3.250 \(\int \frac{1}{\sqrt{a+\frac{b}{x}} (c+\frac{d}{x})^2} \, dx\)

Optimal. Leaf size=172 \[ -\frac{(4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2} c^3}-\frac{d^{3/2} (5 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 (b c-a d)^{3/2}}+\frac{d \sqrt{a+\frac{b}{x}} (b c-2 a d)}{a c^2 \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{x \sqrt{a+\frac{b}{x}}}{a c \left (c+\frac{d}{x}\right )} \]

[Out]

(d*(b*c - 2*a*d)*Sqrt[a + b/x])/(a*c^2*(b*c - a*d)*(c + d/x)) + (Sqrt[a + b/x]*x)/(a*c*(c + d/x)) - (d^(3/2)*(
5*b*c - 4*a*d)*ArcTan[(Sqrt[d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*(b*c - a*d)^(3/2)) - ((b*c + 4*a*d)*ArcTa
nh[Sqrt[a + b/x]/Sqrt[a]])/(a^(3/2)*c^3)

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Rubi [A]  time = 0.218058, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 103, 151, 156, 63, 208, 205} \[ -\frac{(4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2} c^3}-\frac{d^{3/2} (5 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 (b c-a d)^{3/2}}+\frac{d \sqrt{a+\frac{b}{x}} (b c-2 a d)}{a c^2 \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{x \sqrt{a+\frac{b}{x}}}{a c \left (c+\frac{d}{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x]*(c + d/x)^2),x]

[Out]

(d*(b*c - 2*a*d)*Sqrt[a + b/x])/(a*c^2*(b*c - a*d)*(c + d/x)) + (Sqrt[a + b/x]*x)/(a*c*(c + d/x)) - (d^(3/2)*(
5*b*c - 4*a*d)*ArcTan[(Sqrt[d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*(b*c - a*d)^(3/2)) - ((b*c + 4*a*d)*ArcTa
nh[Sqrt[a + b/x]/Sqrt[a]])/(a^(3/2)*c^3)

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 151

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

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 )^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x} (c+d x)^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{a+\frac{b}{x}} x}{a c \left (c+\frac{d}{x}\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (b c+4 a d)+\frac{3 b d x}{2}}{x \sqrt{a+b x} (c+d x)^2} \, dx,x,\frac{1}{x}\right )}{a c}\\ &=\frac{d (b c-2 a d) \sqrt{a+\frac{b}{x}}}{a c^2 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{a c \left (c+\frac{d}{x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} (b c-a d) (b c+4 a d)-\frac{1}{2} b d (b c-2 a d) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{a c^2 (b c-a d)}\\ &=\frac{d (b c-2 a d) \sqrt{a+\frac{b}{x}}}{a c^2 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{a c \left (c+\frac{d}{x}\right )}-\frac{\left (d^2 (5 b c-4 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{2 c^3 (b c-a d)}+\frac{(b c+4 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a c^3}\\ &=\frac{d (b c-2 a d) \sqrt{a+\frac{b}{x}}}{a c^2 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{a c \left (c+\frac{d}{x}\right )}-\frac{\left (d^2 (5 b c-4 a d)\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^3 (b c-a d)}+\frac{(b c+4 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^3}\\ &=\frac{d (b c-2 a d) \sqrt{a+\frac{b}{x}}}{a c^2 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{a c \left (c+\frac{d}{x}\right )}-\frac{d^{3/2} (5 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 (b c-a d)^{3/2}}-\frac{(b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{3/2} c^3}\\ \end{align*}

Mathematica [A]  time = 0.609596, size = 150, normalized size = 0.87 \[ \frac{\frac{a d^{3/2} (4 a d-5 b c) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac{c x \sqrt{a+\frac{b}{x}} (b c (c x+d)-a d (c x+2 d))}{(c x+d) (b c-a d)}-\frac{(4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}}{a c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x]*(c + d/x)^2),x]

[Out]

((c*Sqrt[a + b/x]*x*(b*c*(d + c*x) - a*d*(2*d + c*x)))/((b*c - a*d)*(d + c*x)) + (a*d^(3/2)*(-5*b*c + 4*a*d)*A
rcTan[(Sqrt[d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2) - ((b*c + 4*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[
a]])/Sqrt[a])/(a*c^3)

