3.678 \(\int \frac{(a+b x)^{5/2}}{x^2 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=162 \[ -\frac{a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}-\frac{b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (a d+b c)}{c d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154712, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {98, 154, 157, 63, 217, 206, 93, 208} \[ -\frac{a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}-\frac{b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (a d+b c)}{c d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^2*Sqrt[c + d*x]),x]

[Out]

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

Rule 98

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

Rule 154

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 \frac{(a+b x)^{5/2}}{x^2 \sqrt{c+d x}} \, dx &=-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{\int \frac{\sqrt{a+b x} \left (-\frac{1}{2} a (5 b c-a d)-b (b c+a d) x\right )}{x \sqrt{c+d x}} \, dx}{c}\\ &=\frac{b (b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{c d}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{\int \frac{-\frac{1}{2} a^2 d (5 b c-a d)+\frac{1}{2} b^2 c (b c-5 a d) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{c d}\\ &=\frac{b (b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{c d}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{\left (b^2 (b c-5 a d)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 d}+\frac{\left (a^2 (5 b c-a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c}\\ &=\frac{b (b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{c d}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{(b (b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{d}+\frac{\left (a^2 (5 b c-a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{c}\\ &=\frac{b (b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{c d}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}-\frac{(b (b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{d}\\ &=\frac{b (b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{c d}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{c x}-\frac{a^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}-\frac{b^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.820756, size = 176, normalized size = 1.09 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} \left (b^2 c x-a^2 d\right )}{c d x}+\frac{a^{3/2} (a d-5 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{3/2}}-\frac{(b c-5 a d) (b c-a d)^{3/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{3/2} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^2*Sqrt[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.017, size = 320, normalized size = 2. \begin{align*}{\frac{1}{2\,cdx}\sqrt{bx+a}\sqrt{dx+c} \left ( 5\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) xa{b}^{2}cd\sqrt{ac}-\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) x{b}^{3}{c}^{2}\sqrt{ac}+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) x{a}^{3}{d}^{2}\sqrt{bd}-5\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}bcd\sqrt{bd}+2\,x{b}^{2}c\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-2\,{a}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^2/(d*x+c)^(1/2),x)

[Out]

1/2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(5*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2
))*x*a*b^2*c*d*(a*c)^(1/2)-ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^3*c
^2*(a*c)^(1/2)+ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^3*d^2*(b*d)^(1/2)-5*ln((a*d
*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^2*b*c*d*(b*d)^(1/2)+2*x*b^2*c*((b*x+a)*(d*x+c))^(
1/2)*(b*d)^(1/2)*(a*c)^(1/2)-2*a^2*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/
x/(b*d)^(1/2)/(a*c)^(1/2)/d

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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((b*x+a)^(5/2)/x^2/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 14.1628, size = 2188, normalized size = 13.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((b^2*c^2 - 5*a*b*c*d)*x*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x + b*
c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) + (5*a*b*c*d - a^2*d^2)*x*sqrt(a
/c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(
d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(b^2*c*x - a^2*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d*x),
1/4*(2*(b^2*c^2 - 5*a*b*c*d)*x*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-
b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) - (5*a*b*c*d - a^2*d^2)*x*sqrt(a/c)*log((8*a^2*c^2 + (b^2*c^2 +
6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2
+ a^2*c*d)*x)/x^2) + 4*(b^2*c*x - a^2*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d*x), 1/4*(2*(5*a*b*c*d - a^2*d^2)*x*
sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b
*c + a^2*d)*x)) - (b^2*c^2 - 5*a*b*c*d)*x*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b
*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(b^2*c*x - a^2*d)
*sqrt(b*x + a)*sqrt(d*x + c))/(c*d*x), 1/2*((5*a*b*c*d - a^2*d^2)*x*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)
*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) + (b^2*c^2 - 5*a*b*c*d)*x*
sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c
+ a*b*d)*x)) + 2*(b^2*c*x - a^2*d)*sqrt(b*x + a)*sqrt(d*x + c))/(c*d*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{5}{2}}}{x^{2} \sqrt{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/x**2/(d*x+c)**(1/2),x)

[Out]

Integral((a + b*x)**(5/2)/(x**2*sqrt(c + d*x)), x)

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Giac [B]  time = 1.93411, size = 710, normalized size = 4.38 \begin{align*} \frac{b{\left (\frac{2 \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a} b}{d} + \frac{{\left (\sqrt{b d} b^{2} c - 5 \, \sqrt{b d} a b d\right )} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{d^{2}} - \frac{2 \,{\left (5 \, \sqrt{b d} a^{2} b^{2} c - \sqrt{b d} a^{3} b d\right )} \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b c} - \frac{4 \,{\left (\sqrt{b d} a^{2} b^{4} c^{2} - 2 \, \sqrt{b d} a^{3} b^{3} c d + \sqrt{b d} a^{4} b^{2} d^{2} - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} c - \sqrt{b d}{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \,{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2} a b d +{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} c}\right )}}{2 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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