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

Optimal. Leaf size=224 \[ \frac{b \left (2 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{a^2 c^2 \sqrt{a+\frac{b}{x}} (b c-a d)^2}-\frac{(4 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2} c^3}+\frac{d^{5/2} (7 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 (b c-a d)^{5/2}}+\frac{d (b c-2 a d)}{a c^2 \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )} \]

[Out]

(b*(3*b^2*c^2 - 2*a*b*c*d + 2*a^2*d^2))/(a^2*c^2*(b*c - a*d)^2*Sqrt[a + b/x]) + (d*(b*c - 2*a*d))/(a*c^2*(b*c
- a*d)*Sqrt[a + b/x]*(c + d/x)) + x/(a*c*Sqrt[a + b/x]*(c + d/x)) + (d^(5/2)*(7*b*c - 4*a*d)*ArcTan[(Sqrt[d]*S
qrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*(b*c - a*d)^(5/2)) - ((3*b*c + 4*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(a^
(5/2)*c^3)

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Rubi [A]  time = 0.322693, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {375, 103, 151, 152, 156, 63, 208, 205} \[ \frac{b \left (2 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{a^2 c^2 \sqrt{a+\frac{b}{x}} (b c-a d)^2}-\frac{(4 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2} c^3}+\frac{d^{5/2} (7 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 (b c-a d)^{5/2}}+\frac{d (b c-2 a d)}{a c^2 \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*b^2*c^2 - 2*a*b*c*d + 2*a^2*d^2))/(a^2*c^2*(b*c - a*d)^2*Sqrt[a + b/x]) + (d*(b*c - 2*a*d))/(a*c^2*(b*c
- a*d)*Sqrt[a + b/x]*(c + d/x)) + x/(a*c*Sqrt[a + b/x]*(c + d/x)) + (d^(5/2)*(7*b*c - 4*a*d)*ArcTan[(Sqrt[d]*S
qrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*(b*c - a*d)^(5/2)) - ((3*b*c + 4*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(a^
(5/2)*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 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 )^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2} (c+d x)^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (3 b c+4 a d)+\frac{5 b d x}{2}}{x (a+b x)^{3/2} (c+d x)^2} \, dx,x,\frac{1}{x}\right )}{a c}\\ &=\frac{d (b c-2 a d)}{a c^2 (b c-a d) \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} (b c-a d) (3 b c+4 a d)-\frac{3}{2} b d (b c-2 a d) x}{x (a+b x)^{3/2} (c+d x)} \, dx,x,\frac{1}{x}\right )}{a c^2 (b c-a d)}\\ &=\frac{b \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )}{a^2 c^2 (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{d (b c-2 a d)}{a c^2 (b c-a d) \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{4} (b c-a d)^2 (3 b c+4 a d)-\frac{1}{4} b d \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right ) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{a^2 c^2 (b c-a d)^2}\\ &=\frac{b \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )}{a^2 c^2 (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{d (b c-2 a d)}{a c^2 (b c-a d) \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{\left (d^3 (7 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)^2}+\frac{(3 b c+4 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^2 c^3}\\ &=\frac{b \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )}{a^2 c^2 (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{d (b c-2 a d)}{a c^2 (b c-a d) \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{\left (d^3 (7 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)^2}+\frac{(3 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^2 b c^3}\\ &=\frac{b \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )}{a^2 c^2 (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{d (b c-2 a d)}{a c^2 (b c-a d) \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{x}{a c \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )}+\frac{d^{5/2} (7 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)^{5/2}}-\frac{(3 b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2} c^3}\\ \end{align*}

Mathematica [C]  time = 0.115698, size = 164, normalized size = 0.73 \[ \frac{(b c-a d) \left ((c x+d) \left (-4 a^2 d^2+a b c d+3 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b}{a x}+1\right )+a c x (b c (c x+d)-a d (c x+2 d))\right )+a^2 d^2 (c x+d) (7 b c-4 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d \left (a+\frac{b}{x}\right )}{a d-b c}\right )}{a^2 c^3 \sqrt{a+\frac{b}{x}} (c x+d) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*d^2*(7*b*c - 4*a*d)*(d + c*x)*Hypergeometric2F1[-1/2, 1, 1/2, (d*(a + b/x))/(-(b*c) + a*d)] + (b*c - a*d)
*(a*c*x*(b*c*(d + c*x) - a*d*(2*d + c*x)) + (3*b^2*c^2 + a*b*c*d - 4*a^2*d^2)*(d + c*x)*Hypergeometric2F1[-1/2
, 1, 1/2, 1 + b/(a*x)]))/(a^2*c^3*(b*c - a*d)^2*Sqrt[a + b/x]*(d + c*x))

