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

Optimal. Leaf size=67 \[ \frac{x \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{a}-\frac{b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0405132, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1342, 730, 724, 206} \[ \frac{x \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{a}-\frac{b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + c/x^2 + b/x],x]

[Out]

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

Rule 1342

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

Rule 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, 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]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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}{\sqrt{a+\frac{c}{x^2}+\frac{b}{x}}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x+c x^2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{a}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=\frac{\sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{a}-\frac{b \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+\frac{b}{x}}{\sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}\right )}{a}\\ &=\frac{\sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{a}-\frac{b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}\right )}{2 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0544175, size = 89, normalized size = 1.33 \[ \frac{2 \sqrt{a} (x (a x+b)+c)-b \sqrt{x (a x+b)+c} \tanh ^{-1}\left (\frac{2 a x+b}{2 \sqrt{a} \sqrt{x (a x+b)+c}}\right )}{2 a^{3/2} x \sqrt{a+\frac{b x+c}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + c/x^2 + b/x],x]

[Out]

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

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Maple [A]  time = 0.006, size = 88, normalized size = 1.3 \begin{align*}{\frac{1}{2\,x}\sqrt{a{x}^{2}+bx+c} \left ( 2\,\sqrt{a{x}^{2}+bx+c}{a}^{3/2}-b\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) a \right ){\frac{1}{\sqrt{{\frac{a{x}^{2}+bx+c}{{x}^{2}}}}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x^2+b*x+c)^(1/2)*(2*(a*x^2+b*x+c)^(1/2)*a^(3/2)-b*ln(1/2*(2*(a*x^2+b*x+c)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2
))*a)/((a*x^2+b*x+c)/x^2)^(1/2)/x/a^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + \frac{b}{x} + \frac{c}{x^{2}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(a + b/x + c/x^2), x)

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Fricas [A]  time = 1.85355, size = 410, normalized size = 6.12 \begin{align*} \left [\frac{4 \, a x \sqrt{\frac{a x^{2} + b x + c}{x^{2}}} + \sqrt{a} b \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - b^{2} - 4 \, a c + 4 \,{\left (2 \, a x^{2} + b x\right )} \sqrt{a} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}\right )}{4 \, a^{2}}, \frac{2 \, a x \sqrt{\frac{a x^{2} + b x + c}{x^{2}}} + \sqrt{-a} b \arctan \left (\frac{{\left (2 \, a x^{2} + b x\right )} \sqrt{-a} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}}{2 \,{\left (a^{2} x^{2} + a b x + a c\right )}}\right )}{2 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*a*x*sqrt((a*x^2 + b*x + c)/x^2) + sqrt(a)*b*log(-8*a^2*x^2 - 8*a*b*x - b^2 - 4*a*c + 4*(2*a*x^2 + b*x)
*sqrt(a)*sqrt((a*x^2 + b*x + c)/x^2)))/a^2, 1/2*(2*a*x*sqrt((a*x^2 + b*x + c)/x^2) + sqrt(-a)*b*arctan(1/2*(2*
a*x^2 + b*x)*sqrt(-a)*sqrt((a*x^2 + b*x + c)/x^2)/(a^2*x^2 + a*b*x + a*c)))/a^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + \frac{b}{x} + \frac{c}{x^{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a + b/x + c/x**2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError