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

Optimal. Leaf size=74 \[ -\frac{\sqrt{a+\frac{b}{x}} (2 a d+b c)}{a}+\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{c x \left (a+\frac{b}{x}\right )^{3/2}}{a} \]

[Out]

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

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Rubi [A]  time = 0.0483524, antiderivative size = 74, 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, 78, 50, 63, 208} \[ -\frac{\sqrt{a+\frac{b}{x}} (2 a d+b c)}{a}+\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{c x \left (a+\frac{b}{x}\right )^{3/2}}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]*(c + d/x),x]

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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]

Rubi steps

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

Mathematica [A]  time = 0.0411902, size = 52, normalized size = 0.7 \[ \sqrt{a+\frac{b}{x}} (c x-2 d)+\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]*(c + d/x),x]

[Out]

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

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Maple [B]  time = 0.009, size = 163, normalized size = 2.2 \begin{align*}{\frac{1}{2\,bx}\sqrt{{\frac{ax+b}{x}}} \left ( 2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}abd+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ){x}^{2}{b}^{2}c+4\,{a}^{3/2}\sqrt{a{x}^{2}+bx}{x}^{2}d+2\,\sqrt{a}\sqrt{a{x}^{2}+bx}{x}^{2}bc-4\,\sqrt{a} \left ( a{x}^{2}+bx \right ) ^{3/2}d \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x+b)/x)^(1/2)/x*(2*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a*b*d+ln(1/2*(2*(a*x^2+b*
x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*b^2*c+4*a^(3/2)*(a*x^2+b*x)^(1/2)*x^2*d+2*a^(1/2)*(a*x^2+b*x)^(1/2)*x^2
*b*c-4*a^(1/2)*(a*x^2+b*x)^(3/2)*d)/((a*x+b)*x)^(1/2)/b/a^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.31298, size = 300, normalized size = 4.05 \begin{align*} \left [\frac{{\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 \,{\left (a c x - 2 \, a d\right )} \sqrt{\frac{a x + b}{x}}}{2 \, a}, -\frac{{\left (b c + 2 \, a d\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (a c x - 2 \, a d\right )} \sqrt{\frac{a x + b}{x}}}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 20.0446, size = 87, normalized size = 1.18 \begin{align*} - \frac{2 a d \operatorname{atan}{\left (\frac{\sqrt{a + \frac{b}{x}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + \sqrt{b} c \sqrt{x} \sqrt{\frac{a x}{b} + 1} - 2 d \sqrt{a + \frac{b}{x}} + \frac{b c \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{\sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*d*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + sqrt(b)*c*sqrt(x)*sqrt(a*x/b + 1) - 2*d*sqrt(a + b/x) + b*c*asi
nh(sqrt(a)*sqrt(x)/sqrt(b))/sqrt(a)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError