3.433 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^3 x^2} \, dx\)

Optimal. Leaf size=111 \[ \frac{12 a^2 \tanh ^{-1}\left (\frac{\frac{2 a}{x}+b}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac{3 a \left (\frac{2 a}{x}+b\right )}{\left (b^2-4 a c\right )^2 \left (\frac{a}{x^2}+\frac{b}{x}+c\right )}+\frac{\frac{2 a}{x}+b}{2 \left (b^2-4 a c\right ) \left (\frac{a}{x^2}+\frac{b}{x}+c\right )^2} \]

[Out]

(b + (2*a)/x)/(2*(b^2 - 4*a*c)*(c + a/x^2 + b/x)^2) - (3*a*(b + (2*a)/x))/((b^2 - 4*a*c)^2*(c + a/x^2 + b/x))
+ (12*a^2*ArcTanh[(b + (2*a)/x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.0665031, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1352, 614, 618, 206} \[ \frac{12 a^2 \tanh ^{-1}\left (\frac{\frac{2 a}{x}+b}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac{3 a \left (\frac{2 a}{x}+b\right )}{\left (b^2-4 a c\right )^2 \left (\frac{a}{x^2}+\frac{b}{x}+c\right )}+\frac{\frac{2 a}{x}+b}{2 \left (b^2-4 a c\right ) \left (\frac{a}{x^2}+\frac{b}{x}+c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^2),x]

[Out]

(b + (2*a)/x)/(2*(b^2 - 4*a*c)*(c + a/x^2 + b/x)^2) - (3*a*(b + (2*a)/x))/((b^2 - 4*a*c)^2*(c + a/x^2 + b/x))
+ (12*a^2*ArcTanh[(b + (2*a)/x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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 )^3 x^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (c+b x+a x^2\right )^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b+\frac{2 a}{x}}{2 \left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )^2}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{\left (c+b x+a x^2\right )^2} \, dx,x,\frac{1}{x}\right )}{b^2-4 a c}\\ &=\frac{b+\frac{2 a}{x}}{2 \left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )^2}-\frac{3 a \left (b+\frac{2 a}{x}\right )}{\left (b^2-4 a c\right )^2 \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{c+b x+a x^2} \, dx,x,\frac{1}{x}\right )}{\left (b^2-4 a c\right )^2}\\ &=\frac{b+\frac{2 a}{x}}{2 \left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )^2}-\frac{3 a \left (b+\frac{2 a}{x}\right )}{\left (b^2-4 a c\right )^2 \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}+\frac{\left (12 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+\frac{2 a}{x}\right )}{\left (b^2-4 a c\right )^2}\\ &=\frac{b+\frac{2 a}{x}}{2 \left (b^2-4 a c\right ) \left (c+\frac{a}{x^2}+\frac{b}{x}\right )^2}-\frac{3 a \left (b+\frac{2 a}{x}\right )}{\left (b^2-4 a c\right )^2 \left (c+\frac{a}{x^2}+\frac{b}{x}\right )}+\frac{12 a^2 \tanh ^{-1}\left (\frac{b+\frac{2 a}{x}}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.182434, size = 174, normalized size = 1.57 \[ \frac{1}{2} \left (\frac{a^2 c (2 c x-3 b)+a b^2 (b-4 c x)+b^4 x}{c^3 \left (4 a c-b^2\right ) (a+x (b+c x))^2}+\frac{22 a^2 b c^2-20 a^2 c^3 x+16 a b^2 c^2 x-8 a b^3 c-2 b^4 c x+b^5}{c^3 \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac{24 a^2 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^2),x]

[Out]

((b^5 - 8*a*b^3*c + 22*a^2*b*c^2 - 2*b^4*c*x + 16*a*b^2*c^2*x - 20*a^2*c^3*x)/(c^3*(b^2 - 4*a*c)^2*(a + x*(b +
c*x))) + (b^4*x + a*b^2*(b - 4*c*x) + a^2*c*(-3*b + 2*c*x))/(c^3*(-b^2 + 4*a*c)*(a + x*(b + c*x))^2) + (24*a^
2*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/2

