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]
-(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)
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Rubi [A] time = 0.190133, antiderivative size = 206, normalized size of antiderivative = 1.,
number of steps used = 6, 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} \[ -\frac{2 (b+2 c x) \left (C \left (a c+b^2\right )+5 A c^2\right )}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}-\frac{x \left (C \left (b^2-2 a c\right )+2 A c^2\right )+b c \left (\frac{a C}{c}+A\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac{8 c \left (C \left (a c+b^2\right )+5 A c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}+\frac{(b+2 c x) \left (C \left (a+\frac{b^2}{c}\right )+5 A c\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2} \]
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 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 )^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*}
Mathematica [A] time = 0.399707, 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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Maple [B] time = 0.187, size = 643, normalized size = 3.1 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{3}} \left ( 4\,{\frac{{c}^{3} \left ( 5\,A{c}^{2}+Cac+C{b}^{2} \right ){x}^{5}}{64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6}}}+10\,{\frac{{c}^{2} \left ( 5\,A{c}^{2}+Cac+C{b}^{2} \right ) b{x}^{4}}{64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6}}}+{\frac{ \left ( 32\,ac+22\,{b}^{2} \right ) c \left ( 5\,A{c}^{2}+Cac+C{b}^{2} \right ){x}^{3}}{192\,{c}^{3}{a}^{3}-144\,{a}^{2}{b}^{2}{c}^{2}+36\,a{b}^{4}c-3\,{b}^{6}}}+{\frac{b \left ( 16\,ac+{b}^{2} \right ) \left ( 5\,A{c}^{2}+Cac+C{b}^{2} \right ){x}^{2}}{64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6}}}+{\frac{ \left ( 44\,{a}^{2}A{c}^{3}+18\,Aa{b}^{2}{c}^{2}-A{b}^{4}c-4\,C{a}^{3}{c}^{2}+22\,C{a}^{2}{b}^{2}c+Ca{b}^{4} \right ) x}{64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6}}}+{\frac{ \left ( 66\,{a}^{2}A{c}^{2}-13\,Aa{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}}} \right ) }+40\,{\frac{A{c}^{3}}{ \left ( 64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+8\,{\frac{Ca{c}^{2}}{ \left ( 64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+8\,{\frac{C{b}^{2}c}{ \left ( 64\,{c}^{3}{a}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6} \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)^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., 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)^4,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.75595, size = 4547, normalized size = 22.07 \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)^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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Sympy [B] time = 5.19637, size = 1224, normalized size = 5.94 \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)**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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Giac [B] time = 1.17805, size = 549, normalized size = 2.67 \begin{align*} -\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}} \end{align*}
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)