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

Optimal. Leaf size=47 \[ \frac{a x \sqrt{c+\frac{d}{x^2}}}{c}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{\sqrt{d}} \]

[Out]

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

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Rubi [A]  time = 0.0292151, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {375, 451, 217, 206} \[ \frac{a x \sqrt{c+\frac{d}{x^2}}}{c}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/Sqrt[c + d/x^2],x]

[Out]

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

Rule 375

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

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 217

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

Mathematica [A]  time = 0.0310319, size = 71, normalized size = 1.51 \[ \frac{a \sqrt{d} \left (c x^2+d\right )-b c \sqrt{c x^2+d} \tanh ^{-1}\left (\frac{\sqrt{c x^2+d}}{\sqrt{d}}\right )}{c \sqrt{d} x \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/Sqrt[c + d/x^2],x]

[Out]

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

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Maple [A]  time = 0.01, size = 73, normalized size = 1.6 \begin{align*}{\frac{1}{cx}\sqrt{c{x}^{2}+d} \left ( a\sqrt{c{x}^{2}+d}\sqrt{d}-b\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ) c \right ){\frac{1}{\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}}}}{\frac{1}{\sqrt{d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2+d)^(1/2)*(a*(c*x^2+d)^(1/2)*d^(1/2)-b*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*c)/((c*x^2+d)/x^2)^(1/2)/x/c/
d^(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((a+b/x^2)/(c+d/x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37344, size = 305, normalized size = 6.49 \begin{align*} \left [\frac{2 \, a d x \sqrt{\frac{c x^{2} + d}{x^{2}}} + b c \sqrt{d} \log \left (-\frac{c x^{2} - 2 \, \sqrt{d} x \sqrt{\frac{c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right )}{2 \, c d}, \frac{a d x \sqrt{\frac{c x^{2} + d}{x^{2}}} + b c \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right )}{c d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.611, size = 39, normalized size = 0.83 \begin{align*} \frac{a \sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}}{c} - \frac{b \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{\sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sqrt(d)*sqrt(c*x**2/d + 1)/c - b*asinh(sqrt(d)/(sqrt(c)*x))/sqrt(d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^2)/sqrt(c + d/x^2), x)