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Maple [B]  time = 0.012, size = 260, normalized size = 2.3 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ( -{\frac{ \left ( 10\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ){x}^{3}}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{b \left ( 2\,{a}^{2}{c}^{2}+8\,a{b}^{2}c-{b}^{4} \right ){x}^{2}}{2\,{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-{\frac{a \left ( 6\,{a}^{2}{c}^{2}-10\,a{b}^{2}c+{b}^{4} \right ) x}{{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}+{\frac{{a}^{2}b \left ( 10\,ac-{b}^{2} \right ) }{2\,{c}^{2} \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }} \right ) }+12\,{\frac{{a}^{2}}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{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)^3/x^2,x)

[Out]

(-1/c*(10*a^2*c^2-8*a*b^2*c+b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+1/2*b*(2*a^2*c^2+8*a*b^2*c-b^4)/c^2/(16*a^2*c^
2-8*a*b^2*c+b^4)*x^2-a*(6*a^2*c^2-10*a*b^2*c+b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/c^2*x+1/2*a^2*b*(10*a*c-b^2)/c^2/
(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2+12*a^2/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*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)^3/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.05225, size = 2006, normalized size = 18.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^2,x, algorithm="fricas")

[Out]

[-1/2*(a^2*b^5 - 14*a^3*b^3*c + 40*a^4*b*c^2 + 2*(b^6*c - 12*a*b^4*c^2 + 42*a^2*b^2*c^3 - 40*a^3*c^4)*x^3 + (b
^7 - 12*a*b^5*c + 30*a^2*b^3*c^2 + 8*a^3*b*c^3)*x^2 - 12*(a^2*c^4*x^4 + 2*a^2*b*c^3*x^3 + 2*a^3*b*c^2*x + a^4*
c^2 + (a^2*b^2*c^2 + 2*a^3*c^3)*x^2)*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)) + 2*(a*b^6 - 14*a^2*b^4*c + 46*a^3*b^2*c^2 - 24*a^4*c^3)*x)/(a^2*b^6*c^2 -
12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5 + (b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*x^4 + 2
*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*a^3*b*c^6)*x^3 + (b^8*c^2 - 10*a*b^6*c^3 + 24*a^2*b^4*c^4 + 32*
a^3*b^2*c^5 - 128*a^4*c^6)*x^2 + 2*(a*b^7*c^2 - 12*a^2*b^5*c^3 + 48*a^3*b^3*c^4 - 64*a^4*b*c^5)*x), -1/2*(a^2*
b^5 - 14*a^3*b^3*c + 40*a^4*b*c^2 + 2*(b^6*c - 12*a*b^4*c^2 + 42*a^2*b^2*c^3 - 40*a^3*c^4)*x^3 + (b^7 - 12*a*b
^5*c + 30*a^2*b^3*c^2 + 8*a^3*b*c^3)*x^2 + 24*(a^2*c^4*x^4 + 2*a^2*b*c^3*x^3 + 2*a^3*b*c^2*x + a^4*c^2 + (a^2*
b^2*c^2 + 2*a^3*c^3)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(a*b^6
- 14*a^2*b^4*c + 46*a^3*b^2*c^2 - 24*a^4*c^3)*x)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5 +
(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*x^4 + 2*(b^7*c^3 - 12*a*b^5*c^4 + 48*a^2*b^3*c^5 - 64*
a^3*b*c^6)*x^3 + (b^8*c^2 - 10*a*b^6*c^3 + 24*a^2*b^4*c^4 + 32*a^3*b^2*c^5 - 128*a^4*c^6)*x^2 + 2*(a*b^7*c^2 -
12*a^2*b^5*c^3 + 48*a^3*b^3*c^4 - 64*a^4*b*c^5)*x)]