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Maple [B]  time = 0.014, size = 3119, normalized size = 13.9 \begin{align*} \text{output 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)^2,x)

[Out]

-1/2*((a*x+b)/x)^(1/2)*x*(7*a^(11/2)*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^3*b^2*c^3*d^3-3*a^(13/2)*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*b*c*d^5-15*a^(11/2)*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*b^2*
c^2*d^4+14*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*b^3*c^3*d
^3+2*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*b^2*c^4*d^2-4*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*
x)^(3/2)*b^3*c^5*d-18*a^(11/2)*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*d^5+3*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*b^3*c^2*d
^4+2*a^(13/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^4*c^4*d^2+4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*
x+b)/a^(1/2))*a^7*((a*d-b*c)*d/c^2)^(1/2)*x^3*c^2*d^4-3*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^3*b^4*c^6-2*a^(11/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x^2*c^4*d^2-2*a^(
13/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*c^3*d^3+6*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2
)*x^3*b^3*c^6-11*a^(13/2)*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^3*b*
c^2*d^4-3*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^6*c^5*d+6*a^(3/2)*
((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^5*c^5*d-3*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^6*c^6+6*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^5*c^6+7*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^4*c^3*d^3+4*ln(1/2*(2*((a*x+b)*x
)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^5*((a*d-b*c)*d/c^2)^(1/2)*b^2*c*d^5-9*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)*b^3*c^2*d^4+3*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^4*c^3*d^3+5*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^5*c^4*d^2-4*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^2*c^2*d^4+8*a^(7/2)*((a*d-
b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^3*c^3*d^3-10*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^4*c^4*d
^2-3*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*b^3*c^3*d^3+13*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^4*c^4*d^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^5*c^5*d-8*a^(11/2)*((a*d-b*c)*d/c^2)^(1/2)*(
(a*x+b)*x)^(1/2)*x*b*c^2*d^4+14*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^2*c^3*d^3-12*a^(7/2)*((a
*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^3*c^4*d^2+8*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b
^2*c^4*d^2-14*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b^3*c^5*d+8*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^
(1/2)+2*a*x+b)/a^(1/2))*a^6*((a*d-b*c)*d/c^2)^(1/2)*x*b*c*d^5-14*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*a^5*((a*d-b*c)*d/c^2)^(1/2)*x*b^2*c^2*d^4-4*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x*b*c^4
*d^2+4*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x*b^2*c^5*d+4*a^(11/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+
b)*x)^(1/2)*x^2*b*c^3*d^3+11*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^2*b^3*c^4*d^2+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^2*b^4*c
^5*d-10*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*b^2*c^5*d-ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*
a*x+b)/a^(1/2))*a^6*((a*d-b*c)*d/c^2)^(1/2)*x^2*b*c^2*d^4-15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1
/2))*a^5*((a*d-b*c)*d/c^2)^(1/2)*x^2*b^2*c^3*d^3+5*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^3*b^3*c^5*d+12*a^(11/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*b*c^4*d^2-9*ln(
1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^6*((a*d-b*c)*d/c^2)^(1/2)*x^3*b*c^3*d^3+3*ln(1/2*(2*((a*x
+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^5*((a*d-b*c)*d/c^2)^(1/2)*x^3*b^2*c^4*d^2+2*a^(5/2)*((a*d-b*c)*d/c^2)
^(1/2)*((a*x+b)*x)^(1/2)*x*b^4*c^5*d+4*a^(15/2)*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^6+4*a^(11/2)*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^6+8*a^(13/2)*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^6-11
*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))*b^3*c*d^5+7*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^4*c^2*d^4+4*a^(15/2)*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^3*c*d^5+4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1
/2)+2*a*x+b)/a^(1/2))*a^7*((a*d-b*c)*d/c^2)^(1/2)*x^2*c*d^5-4*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2
)*x*b^3*c^6-4*a^(13/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*c^2*d^4-6*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^2*b^5*c^6+12*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^
(1/2)*x^2*b^4*c^6)/c^4/a^(7/2)/((a*x+b)*x)^(1/2)/(a*d-b*c)^3/(c*x+d)/((a*d-b*c)*d/c^2)^(1/2)/(a*x+b)^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 )}^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 4.01772, size = 4641, normalized size = 20.72 \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)^2,x, algorithm="fricas")

