3.5.17 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x}) x^2} \, dx\) [417]

Optimal. Leaf size=36 \[ \frac {2 \tanh ^{-1}\left (\frac {b+\frac {2 a}{x}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1366, 632, 212} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\frac {2 a}{x}+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)*x^2),x]

[Out]

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

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 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 38, normalized size = 1.06 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 35, normalized size = 0.97

method	result	size
default	\(\frac {2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\)	\(35\)
risch	\(-\frac {\ln \left (b +2 c x +\sqrt {-4 a c +b^{2}}\right )}{\sqrt {-4 a c +b^{2}}}+\frac {\ln \left (-b -2 c x +\sqrt {-4 a c +b^{2}}\right )}{\sqrt {-4 a c +b^{2}}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(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 = 0.00, 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)/x^2,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

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Fricas [A]
time = 0.34, size = 120, normalized size = 3.33 \begin {gather*} \left [\frac {\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 )}{\sqrt {b^{2} - 4 \, a c}}, -\frac {2 \, \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 )}{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)/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (32) = 64\).
time = 0.09, size = 124, normalized size = 3.44 \begin {gather*} - \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {- 4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} + b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )} + \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} - b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2))*log(x + (-4*a*c*sqrt(-1/(4*a*c - b**2)) + b**2*sqrt(-1/(4*a*c - b**2)) + b)/(2*c)) +
sqrt(-1/(4*a*c - b**2))*log(x + (4*a*c*sqrt(-1/(4*a*c - b**2)) - b**2*sqrt(-1/(4*a*c - b**2)) + b)/(2*c))

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Giac [A]
time = 4.10, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 \, \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 46, normalized size = 1.28 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(c + a/x^2 + b/x)),x)

[Out]

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

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