3.679 \(\int \frac{(a+b x)^{5/2}}{x^3 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=177 \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{a \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

[Out]

-(a*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c^2*x) - (a*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*c*x^2) - (Sq
rt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(5/
2)) + (2*b^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d]

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Rubi [A]  time = 0.12347, antiderivative size = 177, 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, 149, 157, 63, 217, 206, 93, 208} \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{a \sqrt{a+b x} \sqrt{c+d x} (7 b c-3 a d)}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

-(a*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c^2*x) - (a*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*c*x^2) - (Sq
rt[a]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(5/
2)) + (2*b^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[d]

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 149

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*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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^3 \sqrt{c+d x}} \, dx &=-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}-\frac{\int \frac{\sqrt{a+b x} \left (-\frac{1}{2} a (7 b c-3 a d)-2 b^2 c x\right )}{x^2 \sqrt{c+d x}} \, dx}{2 c}\\ &=-\frac{a (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}-\frac{\int \frac{-\frac{1}{4} a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )-2 b^3 c^2 x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c^2}\\ &=-\frac{a (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}+b^3 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 c^2}\\ &=-\frac{a (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}+\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )+\frac{\left (a \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 c^2}\\ &=-\frac{a (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}-\frac{\sqrt{a} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )\\ &=-\frac{a (7 b c-3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c^2 x}-\frac{a (a+b x)^{3/2} \sqrt{c+d x}}{2 c x^2}-\frac{\sqrt{a} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 1.09919, size = 189, normalized size = 1.07 \[ -\frac{\sqrt{a} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{5/2}}+\frac{a \sqrt{a+b x} \sqrt{c+d x} (-2 a c+3 a d x-9 b c x)}{4 c^2 x^2}+\frac{2 (b c-a d)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^3*Sqrt[c + d*x]),x]

[Out]

(a*Sqrt[a + b*x]*Sqrt[c + d*x]*(-2*a*c - 9*b*c*x + 3*a*d*x))/(4*c^2*x^2) + (2*(b*c - a*d)^(5/2)*((b*(c + d*x))
/(b*c - a*d))^(5/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*(c + d*x)^(5/2)) - (Sqrt[a]*(15
*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(5/2))

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Maple [B]  time = 0.019, size = 354, normalized size = 2. \begin{align*}{\frac{1}{8\,{c}^{2}{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 8\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{3}{c}^{2}\sqrt{ac}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{3}{d}^{2}\sqrt{bd}+10\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}bcd\sqrt{bd}-15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{2}\sqrt{bd}+6\,x{a}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-18\,xabc\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-4\,{a}^{2}c\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^3/(d*x+c)^(1/2),x)

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2*(8*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^2*b^3*c^2*(a*c)^(1/2)-3*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^2*a^3*d^2*(b*
d)^(1/2)+10*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^2*a^2*b*c*d*(b*d)^(1/2)-15*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^2*a*b^2*c^2*(b*d)^(1/2)+6*x*a^2*d*((b*x+a)*(d*x+c
))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*x*a*b*c*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a^2*c*((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^2/(b*d)^(1/2)/(a*c)^(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((b*x+a)^(5/2)/x^3/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 13.0894, size = 2329, normalized size = 13.16 \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^3/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(8*b^2*c^2*x^2*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) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^
2*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*(2*a^2*c + 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)
*sqrt(d*x + c))/(c^2*x^2), -1/16*(16*b^2*c^2*x^2*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqr
t(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*sqr
t(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)*sq
rt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*a^2*c + 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)*sqrt
(d*x + c))/(c^2*x^2), 1/8*(4*b^2*c^2*x^2*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) + (15*b^2*c^2 - 10*a*b
*c*d + 3*a^2*d^2)*x^2*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)) - 2*(2*a^2*c + 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*
x^2), -1/8*(8*b^2*c^2*x^2*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)) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*x^2*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)) + 2*(2*a^2
*c + 3*(3*a*b*c - a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*x^2)]

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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^{3} \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**3/(d*x+c)**(1/2),x)

[Out]

Integral((a + b*x)**(5/2)/(x**3*sqrt(c + d*x)), x)

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Giac [B]  time = 7.57485, size = 1494, normalized size = 8.44 \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^3/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*sqrt(b*d)*b^2*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/d + (15*sqrt(b*d)
*a*b^3*c^2 - 10*sqrt(b*d)*a^2*b^2*c*d + 3*sqrt(b*d)*a^3*b*d^2)*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^2) + 2*(9*sqrt(b*d)*a
*b^9*c^5 - 39*sqrt(b*d)*a^2*b^8*c^4*d + 66*sqrt(b*d)*a^3*b^7*c^3*d^2 - 54*sqrt(b*d)*a^4*b^6*c^2*d^3 + 21*sqrt(
b*d)*a^5*b^5*c*d^4 - 3*sqrt(b*d)*a^6*b^4*d^5 - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*
b*d - a*b*d))^2*a*b^7*c^4 + 40*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^6*c^3*d + 10*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^5*c^2*d^2 -
32*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*c*d^3 + 9*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^3*d^4 + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^5*c^3 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a^2*b^4*c^2*d + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^4*a^3*b^3*c*d^2 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b
^2*d^3 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c^2 - 10*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^2*c*d + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(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)^2*c^2))*b/
abs(b)