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

Optimal. Leaf size=107 \[ -\frac{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 a b \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

-(x^3*(b + 2*c*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*b*x*(2*a + b*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c
*x^2)) + (6*a*b*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 = 0.0500256, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {1354, 728, 722, 618, 206} \[ -\frac{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 a b \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^3),x]

[Out]

-(x^3*(b + 2*c*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*b*x*(2*a + b*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c
*x^2)) + (6*a*b*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 728

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

Rule 722

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 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*(2*p + 3)*(c*d
^2 - b*d*e + a*e^2))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ
[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 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{1}{\left (c+\frac{a}{x^2}+\frac{b}{x}\right )^3 x^3} \, dx &=\int \frac{x^3}{\left (a+b x+c x^2\right )^3} \, dx\\ &=-\frac{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{(3 b) \int \frac{x^2}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{(3 a b) \int \frac{1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{(6 a b) \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{x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 a b \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 = 0.213499, size = 126, normalized size = 1.18 \[ -\frac{a^2 \left (b^2+10 b c x+16 c^2 x^2\right )+8 a^3 c+a b x \left (2 b^2+b c x+6 c^2 x^2\right )+b^4 x^2}{2 c \left (b^2-4 a c\right )^2 (a+x (b+c x))^2}-\frac{6 a b \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(8*a^3*c + b^4*x^2 + a*b*x*(2*b^2 + b*c*x + 6*c^2*x^2) + a^2*(b^2 + 10*b*c*x + 16*c^2*x^2))/(2*c*(b^2 - 4*a*c
)^2*(a + x*(b + c*x))^2) - (6*a*b*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2)

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Maple [B]  time = 0.009, size = 223, normalized size = 2.1 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ( -3\,{\frac{abc{x}^{3}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}-{\frac{ \left ( 16\,{a}^{2}{c}^{2}+a{b}^{2}c+{b}^{4} \right ){x}^{2}}{2\,c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-{\frac{ \left ( 5\,ac+{b}^{2} \right ) abx}{c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }}-{\frac{{a}^{2} \left ( 8\,ac+{b}^{2} \right ) }{2\,c \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) }} \right ) }-6\,{\frac{ab}{ \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^3,x)

[Out]

(-3*a*b*c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3-1/2*(16*a^2*c^2+a*b^2*c+b^4)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-(5*a*c+
b^2)*a*b/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x-1/2*a^2*(8*a*c+b^2)/c/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2-6*a*
b/(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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.83539, size = 1848, normalized size = 17.27 \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^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.51743, size = 510, normalized size = 4.77 \begin{align*} 3 a b \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{- 192 a^{4} b c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{3} b^{3} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a^{2} b^{5} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{7} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )} - 3 a b \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{192 a^{4} b c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{3} b^{3} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a^{2} b^{5} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 3 a b^{7} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )} - \frac{8 a^{3} c + a^{2} b^{2} + 6 a b c^{2} x^{3} + x^{2} \left (16 a^{2} c^{2} + a b^{2} c + b^{4}\right ) + x \left (10 a^{2} b c + 2 a b^{3}\right )}{32 a^{4} c^{3} - 16 a^{3} b^{2} c^{2} + 2 a^{2} b^{4} c + x^{4} \left (32 a^{2} c^{5} - 16 a b^{2} c^{4} + 2 b^{4} c^{3}\right ) + x^{3} \left (64 a^{2} b c^{4} - 32 a b^{3} c^{3} + 4 b^{5} c^{2}\right ) + x^{2} \left (64 a^{3} c^{4} - 12 a b^{4} c^{2} + 2 b^{6} c\right ) + x \left (64 a^{3} b c^{3} - 32 a^{2} b^{3} c^{2} + 4 a b^{5} c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)**5) + 144*a**3*b**3*c**2*sqr
t(-1/(4*a*c - b**2)**5) - 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) + 3*a*b**7*sqrt(-1/(4*a*c - b**2)**5) + 3*
a*b**2)/(6*a*b*c)) - 3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x + (192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)**5) - 14
4*a**3*b**3*c**2*sqrt(-1/(4*a*c - b**2)**5) + 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) - 3*a*b**7*sqrt(-1/(4*
a*c - b**2)**5) + 3*a*b**2)/(6*a*b*c)) - (8*a**3*c + a**2*b**2 + 6*a*b*c**2*x**3 + x**2*(16*a**2*c**2 + a*b**2
*c + b**4) + x*(10*a**2*b*c + 2*a*b**3))/(32*a**4*c**3 - 16*a**3*b**2*c**2 + 2*a**2*b**4*c + x**4*(32*a**2*c**
5 - 16*a*b**2*c**4 + 2*b**4*c**3) + x**3*(64*a**2*b*c**4 - 32*a*b**3*c**3 + 4*b**5*c**2) + x**2*(64*a**3*c**4
- 12*a*b**4*c**2 + 2*b**6*c) + x*(64*a**3*b*c**3 - 32*a**2*b**3*c**2 + 4*a*b**5*c))

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Giac [A]  time = 1.12866, size = 220, normalized size = 2.06 \begin{align*} -\frac{6 \, a b \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{6 \, a b c^{2} x^{3} + b^{4} x^{2} + a b^{2} c x^{2} + 16 \, a^{2} c^{2} x^{2} + 2 \, a b^{3} x + 10 \, a^{2} b c x + a^{2} b^{2} + 8 \, a^{3} c}{2 \,{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\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^3,x, algorithm="giac")

[Out]

-6*a*b*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*(6*a*b
*c^2*x^3 + b^4*x^2 + a*b^2*c*x^2 + 16*a^2*c^2*x^2 + 2*a*b^3*x + 10*a^2*b*c*x + a^2*b^2 + 8*a^3*c)/((b^4*c - 8*
a*b^2*c^2 + 16*a^2*c^3)*(c*x^2 + b*x + a)^2)