[Out]

[1/2*((3*b^4*c^3*d - 2*a*b^3*c^2*d^2 - 5*a^2*b^2*c*d^3 + 4*a^3*b*d^4 + (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*
b*c^2*d^2 + 4*a^4*c*d^3)*x^2 + (3*b^4*c^4 + a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 - a^3*b*c*d^3 + 4*a^4*d^4)*x)*sqrt
(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - (7*a^3*b^2*c*d^3 - 4*a^4*b*d^4 + (7*a^4*b*c^2*d^2 - 4*a^5
*c*d^3)*x^2 + (7*a^3*b^2*c^2*d^2 + 3*a^4*b*c*d^3 - 4*a^5*d^4)*x)*sqrt(-d/(b*c - a*d))*log(-(2*(b*c - a*d)*x*sq
rt(-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^2*c^4 - 2*a^3*b*c^3*d + a
^4*c^2*d^2)*x^3 + (3*a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 2*a^4*c*d^3)*x^2 + (3*a*b^3*c^3*d - 2*a^2*b^2
*c^2*d^2 + 2*a^3*b*c*d^3)*x)*sqrt((a*x + b)/x))/(a^3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 + a^5*b*c^3*d^3 + (a^4*b^2*
c^6 - 2*a^5*b*c^5*d + a^6*c^4*d^2)*x^2 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x), 1/2*(
2*(7*a^3*b^2*c*d^3 - 4*a^4*b*d^4 + (7*a^4*b*c^2*d^2 - 4*a^5*c*d^3)*x^2 + (7*a^3*b^2*c^2*d^2 + 3*a^4*b*c*d^3 -
4*a^5*d^4)*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)) +
(3*b^4*c^3*d - 2*a*b^3*c^2*d^2 - 5*a^2*b^2*c*d^3 + 4*a^3*b*d^4 + (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*b*c^2
*d^2 + 4*a^4*c*d^3)*x^2 + (3*b^4*c^4 + a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 - a^3*b*c*d^3 + 4*a^4*d^4)*x)*sqrt(a)*l
og(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*((a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*x^3 + (3*a*b^3*
c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 2*a^4*c*d^3)*x^2 + (3*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 2*a^3*b*c*d^3)*x
)*sqrt((a*x + b)/x))/(a^3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 + a^5*b*c^3*d^3 + (a^4*b^2*c^6 - 2*a^5*b*c^5*d + a^6*c
^4*d^2)*x^2 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x), 1/2*(2*(3*b^4*c^3*d - 2*a*b^3*c^
2*d^2 - 5*a^2*b^2*c*d^3 + 4*a^3*b*d^4 + (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2 +
(3*b^4*c^4 + a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 - a^3*b*c*d^3 + 4*a^4*d^4)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x
+ b)/x)/a) - (7*a^3*b^2*c*d^3 - 4*a^4*b*d^4 + (7*a^4*b*c^2*d^2 - 4*a^5*c*d^3)*x^2 + (7*a^3*b^2*c^2*d^2 + 3*a^4
*b*c*d^3 - 4*a^5*d^4)*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^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*x^3 + (3*a*b^3*c^4 - a^2*b^2
*c^3*d - a^3*b*c^2*d^2 + 2*a^4*c*d^3)*x^2 + (3*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 2*a^3*b*c*d^3)*x)*sqrt((a*x +
b)/x))/(a^3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 + a^5*b*c^3*d^3 + (a^4*b^2*c^6 - 2*a^5*b*c^5*d + a^6*c^4*d^2)*x^2 +
(a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x), ((7*a^3*b^2*c*d^3 - 4*a^4*b*d^4 + (7*a^4*b*c^
