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

Optimal. Leaf size=61 \[ \frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 d^{3/2}}-\frac{b \sqrt{c+\frac{d}{x^2}}}{2 d x} \]

[Out]

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

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Rubi [A]  time = 0.0375054, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {459, 335, 217, 206} \[ \frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 d^{3/2}}-\frac{b \sqrt{c+\frac{d}{x^2}}}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 459

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

Rule 335

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

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

Mathematica [A]  time = 0.0525333, size = 80, normalized size = 1.31 \[ \frac{x^2 \sqrt{c x^2+d} (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{c x^2+d}}{\sqrt{d}}\right )-b \sqrt{d} \left (c x^2+d\right )}{2 d^{3/2} x^3 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 105, normalized size = 1.7 \begin{align*} -{\frac{1}{2\,{x}^{3}}\sqrt{c{x}^{2}+d} \left ( 2\,a\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{2}{d}^{2}-\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{2}bcd+{d}^{{\frac{3}{2}}}\sqrt{c{x}^{2}+d}b \right ){\frac{1}{\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}}}}{d}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/x^2/(c+d/x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36996, size = 346, normalized size = 5.67 \begin{align*} \left [-\frac{{\left (b c - 2 \, a d\right )} \sqrt{d} x \log \left (-\frac{c x^{2} - 2 \, \sqrt{d} x \sqrt{\frac{c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, b d \sqrt{\frac{c x^{2} + d}{x^{2}}}}{4 \, d^{2} x}, -\frac{{\left (b c - 2 \, a d\right )} \sqrt{-d} x \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + b d \sqrt{\frac{c x^{2} + d}{x^{2}}}}{2 \, d^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.6894, size = 66, normalized size = 1.08 \begin{align*} - \frac{a \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{\sqrt{d}} - \frac{b \sqrt{c} \sqrt{1 + \frac{d}{c x^{2}}}}{2 d x} + \frac{b c \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2 d^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*asinh(sqrt(d)/(sqrt(c)*x))/sqrt(d) - b*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*d*x) + b*c*asinh(sqrt(d)/(sqrt(c)*x)
)/(2*d**(3/2))

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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}}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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