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

Optimal. Leaf size=94 \[ -\frac{-2 a B+x^2 (-(b B-2 A c))+A b}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0876177, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {1247, 638, 618, 206} \[ -\frac{-2 a B+x^2 (-(b B-2 A c))+A b}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0842634, size = 101, normalized size = 1.07 \[ \frac{\frac{2 (b B-2 A c) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{B \left (2 a+b x^2\right )-A \left (b+2 c x^2\right )}{a+b x^2+c x^4}}{2 \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 127, normalized size = 1.4 \begin{align*}{\frac{ \left ( 2\,Ac-bB \right ){x}^{2}+Ab-2\,aB}{ \left ( 8\,ac-2\,{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{2}+a \right ) }}+2\,{\frac{Ac}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{bB\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*A*c-B*b)*x^2+A*b-2*a*B)/(4*a*c-b^2)/(c*x^4+b*x^2+a)+2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)
^(1/2))*A*c-1/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*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*(B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60384, size = 1026, normalized size = 10.91 \begin{align*} \left [\frac{2 \, B a b^{2} - A b^{3} +{\left (B b^{3} + 8 \, A a c^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{2} +{\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + B a b - 2 \, A a c +{\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c -{\left (2 \, c x^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - 4 \,{\left (2 \, B a^{2} - A a b\right )} c}{2 \,{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}, \frac{2 \, B a b^{2} - A b^{3} +{\left (B b^{3} + 8 \, A a c^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{2} - 2 \,{\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + B a b - 2 \, A a c +{\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{{\left (2 \, c x^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - 4 \,{\left (2 \, B a^{2} - A a b\right )} c}{2 \,{\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} +{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} +{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*B*a*b^2 - A*b^3 + (B*b^3 + 8*A*a*c^2 - 2*(2*B*a*b + A*b^2)*c)*x^2 + ((B*b*c - 2*A*c^2)*x^4 + B*a*b - 2
*A*a*c + (B*b^2 - 2*A*b*c)*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)*sqr
t(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) - 4*(2*B*a^2 - A*a*b)*c)/(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2), 1/2*(2*B*a*b^2 - A*b^3 + (B*b^3 + 8*A*a*c^
2 - 2*(2*B*a*b + A*b^2)*c)*x^2 - 2*((B*b*c - 2*A*c^2)*x^4 + B*a*b - 2*A*a*c + (B*b^2 - 2*A*b*c)*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)) - 4*(2*B*a^2 - A*a*b)*c)/(a*b^4 - 8*a^2*b^2*
c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2)]

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Sympy [B]  time = 3.87652, size = 374, normalized size = 3.98 \begin{align*} \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) \log{\left (x^{2} + \frac{- 2 A b c + B b^{2} - 16 a^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) + 8 a b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) - b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right )}{- 4 A c^{2} + 2 B b c} \right )}}{2} - \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) \log{\left (x^{2} + \frac{- 2 A b c + B b^{2} + 16 a^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) - 8 a b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right ) + b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- 2 A c + B b\right )}{- 4 A c^{2} + 2 B b c} \right )}}{2} - \frac{- A b + 2 B a + x^{2} \left (- 2 A c + B b\right )}{8 a^{2} c - 2 a b^{2} + x^{4} \left (8 a c^{2} - 2 b^{2} c\right ) + x^{2} \left (8 a b c - 2 b^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c + B*b)*log(x**2 + (-2*A*b*c + B*b**2 - 16*a**2*c**2*sqrt(-1/(4*a*c - b**2)*
*3)*(-2*A*c + B*b) + 8*a*b**2*c*sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c + B*b) - b**4*sqrt(-1/(4*a*c - b**2)**3)*(-
2*A*c + B*b))/(-4*A*c**2 + 2*B*b*c))/2 - sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c + B*b)*log(x**2 + (-2*A*b*c + B*b*
*2 + 16*a**2*c**2*sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c + B*b) - 8*a*b**2*c*sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c +
B*b) + b**4*sqrt(-1/(4*a*c - b**2)**3)*(-2*A*c + B*b))/(-4*A*c**2 + 2*B*b*c))/2 - (-A*b + 2*B*a + x**2*(-2*A*c
+ B*b))/(8*a**2*c - 2*a*b**2 + x**4*(8*a*c**2 - 2*b**2*c) + x**2*(8*a*b*c - 2*b**3))

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Giac [A]  time = 20.2415, size = 138, normalized size = 1.47 \begin{align*} \frac{{\left (B b - 2 \, A c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{B b x^{2} - 2 \, A c x^{2} + 2 \, B a - A b}{2 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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