3.230 \(\int \frac{\sqrt{a+\frac{b}{x}}}{(c+\frac{d}{x})^3} \, dx\)

Optimal. Leaf size=213 \[ \frac{\sqrt{d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{4 c^4 (b c-a d)^{3/2}}+\frac{d \sqrt{a+\frac{b}{x}} (11 b c-12 a d)}{4 c^3 \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{(b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^4}+\frac{x \sqrt{a+\frac{b}{x}}}{c \left (c+\frac{d}{x}\right )^2} \]

[Out]

(3*d*Sqrt[a + b/x])/(2*c^2*(c + d/x)^2) + (d*(11*b*c - 12*a*d)*Sqrt[a + b/x])/(4*c^3*(b*c - a*d)*(c + d/x)) +
(Sqrt[a + b/x]*x)/(c*(c + d/x)^2) + (Sqrt[d]*(15*b^2*c^2 - 40*a*b*c*d + 24*a^2*d^2)*ArcTan[(Sqrt[d]*Sqrt[a + b
/x])/Sqrt[b*c - a*d]])/(4*c^4*(b*c - a*d)^(3/2)) + ((b*c - 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(Sqrt[a]*c^4
)

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Rubi [A]  time = 0.339586, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 99, 151, 156, 63, 208, 205} \[ \frac{\sqrt{d} \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{4 c^4 (b c-a d)^{3/2}}+\frac{d \sqrt{a+\frac{b}{x}} (11 b c-12 a d)}{4 c^3 \left (c+\frac{d}{x}\right ) (b c-a d)}+\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{(b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^4}+\frac{x \sqrt{a+\frac{b}{x}}}{c \left (c+\frac{d}{x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]/(c + d/x)^3,x]

[Out]

(3*d*Sqrt[a + b/x])/(2*c^2*(c + d/x)^2) + (d*(11*b*c - 12*a*d)*Sqrt[a + b/x])/(4*c^3*(b*c - a*d)*(c + d/x)) +
(Sqrt[a + b/x]*x)/(c*(c + d/x)^2) + (Sqrt[d]*(15*b^2*c^2 - 40*a*b*c*d + 24*a^2*d^2)*ArcTan[(Sqrt[d]*Sqrt[a + b
/x])/Sqrt[b*c - a*d]])/(4*c^4*(b*c - a*d)^(3/2)) + ((b*c - 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(Sqrt[a]*c^4
)

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 99

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

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{\sqrt{a+\frac{b}{x}}}{\left (c+\frac{d}{x}\right )^3} \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2 (c+d x)^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (b c-6 a d)-\frac{5 b d x}{2}}{x \sqrt{a+b x} (c+d x)^3} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{-(b c-6 a d) (b c-a d)+\frac{9}{2} b d (b c-a d) x}{x \sqrt{a+b x} (c+d x)^2} \, dx,x,\frac{1}{x}\right )}{2 c^2 (b c-a d)}\\ &=\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{d (11 b c-12 a d) \sqrt{a+\frac{b}{x}}}{4 c^3 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{(b c-6 a d) (b c-a d)^2-\frac{1}{4} b d (11 b c-12 a d) (b c-a d) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{2 c^3 (b c-a d)^2}\\ &=\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{d (11 b c-12 a d) \sqrt{a+\frac{b}{x}}}{4 c^3 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}-\frac{(b c-6 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 c^4}+\frac{\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{8 c^4 (b c-a d)}\\ &=\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{d (11 b c-12 a d) \sqrt{a+\frac{b}{x}}}{4 c^3 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}-\frac{(b c-6 a d) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{b c^4}+\frac{\left (d \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right )\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 )}{4 b c^4 (b c-a d)}\\ &=\frac{3 d \sqrt{a+\frac{b}{x}}}{2 c^2 \left (c+\frac{d}{x}\right )^2}+\frac{d (11 b c-12 a d) \sqrt{a+\frac{b}{x}}}{4 c^3 (b c-a d) \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )^2}+\frac{\sqrt{d} \left (15 b^2 c^2-40 a b c d+24 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{4 c^4 (b c-a d)^{3/2}}+\frac{(b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^4}\\ \end{align*}

