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

Optimal. Leaf size=85 \[ -\frac{\sqrt{c+\frac{d}{x^2}} (2 a d+b c)}{2 c x}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 \sqrt{d}}+\frac{a x \left (c+\frac{d}{x^2}\right )^{3/2}}{c} \]

[Out]

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

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Rubi [A]  time = 0.0463275, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {375, 453, 195, 217, 206} \[ -\frac{\sqrt{c+\frac{d}{x^2}} (2 a d+b c)}{2 c x}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 \sqrt{d}}+\frac{a x \left (c+\frac{d}{x^2}\right )^{3/2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2],x]

[Out]

-((b*c + 2*a*d)*Sqrt[c + d/x^2])/(2*c*x) + (a*(c + d/x^2)^(3/2)*x)/c - ((b*c + 2*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c
+ d/x^2]*x)])/(2*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 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.0784238, size = 75, normalized size = 0.88 \[ \frac{\sqrt{c+\frac{d}{x^2}} \left (-\frac{x^2 (2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c x^2+d}}{\sqrt{d}}\right )}{\sqrt{d} \sqrt{c x^2+d}}+2 a x^2-b\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 135, normalized size = 1.6 \begin{align*} -{\frac{1}{2\,xd}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 2\,{d}^{3/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{2}a+\sqrt{d}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{2}bc-2\,\sqrt{c{x}^{2}+d}{x}^{2}ad-\sqrt{c{x}^{2}+d}{x}^{2}bc+ \left ( c{x}^{2}+d \right ) ^{{\frac{3}{2}}}b \right ){\frac{1}{\sqrt{c{x}^{2}+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]

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

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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 = 1.43861, size = 375, normalized size = 4.41 \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 \,{\left (2 \, a d x^{2} - b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{4 \, d 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 ) +{\left (2 \, a d x^{2} - b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{2 \, d x}\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]

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

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Sympy [A]  time = 4.17904, size = 107, normalized size = 1.26 \begin{align*} \frac{a \sqrt{c} x}{\sqrt{1 + \frac{d}{c x^{2}}}} - a \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )} + \frac{a d}{\sqrt{c} x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{b \sqrt{c} \sqrt{1 + \frac{d}{c x^{2}}}}{2 x} - \frac{b c \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2 \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(c)*x/sqrt(1 + d/(c*x**2)) - a*sqrt(d)*asinh(sqrt(d)/(sqrt(c)*x)) + a*d/(sqrt(c)*x*sqrt(1 + d/(c*x**2)))
- b*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*x) - b*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*sqrt(d))

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Giac [A]  time = 1.18739, size = 103, normalized size = 1.21 \begin{align*} \frac{2 \, \sqrt{c x^{2} + d} a c \mathrm{sgn}\left (x\right ) + \frac{{\left (b c^{2} \mathrm{sgn}\left (x\right ) + 2 \, a c d \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + d}}{\sqrt{-d}}\right )}{\sqrt{-d}} - \frac{\sqrt{c x^{2} + d} b c \mathrm{sgn}\left (x\right )}{x^{2}}}{2 \, c} \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]

1/2*(2*sqrt(c*x^2 + d)*a*c*sgn(x) + (b*c^2*sgn(x) + 2*a*c*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/sqrt(-d)
- sqrt(c*x^2 + d)*b*c*sgn(x)/x^2)/c