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

Optimal. Leaf size=139 \[ \frac {3 c (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac {3 \left (b+2 c x^2\right ) (b B-2 A c)}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {-2 a B-\left (x^2 (b B-2 A c)\right )+A b}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1247, 638, 614, 618, 206} \[ -\frac {3 \left (b+2 c x^2\right ) (b B-2 A c)}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {-2 a B+x^2 (-(b B-2 A c))+A b}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 c (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 638

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 142, normalized size = 1.02 \[ \frac {-\frac {12 c (b B-2 A c) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {\left (b^2-4 a c\right ) \left (B \left (2 a+b x^2\right )-A \left (b+2 c x^2\right )\right )}{\left (a+b x^2+c x^4\right )^2}-\frac {3 \left (b+2 c x^2\right ) (b B-2 A c)}{a+b x^2+c x^4}}{4 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(b*B - 2*A*c)*(b + 2*c*x^2))/(a + b*x^2 + c*x^4) + ((b^2 - 4*a*c)*(B*(2*a + b*x^2) - A*(b + 2*c*x^2)))/(a
+ b*x^2 + c*x^4)^2 - (12*c*(b*B - 2*A*c)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(4*(b^
2 - 4*a*c)^2)

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fricas [B]  time = 0.59, size = 1109, normalized size = 7.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*(B*b^3*c^2 + 8*A*a*c^4 - 2*(2*B*a*b + A*b^2)*c^3)*x^6 + B*a*b^4 + A*b^5 + 9*(B*b^4*c + 8*A*a*b*c^3 -
2*(2*B*a*b^2 + A*b^3)*c^2)*x^4 - 8*(4*B*a^3 - 5*A*a^2*b)*c^2 + 2*(B*b^5 + 40*A*a^2*c^3 - 2*(10*B*a^2*b + A*a*b
^2)*c^2 + (B*a*b^3 - 2*A*b^4)*c)*x^2 + 6*((B*b*c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + B*a^2*b*c
- 2*A*a^2*c^2 + (B*b^3*c - 4*A*a*c^3 + 2*(B*a*b - A*b^2)*c^2)*x^4 + 2*(B*a*b^2*c - 2*A*a*b*c^2)*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)) + 2*
(2*B*a^2*b^2 - 7*A*a*b^3)*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*(6*(B*b^3*c^2 + 8*A*a*c^4 - 2*(2*B*a*b + A*b^2)*c^3)*x^6 + B*a*b^4 + A*b^5 + 9*(B*b^4*c +
8*A*a*b*c^3 - 2*(2*B*a*b^2 + A*b^3)*c^2)*x^4 - 8*(4*B*a^3 - 5*A*a^2*b)*c^2 + 2*(B*b^5 + 40*A*a^2*c^3 - 2*(10*B
*a^2*b + A*a*b^2)*c^2 + (B*a*b^3 - 2*A*b^4)*c)*x^2 - 12*((B*b*c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x
^6 + B*a^2*b*c - 2*A*a^2*c^2 + (B*b^3*c - 4*A*a*c^3 + 2*(B*a*b - A*b^2)*c^2)*x^4 + 2*(B*a*b^2*c - 2*A*a*b*c^2)
*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)) + 2*(2*B*a^2*b^2 - 7*A*a*b^3)
*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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giac [A]  time = 5.57, size = 208, normalized size = 1.50 \[ -\frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {6 \, B b c^{2} x^{6} - 12 \, A c^{3} x^{6} + 9 \, B b^{2} c x^{4} - 18 \, A b c^{2} x^{4} + 2 \, B b^{3} x^{2} + 10 \, B a b c x^{2} - 4 \, A b^{2} c x^{2} - 20 \, A a c^{2} x^{2} + B a b^{2} + A b^{3} + 8 \, B a^{2} c - 10 \, A a b c}{4 \, {\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(B*b*c - 2*A*c^2)*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*(6*B*b*c^2*x^6 - 12*A*c^3*x^6 + 9*B*b^2*c*x^4 - 18*A*b*c^2*x^4 + 2*B*b^3*x^2 + 10*B*a*b*c*x^2 - 4*A*
b^2*c*x^2 - 20*A*a*c^2*x^2 + B*a*b^2 + A*b^3 + 8*B*a^2*c - 10*A*a*b*c)/((c*x^4 + b*x^2 + a)^2*(b^4 - 8*a*b^2*c
+ 16*a^2*c^2))

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maple [A]  time = 0.01, size = 262, normalized size = 1.88 \[ \frac {3 A \,c^{2} x^{2}}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}-\frac {3 B b c \,x^{2}}{2 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {6 A \,c^{2} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}-\frac {3 B b c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}+\frac {3 A b c}{2 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}-\frac {3 B \,b^{2}}{4 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {A b -2 B a +\left (2 A c -b B \right ) x^{2}}{4 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(A*b-2*B*a+(2*A*c-B*b)*x^2)/(4*a*c-b^2)/(c*x^4+b*x^2+a)^2+3/(4*a*c-b^2)^2/(c*x^4+b*x^2+a)*c^2*x^2*A-3/2/(4
*a*c-b^2)^2/(c*x^4+b*x^2+a)*c*x^2*b*B+3/2/(4*a*c-b^2)^2/(c*x^4+b*x^2+a)*b*A*c-3/4/(4*a*c-b^2)^2/(c*x^4+b*x^2+a
)*b^2*B+6/(4*a*c-b^2)^(5/2)*c^2*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*A-3/(4*a*c-b^2)^(5/2)*c*arctan((2*c*x^2+
b)/(4*a*c-b^2)^(1/2))*b*B

