3.249 \(\int \frac{1}{\sqrt{a+\frac{b}{x}} (c+\frac{d}{x})} \, dx\)

Optimal. Leaf size=108 \[ -\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2} c^2}-\frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 \sqrt{b c-a d}}+\frac{x \sqrt{a+\frac{b}{x}}}{a c} \]

[Out]

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

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Rubi [A]  time = 0.0963475, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {375, 103, 156, 63, 208, 205} \[ -\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2} c^2}-\frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 \sqrt{b c-a d}}+\frac{x \sqrt{a+\frac{b}{x}}}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x]*(c + d/x)),x]

[Out]

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

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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.176637, size = 104, normalized size = 0.96 \[ \frac{-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{2 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{\sqrt{b c-a d}}+\frac{c x \sqrt{a+\frac{b}{x}}}{a}}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x]*(c + d/x)),x]

[Out]

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

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Maple [B]  time = 0.011, size = 228, normalized size = 2.1 \begin{align*} -{\frac{x}{2\,{c}^{3}} \left ( 2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}acd+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}b{c}^{2}-2\,\sqrt{ \left ( ax+b \right ) x}{c}^{2}\sqrt{a}\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}+2\,\ln \left ({\frac{1}{cx+d} \left ( 2\,\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}c-2\,adx+bcx-bd \right ) } \right ){a}^{3/2}{d}^{2} \right ) \sqrt{{\frac{ax+b}{x}}}{\frac{1}{\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}}}{a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.51129, size = 1215, normalized size = 11.25 \begin{align*} \left [\frac{2 \, a^{2} d \sqrt{-\frac{d}{b c - a d}} \log \left (-\frac{2 \,{\left (b c - a d\right )} x \sqrt{-\frac{d}{b c - a d}} \sqrt{\frac{a x + b}{x}} - b d +{\left (b c - 2 \, a d\right )} x}{c x + d}\right ) + 2 \, a c x \sqrt{\frac{a x + b}{x}} +{\left (b c + 2 \, a d\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right )}{2 \, a^{2} c^{2}}, \frac{a^{2} d \sqrt{-\frac{d}{b c - a d}} \log \left (-\frac{2 \,{\left (b c - a d\right )} x \sqrt{-\frac{d}{b c - a d}} \sqrt{\frac{a x + b}{x}} - b d +{\left (b c - 2 \, a d\right )} x}{c x + d}\right ) + a c x \sqrt{\frac{a x + b}{x}} +{\left (b c + 2 \, a d\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right )}{a^{2} c^{2}}, -\frac{4 \, a^{2} d \sqrt{\frac{d}{b c - a d}} \arctan \left (-\frac{{\left (b c - a d\right )} x \sqrt{\frac{d}{b c - a d}} \sqrt{\frac{a x + b}{x}}}{a d x + b d}\right ) - 2 \, a c x \sqrt{\frac{a x + b}{x}} -{\left (b c + 2 \, a d\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right )}{2 \, a^{2} c^{2}}, -\frac{2 \, a^{2} d \sqrt{\frac{d}{b c - a d}} \arctan \left (-\frac{{\left (b c - a d\right )} x \sqrt{\frac{d}{b c - a d}} \sqrt{\frac{a x + b}{x}}}{a d x + b d}\right ) - a c x \sqrt{\frac{a x + b}{x}} -{\left (b c + 2 \, a d\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right )}{a^{2} c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{a + \frac{b}{x}} \left (c x + d\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(sqrt(a + b/x)*(c*x + d)), x)

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Giac [A]  time = 1.19339, size = 174, normalized size = 1.61 \begin{align*} -b{\left (\frac{2 \, d^{2} \arctan \left (\frac{d \sqrt{\frac{a x + b}{x}}}{\sqrt{b c d - a d^{2}}}\right )}{\sqrt{b c d - a d^{2}} b c^{2}} + \frac{\sqrt{\frac{a x + b}{x}}}{{\left (a - \frac{a x + b}{x}\right )} a c} - \frac{{\left (b c + 2 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a b c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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