3.436 \(\int \frac{1}{(c+\frac{a}{x^2}+\frac{b}{x})^3 x^5} \, dx\)

Optimal. Leaf size=103 \[ \frac{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 b c \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

(2*a + b*x)/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) - (3*b*(b + 2*c*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) +
(6*b*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 = 0.0412466, antiderivative size = 103, 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, 638, 614, 618, 206} \[ \frac{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 b c \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^5),x]

[Out]

(2*a + b*x)/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) - (3*b*(b + 2*c*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) +
(6*b*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 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 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^5} \, dx &=\int \frac{x}{\left (a+b x+c x^2\right )^3} \, dx\\ &=\frac{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{(3 b) \int \frac{1}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=\frac{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{(3 b c) \int \frac{1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=\frac{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{(6 b c) \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{2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{6 b c \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 = 0.103252, size = 102, normalized size = 0.99 \[ \frac{\frac{\left (b^2-4 a c\right ) (2 a+b x)}{(a+x (b+c x))^2}-\frac{12 b c \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}-\frac{3 b (b+2 c x)}{a+x (b+c x)}}{2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^5),x]

[Out]

(((b^2 - 4*a*c)*(2*a + b*x))/(a + x*(b + c*x))^2 - (3*b*(b + 2*c*x))/(a + x*(b + c*x)) - (12*b*c*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 = 0.003, size = 130, normalized size = 1.3 \begin{align*}{\frac{-bx-2\,a}{ \left ( 8\,ac-2\,{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{2}}}-3\,{\frac{bcx}{ \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{3\,{b}^{2}}{2\, \left ( 4\,ac-{b}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-6\,{\frac{bc}{ \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^5,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.79401, size = 1692, normalized size = 16.43 \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^5,x, algorithm="fricas")

[Out]

[-1/2*(a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9*(b^4*c - 4*a*b^2*c^2)*x^2 - 6*(b*c^3
*x^4 + 2*b^2*c^2*x^3 + 2*a*b^2*c*x + a^2*b*c + (b^3*c + 2*a*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*(b^5 + a*b^3*c - 20*a^2*b*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*(a*b^4 + 4*a
^2*b^2*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9*(b^4*c - 4*a*b^2*c^2)*x^2 - 12*(b*c^3*x^4 + 2*b^2*c^2*
x^3 + 2*a*b^2*c*x + a^2*b*c + (b^3*c + 2*a*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*(b^5 + a*b^3*c - 20*a^2*b*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)]

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Sympy [B]  time = 1.35848, size = 479, normalized size = 4.65 \begin{align*} 3 b c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{- 192 a^{3} b c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{2} b^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a b^{5} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{7} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} - 3 b c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} \log{\left (x + \frac{192 a^{3} b c^{4} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{2} b^{3} c^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a b^{5} c^{2} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} - 3 b^{7} c \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} - \frac{8 a^{2} c + a b^{2} + 9 b^{2} c x^{2} + 6 b c^{2} x^{3} + x \left (10 a b c + 2 b^{3}\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**5,x)

[Out]

3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) + 144*a**2*b**3*c**3*sqr
t(-1/(4*a*c - b**2)**5) - 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) + 3*b**7*c*sqrt(-1/(4*a*c - b**2)**5) + 3*
b**2*c)/(6*b*c**2)) - 3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) - 1
44*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2)**5) + 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) - 3*b**7*c*sqrt(-1/(4
*a*c - b**2)**5) + 3*b**2*c)/(6*b*c**2)) - (8*a**2*c + a*b**2 + 9*b**2*c*x**2 + 6*b*c**2*x**3 + x*(10*a*b*c +
2*b**3))/(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 = 1.14007, size = 182, normalized size = 1.77 \begin{align*} -\frac{6 \, b c \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 \, b c^{2} x^{3} + 9 \, b^{2} c x^{2} + 2 \, b^{3} x + 10 \, a b c x + a b^{2} + 8 \, a^{2} 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^5,x, algorithm="giac")

[Out]

-6*b*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*(6*b*c
^2*x^3 + 9*b^2*c*x^2 + 2*b^3*x + 10*a*b*c*x + a*b^2 + 8*a^2*c)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x +
a)^2)