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

Optimal. Leaf size=75 \[ \frac{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

(2*a + b*x^2)/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (b*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 = 0.0612293, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1114, 638, 618, 206} \[ \frac{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

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^3}{\left (a+b x^2+c x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{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{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{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{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{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 = 0.0638087, size = 79, normalized size = 1.05 \[ \frac{2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.171, size = 77, normalized size = 1. \begin{align*}{\frac{-b{x}^{2}-2\,a}{ \left ( 8\,ac-2\,{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{2}+a \right ) }}-{b\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^3/(c*x^4+b*x^2+a)^2,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.74573, size = 778, normalized size = 10.37 \begin{align*} \left [\frac{2 \, a b^{2} - 8 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} x^{2} -{\left (b c x^{4} + b^{2} x^{2} + a b\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 (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 \, a b^{2} - 8 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} x^{2} - 2 \,{\left (b c x^{4} + b^{2} x^{2} + a b\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 (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^3/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/2*(2*a*b^2 - 8*a^2*c + (b^3 - 4*a*b*c)*x^2 - (b*c*x^4 + b^2*x^2 + a*b)*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)))/(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*a*b^2 - 8*a^2*c +
(b^3 - 4*a*b*c)*x^2 - 2*(b*c*x^4 + b^2*x^2 + a*b)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)
/(b^2 - 4*a*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 = 1.74307, size = 267, normalized size = 3.56 \begin{align*} \frac{b \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x^{2} + \frac{- 16 a^{2} b c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )}}{2} - \frac{b \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x^{2} + \frac{16 a^{2} b c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )}}{2} - \frac{2 a + b x^{2}}{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**3/(c*x**4+b*x**2+a)**2,x)

[Out]

b*sqrt(-1/(4*a*c - b**2)**3)*log(x**2 + (-16*a**2*b*c**2*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**3*c*sqrt(-1/(4*a*
c - b**2)**3) - b**5*sqrt(-1/(4*a*c - b**2)**3) + b**2)/(2*b*c))/2 - b*sqrt(-1/(4*a*c - b**2)**3)*log(x**2 + (
16*a**2*b*c**2*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**3*c*sqrt(-1/(4*a*c - b**2)**3) + b**5*sqrt(-1/(4*a*c - b**2
)**3) + b**2)/(2*b*c))/2 - (2*a + b*x**2)/(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 = 21.0379, size = 111, normalized size = 1.48 \begin{align*} \frac{b \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 x^{2} + 2 \, a}{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^3/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

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