3.5.26 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^2 x^3} \, dx\) [426]

Optimal. Leaf size=66 \[ \frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)-2*b*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1368, 652, 632, 212} \begin {gather*} \frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^2*x^3),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 652

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), 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 1368

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 69, normalized size = 1.05 \begin {gather*} \frac {2 a+b x}{\left (b^2-4 a c\right ) (a+x (b+c x))}-\frac {2 b \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 70, normalized size = 1.06

method	result	size
default	\(\frac {-b x -2 a}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}-\frac {2 b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\)	\(70\)
risch	\(\frac {-\frac {b x}{4 a c -b^{2}}-\frac {2 a}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {b \ln \left (\left (-8 c^{2} a +2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right )}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {b \ln \left (\left (8 c^{2} a -2 b^{2} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right )}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

(-b*x-2*a)/(4*a*c-b^2)/(c*x^2+b*x+a)-2*b/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (62) = 124\).
time = 0.35, size = 338, normalized size = 5.12 \begin {gather*} \left [\frac {2 \, a b^{2} - 8 \, a^{2} c - {\left (b c x^{2} + b^{2} x + a b\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (b^{3} - 4 \, a b c\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}, \frac {2 \, a b^{2} - 8 \, a^{2} c - 2 \, {\left (b c x^{2} + b^{2} x + a b\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (b^{3} - 4 \, a b c\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a*b^2 - 8*a^2*c - (b*c*x^2 + b^2*x + a*b)*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)) + (b^3 - 4*a*b*c)*x)/(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c -
8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x), (2*a*b^2 - 8*a^2*c - 2*(b*c*x^2 + b^2*x
+ a*b)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (b^3 - 4*a*b*c)*x)/(a*b^4 -
8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (60) = 120\).
time = 0.29, size = 253, normalized size = 3.83 \begin {gather*} b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {- 16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )} - b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )} + \frac {- 2 a - b x}{4 a^{2} c - a b^{2} + x^{2} \cdot \left (4 a c^{2} - b^{2} c\right ) + x \left (4 a b c - b^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(-1/(4*a*c - b**2)**3)*log(x + (-16*a**2*b*c**2*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**3*c*sqrt(-1/(4*a*c -
b**2)**3) - b**5*sqrt(-1/(4*a*c - b**2)**3) + b**2)/(2*b*c)) - b*sqrt(-1/(4*a*c - b**2)**3)*log(x + (16*a**2*
b*c**2*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**3*c*sqrt(-1/(4*a*c - b**2)**3) + b**5*sqrt(-1/(4*a*c - b**2)**3) +
b**2)/(2*b*c)) + (-2*a - b*x)/(4*a**2*c - a*b**2 + x**2*(4*a*c**2 - b**2*c) + x*(4*a*b*c - b**3))

________________________________________________________________________________________

Giac [A]
time = 3.43, size = 76, normalized size = 1.15 \begin {gather*} \frac {2 \, b \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {b x + 2 \, a}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) + (b*x + 2*a)/((c*x^2 + b*x + a)
*(b^2 - 4*a*c))

________________________________________________________________________________________

Mupad [B]
time = 1.37, size = 110, normalized size = 1.67 \begin {gather*} -\frac {\frac {2\,a}{4\,a\,c-b^2}+\frac {b\,x}{4\,a\,c-b^2}}{c\,x^2+b\,x+a}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (\frac {b^2}{{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {2\,b\,c\,x}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )\,\left (4\,a\,c-b^2\right )}{b}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(c + a/x^2 + b/x)^2),x)

[Out]

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

________________________________________________________________________________________