Mathematica [A]  time = 0.609968, size = 330, normalized size = 1.55 \[ \frac{(c x+d) \left (\frac{1}{2} c d^{5/2} \sqrt{a+\frac{b}{x}} (a x+b) \left (12 a^2 d^2-17 a b c d+4 b^2 c^2\right )+(c x+d) \left (-\frac{1}{2} a d^2 \left (24 a^2 d^2-40 a b c d+15 b^2 c^2\right ) \left (\sqrt{d} \sqrt{a+\frac{b}{x}}-\sqrt{b c-a d} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )\right )-d^{3/2} (b c-6 a d) (b c-a d)^2 \left (2 \sqrt{a+\frac{b}{x}}-2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )\right )\right )\right )+2 c^3 d^{3/2} x^3 \left (a+\frac{b}{x}\right )^{3/2} (b c-a d)^2+c^2 d^{5/2} x \sqrt{a+\frac{b}{x}} (a x+b) (2 b c-3 a d) (b c-a d)}{2 a c^4 d^{3/2} (c x+d)^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]/(c + d/x)^3,x]

[Out]

(2*c^3*d^(3/2)*(b*c - a*d)^2*(a + b/x)^(3/2)*x^3 + c^2*d^(5/2)*(2*b*c - 3*a*d)*(b*c - a*d)*Sqrt[a + b/x]*x*(b
+ a*x) + (d + c*x)*((c*d^(5/2)*(4*b^2*c^2 - 17*a*b*c*d + 12*a^2*d^2)*Sqrt[a + b/x]*(b + a*x))/2 + (d + c*x)*(-
(a*d^2*(15*b^2*c^2 - 40*a*b*c*d + 24*a^2*d^2)*(Sqrt[d]*Sqrt[a + b/x] - Sqrt[b*c - a*d]*ArcTan[(Sqrt[d]*Sqrt[a
+ b/x])/Sqrt[b*c - a*d]]))/2 - d^(3/2)*(b*c - 6*a*d)*(b*c - a*d)^2*(2*Sqrt[a + b/x] - 2*Sqrt[a]*ArcTanh[Sqrt[a
+ b/x]/Sqrt[a]]))))/(2*a*c^4*d^(3/2)*(b*c - a*d)^2*(d + c*x)^2)

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Maple [B]  time = 0.015, size = 1972, normalized size = 9.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a*x+b)/x)^(1/2)*x*(64*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^2+78*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b*c^4*d^2+32*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^2*c^5*d+18*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+
b)*x)^(1/2)*x^2*b*c^5*d-104*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^3-52*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*c^4*d^2
-8*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*d-4*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^2*b^3*c^6-52*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^4+32*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^3+46*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b*c^3*
d^3+24*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^6-4*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^2-22*a^(3/2)*((a*d-b*c)*d/c^2)^
(1/2)*((a*x+b)*x)^(1/2)*b^2*c^4*d^2+12*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*c^5*d-14*a^(5/2)*
((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*b*c^6+24*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*c^3*d^3-12*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x*c^5*d-64*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^2*b*c^3*d^3+55*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^2*b^2*c^4*d^2+48*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^4-36*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((
a*x+b)*x)^(1/2)*x*c^3*d^3-128*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^4+110*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^3-22*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b^2*c^6-15*a^(3/2)*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^5*d-30*a^(3/2)*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^4*d^2+10*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^
(3/2)*b*c^5*d-44*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^2*c^5*d+24*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*c^2*d^4-8*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+
b)*x)^(3/2)*c^4*d^2+24*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^5
-24*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c^2*d^4-64*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^5+55*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^4-15*a^(3/2)*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^3*d^3+48*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^5+14*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x*b*c^6)/c^5/((a*x+b)*x)^(1/2)/(a*d-b*c
)^2/(c*x+d)^2/a^(3/2)/((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{\sqrt{a + \frac{b}{x}}}{{\left (c + \frac{d}{x}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x)/(c + d/x)^3, x)