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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(x*(B*x^2+A)/(c*x^4+b*x^2+a)^3,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 = 0.59, size = 517, normalized size = 3.72 \[ \frac {3\,c\,\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {3\,c\,\left (2\,A\,c-B\,b\right )\,\left (6\,A\,c^4-3\,B\,b\,c^3\right )}{a\,{\left (4\,a\,c-b^2\right )}^{9/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b\,c^2\,{\left (2\,A\,c-B\,b\right )}^2\,\left (32\,a^2\,b\,c^4-16\,a\,b^3\,c^3+2\,b^5\,c^2\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{15/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )+\frac {18\,b\,c^4\,{\left (2\,A\,c-B\,b\right )}^2}{{\left (4\,a\,c-b^2\right )}^{15/2}}\right )\,\left (b^4\,{\left (4\,a\,c-b^2\right )}^5+16\,a^2\,c^2\,{\left (4\,a\,c-b^2\right )}^5-8\,a\,b^2\,c\,{\left (4\,a\,c-b^2\right )}^5\right )}{72\,A^2\,c^6-72\,A\,B\,b\,c^5+18\,B^2\,b^2\,c^4}\right )\,\left (2\,A\,c-B\,b\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {8\,B\,c\,a^2+B\,a\,b^2-10\,A\,c\,a\,b+A\,b^3}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {9\,x^4\,\left (2\,A\,b\,c^2-B\,b^2\,c\right )}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (B\,b^3-2\,A\,b^2\,c+5\,B\,a\,b\,c-10\,A\,a\,c^2\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {3\,c^2\,x^6\,\left (2\,A\,c-B\,b\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^4\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^8+2\,a\,b\,x^2+2\,b\,c\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*c*atan(((x^2*((3*c*(2*A*c - B*b)*(6*A*c^4 - 3*B*b*c^3))/(a*(4*a*c - b^2)^(9/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*
c)) + (9*b*c^2*(2*A*c - B*b)^2*(2*b^5*c^2 - 16*a*b^3*c^3 + 32*a^2*b*c^4))/(2*a*(4*a*c - b^2)^(15/2)*(b^4 + 16*
a^2*c^2 - 8*a*b^2*c))) + (18*b*c^4*(2*A*c - B*b)^2)/(4*a*c - b^2)^(15/2))*(b^4*(4*a*c - b^2)^5 + 16*a^2*c^2*(4
*a*c - b^2)^5 - 8*a*b^2*c*(4*a*c - b^2)^5))/(72*A^2*c^6 + 18*B^2*b^2*c^4 - 72*A*B*b*c^5))*(2*A*c - B*b))/(4*a*
c - b^2)^(5/2) - ((A*b^3 + B*a*b^2 + 8*B*a^2*c - 10*A*a*b*c)/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (9*x^4*(2*A*
b*c^2 - B*b^2*c))/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(B*b^3 - 10*A*a*c^2 - 2*A*b^2*c + 5*B*a*b*c))/(2*(
b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (3*c^2*x^6*(2*A*c - B*b))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^4*(2*a*c + b
^2) + a^2 + c^2*x^8 + 2*a*b*x^2 + 2*b*c*x^6)

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sympy [B]  time = 12.40, size = 661, normalized size = 4.76 \[ \frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) \log {\left (x^{2} + \frac {- 6 A b c^{2} + 3 B b^{2} c - 192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) + 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) - 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) + 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right )}{- 12 A c^{3} + 6 B b c^{2}} \right )}}{2} - \frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) \log {\left (x^{2} + \frac {- 6 A b c^{2} + 3 B b^{2} c + 192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) - 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) + 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right ) - 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (- 2 A c + B b\right )}{- 12 A c^{3} + 6 B b c^{2}} \right )}}{2} + \frac {10 A a b c - A b^{3} - 8 B a^{2} c - B a b^{2} + x^{6} \left (12 A c^{3} - 6 B b c^{2}\right ) + x^{4} \left (18 A b c^{2} - 9 B b^{2} c\right ) + x^{2} \left (20 A a c^{2} + 4 A b^{2} c - 10 B a b c - 2 B b^{3}\right )}{64 a^{4} c^{2} - 32 a^{3} b^{2} c + 4 a^{2} b^{4} + x^{8} \left (64 a^{2} c^{4} - 32 a b^{2} c^{3} + 4 b^{4} c^{2}\right ) + x^{6} \left (128 a^{2} b c^{3} - 64 a b^{3} c^{2} + 8 b^{5} c\right ) + x^{4} \left (128 a^{3} c^{3} - 24 a b^{4} c + 4 b^{6}\right ) + x^{2} \left (128 a^{3} b c^{2} - 64 a^{2} b^{3} c + 8 a b^{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*c*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b)*log(x**2 + (-6*A*b*c**2 + 3*B*b**2*c - 192*a**3*c**4*sqrt(-1/(4*
a*c - b**2)**5)*(-2*A*c + B*b) + 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b) - 36*a*b**4*c**2
*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b) + 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b))/(-12*A*c**3 +
6*B*b*c**2))/2 - 3*c*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b)*log(x**2 + (-6*A*b*c**2 + 3*B*b**2*c + 192*a**
3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b) - 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b
) + 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B*b) - 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(-2*A*c + B
*b))/(-12*A*c**3 + 6*B*b*c**2))/2 + (10*A*a*b*c - A*b**3 - 8*B*a**2*c - B*a*b**2 + x**6*(12*A*c**3 - 6*B*b*c**
2) + x**4*(18*A*b*c**2 - 9*B*b**2*c) + x**2*(20*A*a*c**2 + 4*A*b**2*c - 10*B*a*b*c - 2*B*b**3))/(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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