2*d^2 - 4*a^5*c*d^3)*x^2 + (7*a^3*b^2*c^2*d^2 + 3*a^4*b*c*d^3 - 4*a^5*d^4)*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)) + (3*b^4*c^3*d - 2*a*b^3*c^2*d^2 - 5*a^2*b^2*c*
d^3 + 4*a^3*b*d^4 + (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2 + (3*b^4*c^4 + a*b^3*c
^3*d - 7*a^2*b^2*c^2*d^2 - a^3*b*c*d^3 + 4*a^4*d^4)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + ((a^2*b
^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*x^3 + (3*a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 2*a^4*c*d^3)*x^2 +
(3*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 2*a^3*b*c*d^3)*x)*sqrt((a*x + b)/x))/(a^3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 +
a^5*b*c^3*d^3 + (a^4*b^2*c^6 - 2*a^5*b*c^5*d + a^6*c^4*d^2)*x^2 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^
2 + a^6*c^3*d^3)*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/(a+b/x)**(3/2)/(c+d/x)**2,x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 1.18649, size = 560, normalized size = 2.5 \begin{align*} b{\left (\frac{{\left (7 \, b c d^{3} - 4 \, a d^{4}\right )} \arctan \left (\frac{d \sqrt{\frac{a x + b}{x}}}{\sqrt{b c d - a d^{2}}}\right )}{{\left (b^{3} c^{5} - 2 \, a b^{2} c^{4} d + a^{2} b c^{3} d^{2}\right )} \sqrt{b c d - a d^{2}}} + \frac{2 \, a b^{3} c^{3} - 2 \, a^{2} b^{2} c^{2} d - \frac{3 \,{\left (a x + b\right )} b^{3} c^{3}}{x} + \frac{7 \,{\left (a x + b\right )} a b^{2} c^{2} d}{x} - \frac{3 \,{\left (a x + b\right )} a^{2} b c d^{2}}{x} + \frac{2 \,{\left (a x + b\right )} a^{3} d^{3}}{x} - \frac{3 \,{\left (a x + b\right )}^{2} b^{2} c^{2} d}{x^{2}} + \frac{2 \,{\left (a x + b\right )}^{2} a b c d^{2}}{x^{2}} - \frac{2 \,{\left (a x + b\right )}^{2} a^{2} d^{3}}{x^{2}}}{{\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )}{\left (a b c \sqrt{\frac{a x + b}{x}} - a^{2} d \sqrt{\frac{a x + b}{x}} - \frac{{\left (a x + b\right )} b c \sqrt{\frac{a x + b}{x}}}{x} + \frac{2 \,{\left (a x + b\right )} a d \sqrt{\frac{a x + b}{x}}}{x} - \frac{{\left (a x + b\right )}^{2} d \sqrt{\frac{a x + b}{x}}}{x^{2}}\right )}} + \frac{{\left (3 \, b c + 4 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} b c^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*((7*b*c*d^3 - 4*a*d^4)*arctan(d*sqrt((a*x + b)/x)/sqrt(b*c*d - a*d^2))/((b^3*c^5 - 2*a*b^2*c^4*d + a^2*b*c^3
*d^2)*sqrt(b*c*d - a*d^2)) + (2*a*b^3*c^3 - 2*a^2*b^2*c^2*d - 3*(a*x + b)*b^3*c^3/x + 7*(a*x + b)*a*b^2*c^2*d/
x - 3*(a*x + b)*a^2*b*c*d^2/x + 2*(a*x + b)*a^3*d^3/x - 3*(a*x + b)^2*b^2*c^2*d/x^2 + 2*(a*x + b)^2*a*b*c*d^2/
x^2 - 2*(a*x + b)^2*a^2*d^3/x^2)/((a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2)*(a*b*c*sqrt((a*x + b)/x) - a^2*d
*sqrt((a*x + b)/x) - (a*x + b)*b*c*sqrt((a*x + b)/x)/x + 2*(a*x + b)*a*d*sqrt((a*x + b)/x)/x - (a*x + b)^2*d*s
qrt((a*x + b)/x)/x^2)) + (3*b*c + 4*a*d)*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^2*b*c^3))