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Maple [B]  time = 0.017, size = 1135, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*x+b)/x)^(1/2)*x*(4*a^(9/2)*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))*x*c*d^4+2*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*c^4*d+4*a^(9/2)*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))*d^5-2*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*
c^3*d^2-9*a^(7/2)*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))*x*b*c^2*d^3-4*
a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c^2*d^3-9*a^(7/2)*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))*b*c*d^4-2*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*c^4*d+6*a^(5/2
)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b*c^4*d+5*a^(5/2)*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))*x*b^2*c^3*d^2+6*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b*c^3*d^2+5*
a^(5/2)*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))*b^2*c^2*d^3-2*a^(3/2)*((
a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^2*c^5+4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^4*
((a*d-b*c)*d/c^2)^(1/2)*x*c^2*d^3-7*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*((a*d-b*c)*d/c^2
)^(1/2)*x*b*c^3*d^2+2*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^2*((a*d-b*c)*d/c^2)^(1/2)*x*b^2*
c^4*d+ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*((a*d-b*c)*d/c^2)^(1/2)*x*b^3*c^5-2*a^(3/2)*((a*
d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^2*c^4*d+4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^4*((
a*d-b*c)*d/c^2)^(1/2)*c*d^4-7*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*((a*d-b*c)*d/c^2)^(1/2
)*b*c^2*d^3+2*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^2*((a*d-b*c)*d/c^2)^(1/2)*b^2*c^3*d^2+ln
(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*((a*d-b*c)*d/c^2)^(1/2)*b^3*c^4*d)/c^4/((a*x+b)*x)^(1/2)
/(a*d-b*c)^2/(c*x+d)/a^(5/2)/((a*d-b*c)*d/c^2)^(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 )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.89441, size = 2408, normalized size = 14. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(a)*log(2*a*x - 2*sq
rt(a)*x*sqrt((a*x + b)/x) + b) + (5*a^2*b*c*d^2 - 4*a^3*d^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x)*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
*b*c^3 - a^2*c^2*d)*x^2 + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^2*b*c^4*d - a^3*c^3*d^2 + (a^2*b*
c^5 - a^3*c^4*d)*x), -1/2*(2*(5*a^2*b*c*d^2 - 4*a^3*d^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x)*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)) - (b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2
*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*((a
*b*c^3 - a^2*c^2*d)*x^2 + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^2*b*c^4*d - a^3*c^3*d^2 + (a^2*b*
c^5 - a^3*c^4*d)*x), 1/2*(2*(b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sq
rt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (5*a^2*b*c*d^2 - 4*a^3*d^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x)*sq
rt(-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*b*c^3 - a^2*c^2*d)*x^2 + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^2*b*c^4*d - a^3*c^3*
d^2 + (a^2*b*c^5 - a^3*c^4*d)*x), -((5*a^2*b*c*d^2 - 4*a^3*d^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x)*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)) - (b^2*c^2*d + 3*a*b*c*d^2
- 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - ((a*b*c
^3 - a^2*c^2*d)*x^2 + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^2*b*c^4*d - a^3*c^3*d^2 + (a^2*b*c^5
- a^3*c^4*d)*x)]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.21074, size = 393, normalized size = 2.28 \begin{align*} -b{\left (\frac{{\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} \arctan \left (\frac{d \sqrt{\frac{a x + b}{x}}}{\sqrt{b c d - a d^{2}}}\right )}{{\left (b^{2} c^{4} - a b c^{3} d\right )} \sqrt{b c d - a d^{2}}} + \frac{b^{2} c^{2} \sqrt{\frac{a x + b}{x}} - 2 \, a b c d \sqrt{\frac{a x + b}{x}} + 2 \, a^{2} d^{2} \sqrt{\frac{a x + b}{x}} + \frac{{\left (a x + b\right )} b c d \sqrt{\frac{a x + b}{x}}}{x} - \frac{2 \,{\left (a x + b\right )} a d^{2} \sqrt{\frac{a x + b}{x}}}{x}}{{\left (a b c^{3} - a^{2} c^{2} d\right )}{\left (a b c - a^{2} d - \frac{{\left (a x + b\right )} b c}{x} + \frac{2 \,{\left (a x + b\right )} a d}{x} - \frac{{\left (a x + b\right )}^{2} d}{x^{2}}\right )}} - \frac{{\left (b c + 4 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a b c^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*((5*b*c*d^2 - 4*a*d^3)*arctan(d*sqrt((a*x + b)/x)/sqrt(b*c*d - a*d^2))/((b^2*c^4 - a*b*c^3*d)*sqrt(b*c*d -
a*d^2)) + (b^2*c^2*sqrt((a*x + b)/x) - 2*a*b*c*d*sqrt((a*x + b)/x) + 2*a^2*d^2*sqrt((a*x + b)/x) + (a*x + b)*b
*c*d*sqrt((a*x + b)/x)/x - 2*(a*x + b)*a*d^2*sqrt((a*x + b)/x)/x)/((a*b*c^3 - a^2*c^2*d)*(a*b*c - a^2*d - (a*x
+ b)*b*c/x + 2*(a*x + b)*a*d/x - (a*x + b)^2*d/x^2)) - (b*c + 4*a*d)*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt
(-a)*a*b*c^3))