^{2}\right ) \left (5 A \,c^{2}+C a c +C \,b^{2}\right ) b \,x^{2}}{64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}+\frac {\left (66 A \,a^{2} c^{2}-13 A a \,b^{2} c +A \,b^{4}+26 C \,a^{3} c +C \,a^{2} b^{2}\right ) b}{192 c^{3} a^{3}-144 a^{2} b^{2} c^{2}+36 a \,b^{4} c -3 b^{6}}+\frac {\left (44 A \,a^{2} c^{3}+18 A a \,b^{2} c^{2}-A \,b^{4} c -4 C \,a^{3} c^{2}+22 C \,a^{2} b^{2} c +C a \,b^{4}\right ) x}{64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}}{\left (c \,x^{2}+b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+A)/(c*x^2+b*x+a)^4,x)

[Out]

(4*c^3*(5*A*c^2+C*a*c+C*b^2)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)*x^5+10*c^2*(5*A*c^2+C*a*c+C*b^2)/(64*a
^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)*b*x^4+2/3*(16*a*c+11*b^2)*c*(5*A*c^2+C*a*c+C*b^2)/(64*a^3*c^3-48*a^2*b^2
*c^2+12*a*b^4*c-b^6)*x^3+b*(16*a*c+b^2)*(5*A*c^2+C*a*c+C*b^2)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)*x^2+(
44*A*a^2*c^3+18*A*a*b^2*c^2-A*b^4*c-4*C*a^3*c^2+22*C*a^2*b^2*c+C*a*b^4)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-
b^6)*x+1/3*(66*A*a^2*c^2-13*A*a*b^2*c+A*b^4+26*C*a^3*c+C*a^2*b^2)*b/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)
)/(c*x^2+b*x+a)^3+40*c^3/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-
b^2)^(1/2))*A+8*c^2/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^
(1/2))*C*a+8*c/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+A)/(c*x^2+b*x+a)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 4.36, size = 698, normalized size = 3.39 \[ -\frac {\frac {26\,C\,a^3\,b\,c+C\,a^2\,b^3+66\,A\,a^2\,b\,c^2-13\,A\,a\,b^3\,c+A\,b^5}{3\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {x\,\left (-4\,C\,a^3\,c^2+22\,C\,a^2\,b^2\,c+44\,A\,a^2\,c^3+C\,a\,b^4+18\,A\,a\,b^2\,c^2-A\,b^4\,c\right )}{-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}+\frac {2\,x^3\,\left (11\,b^2\,c+16\,a\,c^2\right )\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{3\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}+\frac {x^2\,\left (b^3+16\,a\,c\,b\right )\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}+\frac {4\,c^3\,x^5\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}+\frac {10\,b\,c^2\,x^4\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}}{x^2\,\left (3\,c\,a^2+3\,a\,b^2\right )+x^4\,\left (3\,b^2\,c+3\,a\,c^2\right )+a^3+x^3\,\left (b^3+6\,a\,c\,b\right )+c^3\,x^6+3\,b\,c^2\,x^5+3\,a^2\,b\,x}-\frac {8\,c\,\mathrm {atan}\left (\frac {\left (\frac {8\,c^2\,x\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {4\,c\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )\,\left (-64\,a^3\,b\,c^3+48\,a^2\,b^3\,c^2-12\,a\,b^5\,c+b^7\right )}{{\left (4\,a\,c-b^2\right )}^{7/2}\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}\right )\,\left (-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6\right )}{4\,C\,b^2\,c+20\,A\,c^3+4\,C\,a\,c^2}\right )\,\left (C\,b^2+5\,A\,c^2+C\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*x^2)/(a + b*x + c*x^2)^4,x)

[Out]