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Fricas [B]  time = 2.02808, size = 3664, normalized size = 17.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*(b^2*c^2*d^2 - 7*a*b*c*d^3 + 6*a^2*d^4 + (b^2*c^4 - 7*a*b*c^3*d + 6*a^2*c^2*d^2)*x^2 + 2*(b^2*c^3*d -
7*a*b*c^2*d^2 + 6*a^2*c*d^3)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (15*a*b^2*c^2*d^2 -
40*a^2*b*c*d^3 + 24*a^3*d^4 + (15*a*b^2*c^4 - 40*a^2*b*c^3*d + 24*a^3*c^2*d^2)*x^2 + 2*(15*a*b^2*c^3*d - 40*a^
2*b*c^2*d^2 + 24*a^3*c*d^3)*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*(4*(a*b*c^4 - a^2*c^3*d)*x^3 + (17*a*b*c^3*d - 18*a^2*c^2*d^2)*x^2
+ (11*a*b*c^2*d^2 - 12*a^2*c*d^3)*x)*sqrt((a*x + b)/x))/(a*b*c^5*d^2 - a^2*c^4*d^3 + (a*b*c^7 - a^2*c^6*d)*x^2
+ 2*(a*b*c^6*d - a^2*c^5*d^2)*x), 1/4*((15*a*b^2*c^2*d^2 - 40*a^2*b*c*d^3 + 24*a^3*d^4 + (15*a*b^2*c^4 - 40*a
^2*b*c^3*d + 24*a^3*c^2*d^2)*x^2 + 2*(15*a*b^2*c^3*d - 40*a^2*b*c^2*d^2 + 24*a^3*c*d^3)*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)) - 2*(b^2*c^2*d^2 - 7*a*b*c*d^3 + 6
*a^2*d^4 + (b^2*c^4 - 7*a*b*c^3*d + 6*a^2*c^2*d^2)*x^2 + 2*(b^2*c^3*d - 7*a*b*c^2*d^2 + 6*a^2*c*d^3)*x)*sqrt(a
)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (4*(a*b*c^4 - a^2*c^3*d)*x^3 + (17*a*b*c^3*d - 18*a^2*c^2*d
^2)*x^2 + (11*a*b*c^2*d^2 - 12*a^2*c*d^3)*x)*sqrt((a*x + b)/x))/(a*b*c^5*d^2 - a^2*c^4*d^3 + (a*b*c^7 - a^2*c^
6*d)*x^2 + 2*(a*b*c^6*d - a^2*c^5*d^2)*x), -1/8*(8*(b^2*c^2*d^2 - 7*a*b*c*d^3 + 6*a^2*d^4 + (b^2*c^4 - 7*a*b*c
^3*d + 6*a^2*c^2*d^2)*x^2 + 2*(b^2*c^3*d - 7*a*b*c^2*d^2 + 6*a^2*c*d^3)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x
+ b)/x)/a) + (15*a*b^2*c^2*d^2 - 40*a^2*b*c*d^3 + 24*a^3*d^4 + (15*a*b^2*c^4 - 40*a^2*b*c^3*d + 24*a^3*c^2*d^2
)*x^2 + 2*(15*a*b^2*c^3*d - 40*a^2*b*c^2*d^2 + 24*a^3*c*d^3)*x)*sqrt(-d/(b*c - a*d))*log(-(2*(b*c - a*d)*x*sqr
t(-d/(b*c - a*d))*sqrt((a*x + b)/x) - b*d + (b*c - 2*a*d)*x)/(c*x + d)) - 2*(4*(a*b*c^4 - a^2*c^3*d)*x^3 + (17
*a*b*c^3*d - 18*a^2*c^2*d^2)*x^2 + (11*a*b*c^2*d^2 - 12*a^2*c*d^3)*x)*sqrt((a*x + b)/x))/(a*b*c^5*d^2 - a^2*c^
4*d^3 + (a*b*c^7 - a^2*c^6*d)*x^2 + 2*(a*b*c^6*d - a^2*c^5*d^2)*x), 1/4*((15*a*b^2*c^2*d^2 - 40*a^2*b*c*d^3 +
