3.435 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^3 x^4} \, dx\)

Optimal. Leaf size=115 \[ \frac{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{2 \left (2 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

(x*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*a*b + (b^2 + 2*a*c)*x)/((b^2 - 4*a*c)^2*(a + b*x +
c*x^2)) - (2*(b^2 + 2*a*c)*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 = 0.0682724, antiderivative size = 115, 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, 738, 638, 618, 206} \[ \frac{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{x \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{2 \left (2 a c+b^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^4),x]

[Out]

(x*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*a*b + (b^2 + 2*a*c)*x)/((b^2 - 4*a*c)^2*(a + b*x +
c*x^2)) - (2*(b^2 + 2*a*c)*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 738

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[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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{1}{\left (c+\frac{a}{x^2}+\frac{b}{x}\right )^3 x^4} \, dx &=\int \frac{x^2}{\left (a+b x+c x^2\right )^3} \, dx\\ &=\frac{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{\int \frac{2 a-2 b x}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=\frac{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 a b+\left (b^2+2 a c\right ) x}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\left (b^2+2 a c\right ) \int \frac{1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=\frac{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 a b+\left (b^2+2 a c\right ) x}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{\left (2 \left (b^2+2 a c\right )\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{x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{3 a b+\left (b^2+2 a c\right ) x}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{2 \left (b^2+2 a c\right ) \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 = 0.14782, size = 131, normalized size = 1.14 \[ \frac{1}{2} \left (\frac{\left (2 a c+b^2\right ) (b+2 c x)}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac{a (b-2 c x)+b^2 x}{c \left (4 a c-b^2\right ) (a+x (b+c x))^2}+\frac{4 \left (2 a c+b^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^4),x]

[Out]

((b^2*x + a*(b - 2*c*x))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))^2) + ((b^2 + 2*a*c)*(b + 2*c*x))/(c*(b^2 - 4*a*c)
^2*(a + x*(b + c*x))) + (4*(b^2 + 2*a*c)*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 = 0.009, size = 262, normalized size = 2.3 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ({\frac{c \left ( 2\,ac+{b}^{2} \right ){x}^{3}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+{\frac{3\,b \left ( 2\,ac+{b}^{2} \right ){x}^{2}}{32\,{a}^{2}{c}^{2}-16\,a{b}^{2}c+2\,{b}^{4}}}-{\frac{a \left ( 2\,ac-5\,{b}^{2} \right ) x}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+3\,{\frac{{a}^{2}b}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}} \right ) }+4\,{\frac{ac}{ \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 ) }+2\,{\frac{{b}^{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^4,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.79526, size = 1893, normalized size = 16.46 \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^4,x, algorithm="fricas")

[Out]

[1/2*(6*a^2*b^3 - 24*a^3*b*c + 2*(b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*x^3 + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^2
+ 2*((b^2*c^2 + 2*a*c^3)*x^4 + a^2*b^2 + 2*a^3*c + 2*(b^3*c + 2*a*b*c^2)*x^3 + (b^4 + 4*a*b^2*c + 4*a^2*c^2)*
x^2 + 2*(a*b^3 + 2*a^2*b*c)*x)*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*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)*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*(6*a^2*b^3 - 24*a^3*b*c + 2*(b^4*c - 2*a*b^2*c^2
- 8*a^2*c^3)*x^3 + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^2 - 4*((b^2*c^2 + 2*a*c^3)*x^4 + a^2*b^2 + 2*a^3*c + 2
*(b^3*c + 2*a*b*c^2)*x^3 + (b^4 + 4*a*b^2*c + 4*a^2*c^2)*x^2 + 2*(a*b^3 + 2*a^2*b*c)*x)*sqrt(-b^2 + 4*a*c)*arc
tan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)*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 - 1
28*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 = 1.4503, size = 570, normalized size = 4.96 \begin{align*} - \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) \log{\left (x + \frac{- 64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) - 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 2 a b c + b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )} + \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) \log{\left (x + \frac{64 a^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) - 48 a^{2} b^{2} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 12 a b^{4} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 2 a b c - b^{6} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )} + \frac{6 a^{2} b + x^{3} \left (4 a c^{2} + 2 b^{2} c\right ) + x^{2} \left (6 a b c + 3 b^{3}\right ) + x \left (- 4 a^{2} c + 10 a b^{2}\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**4,x)

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2)*log(x + (-64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) +
48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c +
b**2) + 2*a*b*c + b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c)) + sqrt(-1/(4*a
*c - b**2)**5)*(2*a*c + b**2)*log(x + (64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 48*a**2*b**2*c
**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + 2*a*b*
c - b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c)) + (6*a**2*b + x**3*(4*a*c**2
+ 2*b**2*c) + x**2*(6*a*b*c + 3*b**3) + x*(-4*a**2*c + 10*a*b**2))/(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 = 1.15729, size = 208, normalized size = 1.81 \begin{align*} \frac{2 \,{\left (b^{2} + 2 \, a c\right )} \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^{2} c x^{3} + 4 \, a c^{2} x^{3} + 3 \, b^{3} x^{2} + 6 \, a b c x^{2} + 10 \, a b^{2} x - 4 \, a^{2} c x + 6 \, a^{2} b}{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^4,x, algorithm="giac")

[Out]

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