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

Optimal. Leaf size=91 \[ \frac{c (b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{3/2}}+\frac{\sqrt{c+\frac{d}{x^2}} (b c-4 a d)}{8 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x} \]

[Out]

((b*c - 4*a*d)*Sqrt[c + d/x^2])/(8*d*x) - (b*(c + d/x^2)^(3/2))/(4*d*x) + (c*(b*c - 4*a*d)*ArcTanh[Sqrt[d]/(Sq
rt[c + d/x^2]*x)])/(8*d^(3/2))

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Rubi [A]  time = 0.0507795, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 335, 195, 217, 206} \[ \frac{c (b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{3/2}}+\frac{\sqrt{c+\frac{d}{x^2}} (b c-4 a d)}{8 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 4*a*d)*Sqrt[c + d/x^2])/(8*d*x) - (b*(c + d/x^2)^(3/2))/(4*d*x) + (c*(b*c - 4*a*d)*ArcTanh[Sqrt[d]/(Sq
rt[c + d/x^2]*x)])/(8*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 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 \frac{\left (a+\frac{b}{x^2}\right ) \sqrt{c+\frac{d}{x^2}}}{x^2} \, dx &=-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x}+\frac{(-b c+4 a d) \int \frac{\sqrt{c+\frac{d}{x^2}}}{x^2} \, dx}{4 d}\\ &=-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x}-\frac{(-b c+4 a d) \operatorname{Subst}\left (\int \sqrt{c+d x^2} \, dx,x,\frac{1}{x}\right )}{4 d}\\ &=\frac{(b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x}+\frac{(c (b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{8 d}\\ &=\frac{(b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x}+\frac{(c (b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{1}{\sqrt{c+\frac{d}{x^2}} x}\right )}{8 d}\\ &=\frac{(b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{4 d x}+\frac{c (b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{8 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.115232, size = 100, normalized size = 1.1 \[ -\frac{\sqrt{c+\frac{d}{x^2}} \left (\left (c x^2+d\right ) \left (4 a d x^2+b c x^2+2 b d\right )+c x^4 \sqrt{\frac{c x^2}{d}+1} (4 a d-b c) \tanh ^{-1}\left (\sqrt{\frac{c x^2}{d}+1}\right )\right )}{8 d x^3 \left (c x^2+d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d/x^2]*((d + c*x^2)*(2*b*d + b*c*x^2 + 4*a*d*x^2) + c*(-(b*c) + 4*a*d)*x^4*Sqrt[1 + (c*x^2)/d]*ArcT
anh[Sqrt[1 + (c*x^2)/d]]))/(8*d*x^3*(d + c*x^2))

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Maple [B]  time = 0.012, size = 175, normalized size = 1.9 \begin{align*} -{\frac{1}{8\,{x}^{3}{d}^{2}}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 4\,{d}^{3/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}ac-\sqrt{d}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}b{c}^{2}-4\,\sqrt{c{x}^{2}+d}{x}^{4}acd+\sqrt{c{x}^{2}+d}{x}^{4}b{c}^{2}+4\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}ad- \left ( c{x}^{2}+d \right ) ^{{\frac{3}{2}}}{x}^{2}bc+2\, \left ( c{x}^{2}+d \right ) ^{3/2}bd \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^2,x)

[Out]

-1/8*((c*x^2+d)/x^2)^(1/2)/x^3*(4*d^(3/2)*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*x^4*a*c-d^(1/2)*ln(2*(d^(1/2)*(c
*x^2+d)^(1/2)+d)/x)*x^4*b*c^2-4*(c*x^2+d)^(1/2)*x^4*a*c*d+(c*x^2+d)^(1/2)*x^4*b*c^2+4*(c*x^2+d)^(3/2)*x^2*a*d-
(c*x^2+d)^(3/2)*x^2*b*c+2*(c*x^2+d)^(3/2)*b*d)/(c*x^2+d)^(1/2)/d^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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.40902, size = 450, normalized size = 4.95 \begin{align*} \left [-\frac{{\left (b c^{2} - 4 \, a c d\right )} \sqrt{d} x^{3} \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 \, b d^{2} +{\left (b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \, d^{2} x^{3}}, -\frac{{\left (b c^{2} - 4 \, a c d\right )} \sqrt{-d} x^{3} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left (2 \, b d^{2} +{\left (b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \, d^{2} x^{3}}\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^2,x, algorithm="fricas")

[Out]

[-1/16*((b*c^2 - 4*a*c*d)*sqrt(d)*x^3*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) + 2*(2*b*d^2
+ (b*c*d + 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^3), -1/8*((b*c^2 - 4*a*c*d)*sqrt(-d)*x^3*arctan(sqrt(-
d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (2*b*d^2 + (b*c*d + 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^3)]

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Sympy [A]  time = 6.51182, size = 144, normalized size = 1.58 \begin{align*} - \frac{a \sqrt{c} \sqrt{1 + \frac{d}{c x^{2}}}}{2 x} - \frac{a c \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2 \sqrt{d}} - \frac{b c^{\frac{3}{2}}}{8 d x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 b \sqrt{c}}{8 x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{b c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 d^{\frac{3}{2}}} - \frac{b d}{4 \sqrt{c} x^{5} \sqrt{1 + \frac{d}{c x^{2}}}} \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**2,x)

[Out]

-a*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*x) - a*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*sqrt(d)) - b*c**(3/2)/(8*d*x*sqrt(1
+ d/(c*x**2))) - 3*b*sqrt(c)/(8*x**3*sqrt(1 + d/(c*x**2))) + b*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d**(3/2)) -
b*d/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2)))

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Giac [A]  time = 1.23753, size = 176, normalized size = 1.93 \begin{align*} -\frac{\frac{{\left (b c^{3} \mathrm{sgn}\left (x\right ) - 4 \, a c^{2} d \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + d}}{\sqrt{-d}}\right )}{\sqrt{-d} d} + \frac{{\left (c x^{2} + d\right )}^{\frac{3}{2}} b c^{3} \mathrm{sgn}\left (x\right ) + 4 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} a c^{2} d \mathrm{sgn}\left (x\right ) + \sqrt{c x^{2} + d} b c^{3} d \mathrm{sgn}\left (x\right ) - 4 \, \sqrt{c x^{2} + d} a c^{2} d^{2} \mathrm{sgn}\left (x\right )}{c^{2} d x^{4}}}{8 \, 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^2,x, algorithm="giac")

[Out]

-1/8*((b*c^3*sgn(x) - 4*a*c^2*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d) + ((c*x^2 + d)^(3/2)*b*c
^3*sgn(x) + 4*(c*x^2 + d)^(3/2)*a*c^2*d*sgn(x) + sqrt(c*x^2 + d)*b*c^3*d*sgn(x) - 4*sqrt(c*x^2 + d)*a*c^2*d^2*
sgn(x))/(c^2*d*x^4))/c