24*a^3*d^4 + (15*a*b^2*c^4 - 40*a^2*b*c^3*d + 24*a^3*c^2*d^2)*x^2 + 2*(15*a*b^2*c^3*d - 40*a^2*b*c^2*d^2 + 24*
a^3*c*d^3)*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)) -
4*(b^2*c^2*d^2 - 7*a*b*c*d^3 + 6*a^2*d^4 + (b^2*c^4 - 7*a*b*c^3*d + 6*a^2*c^2*d^2)*x^2 + 2*(b^2*c^3*d - 7*a*b
*c^2*d^2 + 6*a^2*c*d^3)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (4*(a*b*c^4 - a^2*c^3*d)*x^3 + (17*
a*b*c^3*d - 18*a^2*c^2*d^2)*x^2 + (11*a*b*c^2*d^2 - 12*a^2*c*d^3)*x)*sqrt((a*x + b)/x))/(a*b*c^5*d^2 - a^2*c^4
*d^3 + (a*b*c^7 - a^2*c^6*d)*x^2 + 2*(a*b*c^6*d - a^2*c^5*d^2)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.33123, size = 1107, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*sqrt(a)*b^2*c^2*d*arctan(sqrt(a)*d/sqrt(b*c*d - a*d^2)) - 40*a^(3/2)*b*c*d^2*arctan(sqrt(a)*d/sqrt(b*
c*d - a*d^2)) + 24*a^(5/2)*d^3*arctan(sqrt(a)*d/sqrt(b*c*d - a*d^2)) - 2*sqrt(b*c*d - a*d^2)*b^2*c^2*log(abs(b
)) + 14*sqrt(b*c*d - a*d^2)*a*b*c*d*log(abs(b)) - 12*sqrt(b*c*d - a*d^2)*a^2*d^2*log(abs(b)) + 9*sqrt(b*c*d -
a*d^2)*a*b*c*d - 10*sqrt(b*c*d - a*d^2)*a^2*d^2)*sgn(x)/(sqrt(b*c*d - a*d^2)*sqrt(a)*b*c^5 - sqrt(b*c*d - a*d^
2)*a^(3/2)*c^4*d) - 1/4*(15*b^2*c^2*d*sgn(x) - 40*a*b*c*d^2*sgn(x) + 24*a^2*d^3*sgn(x))*arctan(-((sqrt(a)*x -
sqrt(a*x^2 + b*x))*c + sqrt(a)*d)/sqrt(b*c*d - a*d^2))/((b*c^5 - a*c^4*d)*sqrt(b*c*d - a*d^2)) + sqrt(a*x^2 +
b*x)*sgn(x)/c^3 - 1/4*(9*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*sqrt(a)*b^2*c^3*d*sgn(x) - 32*(sqrt(a)*x - sqrt(a*x
^2 + b*x))^3*a^(3/2)*b*c^2*d^2*sgn(x) + 24*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*a^(5/2)*c*d^3*sgn(x) + 3*(sqrt(a)
*x - sqrt(a*x^2 + b*x))^2*a*b^2*c^2*d^2*sgn(x) - 40*(sqrt(a)*x - sqrt(a*x^2 + b*x))^2*a^2*b*c*d^3*sgn(x) + 40*
(sqrt(a)*x - sqrt(a*x^2 + b*x))^2*a^3*d^4*sgn(x) + 7*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^3*c^2*d^2*sgn(x
) - 44*(sqrt(a)*x - sqrt(a*x^2 + b*x))*a^(3/2)*b^2*c*d^3*sgn(x) + 40*(sqrt(a)*x - sqrt(a*x^2 + b*x))*a^(5/2)*b
*d^4*sgn(x) - 9*a*b^3*c*d^3*sgn(x) + 10*a^2*b^2*d^4*sgn(x))/((sqrt(a)*b*c^5 - a^(3/2)*c^4*d)*((sqrt(a)*x - sqr
t(a*x^2 + b*x))^2*c + 2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*d + b*d)^2) - 1/2*(b*c*sgn(x) - 6*a*d*sgn(x))*
log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))/(sqrt(a)*c^4)