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

Optimal. Leaf size=107 \[ -\frac{x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a B) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.113441, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1251, 777, 618, 206} \[ -\frac{x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a B) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1251

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

Rule 777

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

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^3 \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{x (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{(A b-2 a B) \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 B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a B) \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 B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a B) \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 = 0.0869724, size = 111, normalized size = 1.04 \[ \frac{-2 a c \left (A+B x^2\right )+a b B+b x^2 (b B-A c)}{2 c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a B) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 158, normalized size = 1.5 \begin{align*}{\frac{1}{2\,c{x}^{4}+2\,b{x}^{2}+2\,a} \left ( -{\frac{ \left ( Abc+2\,aBc-{b}^{2}B \right ){x}^{2}}{c \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{a \left ( 2\,Ac-bB \right ) }{c \left ( 4\,ac-{b}^{2} \right ) }} \right ) }-{Ab\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{aB}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.52872, size = 1131, normalized size = 10.57 \begin{align*} \left [-\frac{B a b^{3} + 8 \, A a^{2} c^{2} +{\left (B b^{4} + 4 \,{\left (2 \, B a^{2} + A a b\right )} c^{2} -{\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} x^{2} -{\left ({\left (2 \, B a - A b\right )} c^{2} x^{4} +{\left (2 \, B a b - A b^{2}\right )} c x^{2} +{\left (2 \, B a^{2} - A a b\right )} c\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 ) - 2 \,{\left (2 \, B a^{2} b + A a b^{2}\right )} c}{2 \,{\left (a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} +{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{2}\right )}}, -\frac{B a b^{3} + 8 \, A a^{2} c^{2} +{\left (B b^{4} + 4 \,{\left (2 \, B a^{2} + A a b\right )} c^{2} -{\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} x^{2} - 2 \,{\left ({\left (2 \, B a - A b\right )} c^{2} x^{4} +{\left (2 \, B a b - A b^{2}\right )} c x^{2} +{\left (2 \, B a^{2} - A a b\right )} c\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 ) - 2 \,{\left (2 \, B a^{2} b + A a b^{2}\right )} c}{2 \,{\left (a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} +{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(B*a*b^3 + 8*A*a^2*c^2 + (B*b^4 + 4*(2*B*a^2 + A*a*b)*c^2 - (6*B*a*b^2 + A*b^3)*c)*x^2 - ((2*B*a - A*b)*
c^2*x^4 + (2*B*a*b - A*b^2)*c*x^2 + (2*B*a^2 - A*a*b)*c)*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 + A*a*b^2)*c)/(a*b^4*c - 8*a^2*b^
2*c^2 + 16*a^3*c^3 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^2), -1/
2*(B*a*b^3 + 8*A*a^2*c^2 + (B*b^4 + 4*(2*B*a^2 + A*a*b)*c^2 - (6*B*a*b^2 + A*b^3)*c)*x^2 - 2*((2*B*a - A*b)*c^
2*x^4 + (2*B*a*b - A*b^2)*c*x^2 + (2*B*a^2 - A*a*b)*c)*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 + A*a*b^2)*c)/(a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3 + (b^4*c^2 - 8*a*b^2*c^
3 + 16*a^2*c^4)*x^4 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^2)]

________________________________________________________________________________________

Sympy [B]  time = 6.08602, size = 394, normalized size = 3.68 \begin{align*} - \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) \log{\left (x^{2} + \frac{- A b^{2} + 2 B a b - 16 a^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) + 8 a b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) - b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right )}{- 2 A b c + 4 B a c} \right )}}{2} + \frac{\sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) \log{\left (x^{2} + \frac{- A b^{2} + 2 B a b + 16 a^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) - 8 a b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right ) + b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (- A b + 2 B a\right )}{- 2 A b c + 4 B a c} \right )}}{2} - \frac{2 A a c - B a b + x^{2} \left (A b c + 2 B a c - B b^{2}\right )}{8 a^{2} c^{2} - 2 a b^{2} c + x^{4} \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{2} \left (8 a b c^{2} - 2 b^{3} c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 19.6521, size = 162, normalized size = 1.51 \begin{align*} -\frac{{\left (2 \, B a - A b\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^{2} x^{2} - 2 \, B a c x^{2} - A b c x^{2} + B a b - 2 \, A a c}{2 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (b^{2} c - 4 \, a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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