3.437 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^3 x^6} \, dx\)

Optimal. Leaf size=101 \[ -\frac{12 c^2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac{3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0412279, antiderivative size = 101, 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 = {1354, 614, 618, 206} \[ -\frac{12 c^2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac{3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^6),x]

[Out]

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

Rule 1354

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

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

Mathematica [A]  time = 0.101889, size = 97, normalized size = 0.96 \[ \frac{\frac{24 c^2 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}-\frac{(b+2 c x) \left (-2 c \left (5 a+3 c x^2\right )+b^2-6 b c x\right )}{(a+x (b+c x))^2}}{2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^6),x]

[Out]

(-(((b + 2*c*x)*(b^2 - 6*b*c*x - 2*c*(5*a + 3*c*x^2)))/(a + x*(b + c*x))^2) + (24*c^2*ArcTan[(b + 2*c*x)/Sqrt[
-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c)^2)

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Maple [A]  time = 0.004, size = 129, normalized size = 1.3 \begin{align*}{\frac{2\,cx+b}{ \left ( 8\,ac-2\,{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}+6\,{\frac{{c}^{2}x}{ \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+3\,{\frac{bc}{ \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+12\,{\frac{{c}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) ^{5/2}}\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^6,x)

[Out]

1/2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^2+6*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*x+3*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)*
b+12*c^2/(4*a*c-b^2)^(5/2)*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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.90834, size = 1685, normalized size = 16.68 \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^6,x, algorithm="fricas")

[Out]

[-1/2*(b^5 - 14*a*b^3*c + 40*a^2*b*c^2 - 12*(b^2*c^3 - 4*a*c^4)*x^3 - 18*(b^3*c^2 - 4*a*b*c^3)*x^2 - 12*(c^4*x
^4 + 2*b*c^3*x^3 + 2*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*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)) - 4*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*x)/(a^
2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^
4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3
*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x), -1/2*(b^5 - 14*a*b^
3*c + 40*a^2*b*c^2 - 12*(b^2*c^3 - 4*a*c^4)*x^3 - 18*(b^3*c^2 - 4*a*b*c^3)*x^2 + 24*(c^4*x^4 + 2*b*c^3*x^3 + 2
*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*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)) - 4*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^
6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*
c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c +
48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]

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Sympy [B]  time = 1.37773, size = 474, normalized size = 4.69 \begin{align*} - 6 c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{- 384 a^{3} c^{5} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 288 a^{2} b^{2} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 72 a b^{4} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b^{6} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c^{2}}{12 c^{3}} \right )} + 6 c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{384 a^{3} c^{5} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 288 a^{2} b^{2} c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 72 a b^{4} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 6 b^{6} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c^{2}}{12 c^{3}} \right )} + \frac{10 a b c - b^{3} + 18 b c^{2} x^{2} + 12 c^{3} x^{3} + x \left (20 a c^{2} + 4 b^{2} c\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**3/x**6,x)

[Out]

-6*c**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-384*a**3*c**5*sqrt(-1/(4*a*c - b**2)**5) + 288*a**2*b**2*c**4*sqr
t(-1/(4*a*c - b**2)**5) - 72*a*b**4*c**3*sqrt(-1/(4*a*c - b**2)**5) + 6*b**6*c**2*sqrt(-1/(4*a*c - b**2)**5) +
6*b*c**2)/(12*c**3)) + 6*c**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (384*a**3*c**5*sqrt(-1/(4*a*c - b**2)**5) -
288*a**2*b**2*c**4*sqrt(-1/(4*a*c - b**2)**5) + 72*a*b**4*c**3*sqrt(-1/(4*a*c - b**2)**5) - 6*b**6*c**2*sqrt(-
1/(4*a*c - b**2)**5) + 6*b*c**2)/(12*c**3)) + (10*a*b*c - b**3 + 18*b*c**2*x**2 + 12*c**3*x**3 + x*(20*a*c**2
+ 4*b**2*c))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a**2*b**4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 + 2*b**4*c**2)
+ x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) + x**2*(64*a**3*c**3 - 12*a*b**4*c + 2*b**6) + x*(64*a**3
*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))

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Giac [A]  time = 1.16977, size = 184, normalized size = 1.82 \begin{align*} \frac{12 \, c^{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{12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x + 20 \, a c^{2} x - b^{3} + 10 \, a b c}{2 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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^6,x, algorithm="giac")

[Out]

12*c^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*(12*c^
3*x^3 + 18*b*c^2*x^2 + 4*b^2*c*x + 20*a*c^2*x - b^3 + 10*a*b*c)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x +
a)^2)