3.256 \(\int \frac{1}{(a+\frac{b}{x})^{3/2} (c+\frac{d}{x})} \, dx\)

Optimal. Leaf size=147 \[ -\frac{(2 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2} c^2}+\frac{b (3 b c-a d)}{a^2 c \sqrt{a+\frac{b}{x}} (b c-a d)}+\frac{2 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 (b c-a d)^{3/2}}+\frac{x}{a c \sqrt{a+\frac{b}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.194021, antiderivative size = 147, 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, 152, 156, 63, 208, 205} \[ -\frac{(2 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2} c^2}+\frac{b (3 b c-a d)}{a^2 c \sqrt{a+\frac{b}{x}} (b c-a d)}+\frac{2 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 (b c-a d)^{3/2}}+\frac{x}{a c \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(3/2)*(c + d/x)),x]

[Out]

(b*(3*b*c - a*d))/(a^2*c*(b*c - a*d)*Sqrt[a + b/x]) + x/(a*c*Sqrt[a + b/x]) + (2*d^(5/2)*ArcTan[(Sqrt[d]*Sqrt[
a + b/x])/Sqrt[b*c - a*d]])/(c^2*(b*c - a*d)^(3/2)) - ((3*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(a^(5/2
)*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 152

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

Mathematica [C]  time = 0.0546908, size = 106, normalized size = 0.72 \[ \frac{(a d-b c) \left ((2 a d+3 b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b}{a x}+1\right )+a c x\right )-2 a^2 d^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d \left (a+\frac{b}{x}\right )}{a d-b c}\right )}{a^2 c^2 \sqrt{a+\frac{b}{x}} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(3/2)*(c + d/x)),x]

[Out]

(-2*a^2*d^2*Hypergeometric2F1[-1/2, 1, 1/2, (d*(a + b/x))/(-(b*c) + a*d)] + (-(b*c) + a*d)*(a*c*x + (3*b*c + 2
*a*d)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + b/(a*x)]))/(a^2*c^2*(-(b*c) + a*d)*Sqrt[a + b/x])

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Maple [B]  time = 0.016, size = 962, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(3/2)/(c+d/x),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)*x^2*a^4*c*d^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)*x^2*a^3*b*c^2*d-3*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)*x^2*a^2*b^2*c^3-2*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+
b)*x)^(1/2)*x^2*c^2*d+6*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b*c^3+2*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^2*d^3+4*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)*x*a^3*b*c*d^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)*x*a^2*b^2*c^2*d-6*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)*x*a*b^3*c^3-4*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*b*c^3-4*a^(5/2)*((a*d-b*c)*d/c^2)^(1
/2)*((a*x+b)*x)^(1/2)*x*b*c^2*d+12*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^2*c^3+4*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*d^3+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^2*b^2*c*d^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*b^3*c^2*d-3*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^4*c^3-2*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^2*c^2*d+6*a^(1/2)*((a*d-b*c)*d/c^
2)^(1/2)*((a*x+b)*x)^(1/2)*b^3*c^3+2*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*d^3)/a^(5/2)*x*((a*x+b)/x)^(1/2)/((a*d-b*c)*d/c^2)^(1/2)/c^3/(a*x+b)^2/(a*d-b*c)/((a*x+b)*x)^
(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x}\right )}^{\frac{3}{2}}{\left (c + \frac{d}{x}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/(c+d/x),x, algorithm="maxima")

[Out]

integrate(1/((a + b/x)^(3/2)*(c + d/x)), x)

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Fricas [B]  time = 2.03151, size = 2195, normalized size = 14.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/(c+d/x),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a + \frac{b}{x}\right )^{\frac{3}{2}} \left (c x + d\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(3/2)/(c+d/x),x)

[Out]

Integral(x/((a + b/x)**(3/2)*(c*x + d)), x)

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Giac [A]  time = 1.18955, size = 261, normalized size = 1.78 \begin{align*}{\left (\frac{2 \, d^{3} \arctan \left (\frac{d \sqrt{\frac{a x + b}{x}}}{\sqrt{b c d - a d^{2}}}\right )}{{\left (b^{2} c^{3} - a b c^{2} d\right )} \sqrt{b c d - a d^{2}}} + \frac{2 \, a b c - \frac{3 \,{\left (a x + b\right )} b c}{x} + \frac{{\left (a x + b\right )} a d}{x}}{{\left (a^{2} b c^{2} - a^{3} c d\right )}{\left (a \sqrt{\frac{a x + b}{x}} - \frac{{\left (a x + b\right )} \sqrt{\frac{a x + b}{x}}}{x}\right )}} + \frac{{\left (3 \, b c + 2 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} b c^{2}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/(c+d/x),x, algorithm="giac")

[Out]

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