- ((A*b^5 + C*a^2*b^3 - 13*A*a*b^3*c + 26*C*a^3*b*c + 66*A*a^2*b*c^2)/(3*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 -
12*a*b^4*c)) + (x*(44*A*a^2*c^3 - 4*C*a^3*c^2 - A*b^4*c + C*a*b^4 + 18*A*a*b^2*c^2 + 22*C*a^2*b^2*c))/(b^6 - 6
4*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c) + (2*x^3*(16*a*c^2 + 11*b^2*c)*(5*A*c^2 + C*b^2 + C*a*c))/(3*(b^6 - 6
4*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)) + (x^2*(b^3 + 16*a*b*c)*(5*A*c^2 + C*b^2 + C*a*c))/(b^6 - 64*a^3*c^3
+ 48*a^2*b^2*c^2 - 12*a*b^4*c) + (4*c^3*x^5*(5*A*c^2 + C*b^2 + C*a*c))/(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 1
2*a*b^4*c) + (10*b*c^2*x^4*(5*A*c^2 + C*b^2 + C*a*c))/(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c))/(x^2*(
3*a*b^2 + 3*a^2*c) + x^4*(3*a*c^2 + 3*b^2*c) + a^3 + x^3*(b^3 + 6*a*b*c) + c^3*x^6 + 3*b*c^2*x^5 + 3*a^2*b*x)
- (8*c*atan((((8*c^2*x*(5*A*c^2 + C*b^2 + C*a*c))/(4*a*c - b^2)^(7/2) + (4*c*(5*A*c^2 + C*b^2 + C*a*c)*(b^7 -
64*a^3*b*c^3 + 48*a^2*b^3*c^2 - 12*a*b^5*c))/((4*a*c - b^2)^(7/2)*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^
4*c)))*(b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c))/(20*A*c^3 + 4*C*a*c^2 + 4*C*b^2*c))*(5*A*c^2 + C*b^2
+ C*a*c))/(4*a*c - b^2)^(7/2)

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sympy [B]  time = 4.22, size = 1224, normalized size = 5.94 \[ - 4 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) \log {\left (x + \frac {20 A b c^{3} + 4 C a b c^{2} + 4 C b^{3} c - 1024 a^{4} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) + 1024 a^{3} b^{2} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) - 384 a^{2} b^{4} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) + 64 a b^{6} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) - 4 b^{8} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right )}{40 A c^{4} + 8 C a c^{3} + 8 C b^{2} c^{2}} \right )} + 4 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) \log {\left (x + \frac {20 A b c^{3} + 4 C a b c^{2} + 4 C b^{3} c + 1024 a^{4} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) - 1024 a^{3} b^{2} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) + 384 a^{2} b^{4} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) - 64 a b^{6} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right ) + 4 b^{8} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \left (5 A c^{2} + C a c + C b^{2}\right )}{40 A c^{4} + 8 C a c^{3} + 8 C b^{2} c^{2}} \right )} + \frac {66 A a^{2} b c^{2} - 13 A a b^{3} c + A b^{5} + 26 C a^{3} b c + C a^{2} b^{3} + x^{5} \left (60 A c^{5} + 12 C a c^{4} + 12 C b^{2} c^{3}\right ) + x^{4} \left (150 A b c^{4} + 30 C a b c^{3} + 30 C b^{3} c^{2}\right ) + x^{3} \left (160 A a c^{4} + 110 A b^{2} c^{3} + 32 C a^{2} c^{3} + 54 C a b^{2} c^{2} + 22 C b^{4} c\right ) + x^{2} \left (240 A a b c^{3} + 15 A b^{3} c^{2} + 48 C a^{2} b c^{2} + 51 C a b^{3} c + 3 C b^{5}\right ) + x \left (132 A a^{2} c^{3} + 54 A a b^{2} c^{2} - 3 A b^{4} c - 12 C a^{3} c^{2} + 66 C a^{2} b^{2} c + 3 C a b^{4}\right )}{192 a^{6} c^{3} - 144 a^{5} b^{2} c^{2} + 36 a^{4} b^{4} c - 3 a^{3} b^{6} + x^{6} \left (192 a^{3} c^{6} - 144 a^{2} b^{2} c^{5} + 36 a b^{4} c^{4} - 3 b^{6} c^{3}\right ) + x^{5} \left (576 a^{3} b c^{5} - 432 a^{2} b^{3} c^{4} + 108 a b^{5} c^{3} - 9 b^{7} c^{2}\right ) + x^{4} \left (576 a^{4} c^{5} + 144 a^{3} b^{2} c^{4} - 324 a^{2} b^{4} c^{3} + 99 a b^{6} c^{2} - 9 b^{8} c\right ) + x^{3} \left (1152 a^{4} b c^{4} - 672 a^{3} b^{3} c^{3} + 72 a^{2} b^{5} c^{2} + 18 a b^{7} c - 3 b^{9}\right ) + x^{2} \left (576 a^{5} c^{4} + 144 a^{4} b^{2} c^{3} - 324 a^{3} b^{4} c^{2} + 99 a^{2} b^{6} c - 9 a b^{8}\right ) + x \left (576 a^{5} b c^{3} - 432 a^{4} b^{3} c^{2} + 108 a^{3} b^{5} c - 9 a^{2} b^{7}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2)*log(x + (20*A*b*c**3 + 4*C*a*b*c**2 + 4*C*b**3*c -
1024*a**4*c**5*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2) + 1024*a**3*b**2*c**4*sqrt(-1/(4*a*c -
b**2)**7)*(5*A*c**2 + C*a*c + C*b**2) - 384*a**2*b**4*c**3*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b*
*2) + 64*a*b**6*c**2*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2) - 4*b**8*c*sqrt(-1/(4*a*c - b**2)*
*7)*(5*A*c**2 + C*a*c + C*b**2))/(40*A*c**4 + 8*C*a*c**3 + 8*C*b**2*c**2)) + 4*c*sqrt(-1/(4*a*c - b**2)**7)*(5
*A*c**2 + C*a*c + C*b**2)*log(x + (20*A*b*c**3 + 4*C*a*b*c**2 + 4*C*b**3*c + 1024*a**4*c**5*sqrt(-1/(4*a*c - b
**2)**7)*(5*A*c**2 + C*a*c + C*b**2) - 1024*a**3*b**2*c**4*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b*
*2) + 384*a**2*b**4*c**3*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2) - 64*a*b**6*c**2*sqrt(-1/(4*a*
c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2) + 4*b**8*c*sqrt(-1/(4*a*c - b**2)**7)*(5*A*c**2 + C*a*c + C*b**2))/(
40*A*c**4 + 8*C*a*c**3 + 8*C*b**2*c**2)) + (66*A*a**2*b*c**2 - 13*A*a*b**3*c + A*b**5 + 26*C*a**3*b*c + C*a**2
*b**3 + x**5*(60*A*c**5 + 12*C*a*c**4 + 12*C*b**2*c**3) + x**4*(150*A*b*c**4 + 30*C*a*b*c**3 + 30*C*b**3*c**2)
+ x**3*(160*A*a*c**4 + 110*A*b**2*c**3 + 32*C*a**2*c**3 + 54*C*a*b**2*c**2 + 22*C*b**4*c) + x**2*(240*A*a*b*c
**3 + 15*A*b**3*c**2 + 48*C*a**2*b*c**2 + 51*C*a*b**3*c + 3*C*b**5) + x*(132*A*a**2*c**3 + 54*A*a*b**2*c**2 -
3*A*b**4*c - 12*C*a**3*c**2 + 66*C*a**2*b**2*c + 3*C*a*b**4))/(192*a**6*c**3 - 144*a**5*b**2*c**2 + 36*a**4*b*
*4*c - 3*a**3*b**6 + x**6*(192*a**3*c**6 - 144*a**2*b**2*c**5 + 36*a*b**4*c**4 - 3*b**6*c**3) + x**5*(576*a**3
*b*c**5 - 432*a**2*b**3*c**4 + 108*a*b**5*c**3 - 9*b**7*c**2) + x**4*(576*a**4*c**5 + 144*a**3*b**2*c**4 - 324
*a**2*b**4*c**3 + 99*a*b**6*c**2 - 9*b**8*c) + x**3*(1152*a**4*b*c**4 - 672*a**3*b**3*c**3 + 72*a**2*b**5*c**2
+ 18*a*b**7*c - 3*b**9) + x**2*(576*a**5*c**4 + 144*a**4*b**2*c**3 - 324*a**3*b**4*c**2 + 99*a**2*b**6*c - 9*
a*b**8) + x*(576*a**5*b*c**3 - 432*a**4*b**3*c**2 + 108*a**3*b**5*c - 9*a**2*b**7))

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