3.425 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^2 x^2} \, dx\)

Optimal. Leaf size=71 \[ \frac{\frac{2 a}{x}+b}{\left (b^2-4 a c\right ) \left (\frac{a}{x^2}+\frac{b}{x}+c\right )}-\frac{4 a \tanh ^{-1}\left (\frac{\frac{2 a}{x}+b}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0421379, antiderivative size = 71, 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 = {1352, 614, 618, 206} \[ \frac{\frac{2 a}{x}+b}{\left (b^2-4 a c\right ) \left (\frac{a}{x^2}+\frac{b}{x}+c\right )}-\frac{4 a \tanh ^{-1}\left (\frac{\frac{2 a}{x}+b}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^2*x^2),x]

[Out]

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

Rule 1352

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 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 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{1}{\left (c+\frac{a}{x^2}+\frac{b}{x}\right )^2 x^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (c+b x+a x^2\right )^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b+\frac{2 a}{x}}{\left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{c+b x+a x^2} \, dx,x,\frac{1}{x}\right )}{b^2-4 a c}\\ &=\frac{b+\frac{2 a}{x}}{\left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}-\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+\frac{2 a}{x}\right )}{b^2-4 a c}\\ &=\frac{b+\frac{2 a}{x}}{\left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}-\frac{4 a \tanh ^{-1}\left (\frac{b+\frac{2 a}{x}}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0969133, size = 81, normalized size = 1.14 \[ \frac{a (b-2 c x)+b^2 x}{c \left (4 a c-b^2\right ) (a+x (b+c x))}+\frac{4 a \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^2*x^2),x]

[Out]

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

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Maple [A]  time = 0.007, size = 97, normalized size = 1.4 \begin{align*}{\frac{1}{c{x}^{2}+bx+a} \left ( -{\frac{ \left ( 2\,ac-{b}^{2} \right ) x}{c \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{ab}{c \left ( 4\,ac-{b}^{2} \right ) }} \right ) }+4\,{\frac{a}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\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(1/(c+a/x^2+b/x)^2/x^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.80585, size = 826, normalized size = 11.63 \begin{align*} \left [-\frac{a b^{3} - 4 \, a^{2} b c + 2 \,{\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt{b^{2} - 4 \, a c}{\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) +{\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x}{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^{2} +{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}, -\frac{a b^{3} - 4 \, a^{2} b c - 4 \,{\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) +{\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x}{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^{2} +{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^2/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.752775, size = 280, normalized size = 3.94 \begin{align*} - 2 a \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{- 32 a^{3} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 16 a^{2} b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 2 a b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b}{4 a c} \right )} + 2 a \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{32 a^{3} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 16 a^{2} b^{2} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b}{4 a c} \right )} - \frac{- a b + x \left (2 a c - b^{2}\right )}{4 a^{2} c^{2} - a b^{2} c + x^{2} \left (4 a c^{3} - b^{2} c^{2}\right ) + x \left (4 a b c^{2} - b^{3} c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**2/x**2,x)

[Out]

-2*a*sqrt(-1/(4*a*c - b**2)**3)*log(x + (-32*a**3*c**2*sqrt(-1/(4*a*c - b**2)**3) + 16*a**2*b**2*c*sqrt(-1/(4*
a*c - b**2)**3) - 2*a*b**4*sqrt(-1/(4*a*c - b**2)**3) + 2*a*b)/(4*a*c)) + 2*a*sqrt(-1/(4*a*c - b**2)**3)*log(x
+ (32*a**3*c**2*sqrt(-1/(4*a*c - b**2)**3) - 16*a**2*b**2*c*sqrt(-1/(4*a*c - b**2)**3) + 2*a*b**4*sqrt(-1/(4*
a*c - b**2)**3) + 2*a*b)/(4*a*c)) - (-a*b + x*(2*a*c - b**2))/(4*a**2*c**2 - a*b**2*c + x**2*(4*a*c**3 - b**2*
c**2) + x*(4*a*b*c**2 - b**3*c))

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Giac [A]  time = 1.11492, size = 119, normalized size = 1.68 \begin{align*} -\frac{4 \, a \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{b^{2} x - 2 \, a c x + a b}{{\left (b^{2} c - 4 \, a c^{2}\right )}{\left (c x^{2} + b x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^2/x^2,x, algorithm="giac")

[Out]

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