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Sympy [B]  time = 1.7504, size = 547, normalized size = 4.93 \begin{align*} - 6 a^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{- 384 a^{5} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 288 a^{4} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 72 a^{3} b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 a^{2} b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 a^{2} b}{12 a^{2} c} \right )} + 6 a^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{384 a^{5} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 288 a^{4} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 72 a^{3} b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 6 a^{2} b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 a^{2} b}{12 a^{2} c} \right )} - \frac{- 10 a^{3} b c + a^{2} b^{3} + x^{3} \left (20 a^{2} c^{3} - 16 a b^{2} c^{2} + 2 b^{4} c\right ) + x^{2} \left (- 2 a^{2} b c^{2} - 8 a b^{3} c + b^{5}\right ) + x \left (12 a^{3} c^{2} - 20 a^{2} b^{2} c + 2 a b^{4}\right )}{32 a^{4} c^{4} - 16 a^{3} b^{2} c^{3} + 2 a^{2} b^{4} c^{2} + x^{4} \left (32 a^{2} c^{6} - 16 a b^{2} c^{5} + 2 b^{4} c^{4}\right ) + x^{3} \left (64 a^{2} b c^{5} - 32 a b^{3} c^{4} + 4 b^{5} c^{3}\right ) + x^{2} \left (64 a^{3} c^{5} - 12 a b^{4} c^{3} + 2 b^{6} c^{2}\right ) + x \left (64 a^{3} b c^{4} - 32 a^{2} b^{3} c^{3} + 4 a b^{5} c^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**3/x**2,x)

[Out]

-6*a**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-384*a**5*c**3*sqrt(-1/(4*a*c - b**2)**5) + 288*a**4*b**2*c**2*sqr
t(-1/(4*a*c - b**2)**5) - 72*a**3*b**4*c*sqrt(-1/(4*a*c - b**2)**5) + 6*a**2*b**6*sqrt(-1/(4*a*c - b**2)**5) +
6*a**2*b)/(12*a**2*c)) + 6*a**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (384*a**5*c**3*sqrt(-1/(4*a*c - b**2)**5)
- 288*a**4*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5) + 72*a**3*b**4*c*sqrt(-1/(4*a*c - b**2)**5) - 6*a**2*b**6*sqrt
(-1/(4*a*c - b**2)**5) + 6*a**2*b)/(12*a**2*c)) - (-10*a**3*b*c + a**2*b**3 + x**3*(20*a**2*c**3 - 16*a*b**2*c
**2 + 2*b**4*c) + x**2*(-2*a**2*b*c**2 - 8*a*b**3*c + b**5) + x*(12*a**3*c**2 - 20*a**2*b**2*c + 2*a*b**4))/(3
2*a**4*c**4 - 16*a**3*b**2*c**3 + 2*a**2*b**4*c**2 + x**4*(32*a**2*c**6 - 16*a*b**2*c**5 + 2*b**4*c**4) + x**3
*(64*a**2*b*c**5 - 32*a*b**3*c**4 + 4*b**5*c**3) + x**2*(64*a**3*c**5 - 12*a*b**4*c**3 + 2*b**6*c**2) + x*(64*
a**3*b*c**4 - 32*a**2*b**3*c**3 + 4*a*b**5*c**2))

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Giac [A]  time = 1.15053, size = 273, normalized size = 2.46 \begin{align*} \frac{12 \, a^{2} \arctan \left (\frac{2 \, c x + 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{2 \, b^{4} c x^{3} - 16 \, a b^{2} c^{2} x^{3} + 20 \, a^{2} c^{3} x^{3} + b^{5} x^{2} - 8 \, a b^{3} c x^{2} - 2 \, a^{2} b c^{2} x^{2} + 2 \, a b^{4} x - 20 \, a^{2} b^{2} c x + 12 \, a^{3} c^{2} x + a^{2} b^{3} - 10 \, a^{3} b c}{2 \,{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )}{\left (c x^{2} + b x + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^2,x, algorithm="giac")

[Out]

12*a^2*arctan((2*c*x + 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)) - 1/2*(2*b^4
*c*x^3 - 16*a*b^2*c^2*x^3 + 20*a^2*c^3*x^3 + b^5*x^2 - 8*a*b^3*c*x^2 - 2*a^2*b*c^2*x^2 + 2*a*b^4*x - 20*a^2*b^
2*c*x + 12*a^3*c^2*x + a^2*b^3 - 10*a^3*b*c)/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*(c*x^2 + b*x + a)^2)