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

Optimal. Leaf size=171 \[ -\frac{\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{3/2}}+2 b^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}-\frac{\sqrt{a+b x} \sqrt{c+d x} (a d+3 b c)}{4 c x} \]

[Out]

-((3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c*x) - ((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*x^2) - ((3*b^2*c^2 +
6*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]*c^(3/2)) + 2*b^(3/2
)*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]

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Rubi [A]  time = 0.105918, antiderivative size = 171, 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 = {97, 149, 157, 63, 217, 206, 93, 208} \[ -\frac{\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{3/2}}+2 b^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}-\frac{\sqrt{a+b x} \sqrt{c+d x} (a d+3 b c)}{4 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c*x) - ((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*x^2) - ((3*b^2*c^2 +
6*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*Sqrt[a]*c^(3/2)) + 2*b^(3/2
)*Sqrt[d]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]

Rule 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (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)^{3/2} \sqrt{c+d x}}{x^3} \, dx &=-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}+\frac{1}{2} \int \frac{\sqrt{a+b x} \left (\frac{1}{2} (3 b c+a d)+2 b d x\right )}{x^2 \sqrt{c+d x}} \, dx\\ &=-\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}+\frac{\int \frac{\frac{1}{4} \left (3 b^2 c^2+6 a b c d-a^2 d^2\right )+2 b^2 c d x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c}\\ &=-\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}+\left (b^2 d\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 c}\\ &=-\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}+(2 b 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 )+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\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}\\ &=-\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}-\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{3/2}}+(2 b 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 )\\ &=-\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 x^2}-\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{3/2}}+2 b^{3/2} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )\\ \end{align*}

Mathematica [A]  time = 1.19271, size = 185, normalized size = 1.08 \[ \frac{1}{4} \left (\frac{\left (a^2 d^2-6 a b c d-3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} (2 a c+a d x+5 b c x)}{c x^2}+\frac{8 \sqrt{d} (b c-a d)^{3/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{(c+d x)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c + 5*b*c*x + a*d*x))/(c*x^2)) + (8*Sqrt[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]])/(c + d*x)^(3/2) + ((-3*b^2*c^2 - 6*a*
b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(3/2)))/4

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Maple [B]  time = 0.014, size = 400, normalized size = 2.3 \begin{align*}{\frac{1}{8\,c{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 8\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{2}cd\sqrt{ac}+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac \right ) } \right ){x}^{2}{a}^{2}{d}^{2}\sqrt{bd}-6\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{2}abcd\sqrt{bd}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{2}{b}^{2}{c}^{2}\sqrt{bd}-2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xad-10\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xbc-4\,\sqrt{d{x}^{2}b+adx+bcx+ac}ac\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\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)^(3/2)*(d*x+c)^(1/2)/x^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.93416, size = 2350, normalized size = 13.74 \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)^(3/2)*(d*x+c)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(b*d)*a*b*c^2*x^2*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*x + b*c + a*d)*sqr
t(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^
2*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 + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqr
t(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/(a*c^2*x^2), -1/16*(16*sqrt(-b*d)*a*b*c^2*x^2*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sq
rt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^2
*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 + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt
(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(a*c^2*x^2), 1/8*(4*sqrt(b*d)*a*b*c^2*x^2*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*x +
b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2
)*sqrt(-a*c)*x^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*
c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^2*x^
2), -1/8*(8*sqrt(-b*d)*a*b*c^2*x^2*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^
2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(-a*c)*x^2*arctan(1/2*(2
*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x))
+ 2*(2*a^2*c^2 + (5*a*b*c^2 + a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*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{3}{2}} \sqrt{c + d x}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 7.6229, size = 1521, normalized size = 8.89 \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)^(3/2)*(d*x+c)^(1/2)/x^3,x, algorithm="giac")

[Out]

-1/4*(4*sqrt(b*d)*b*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) + (3*sqrt(b*
d)*b^3*c^2*abs(b) + 6*sqrt(b*d)*a*b^2*c*d*abs(b) - sqrt(b*d)*a^2*b*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (s
qrt(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*
(5*sqrt(b*d)*b^9*c^5*abs(b) - 19*sqrt(b*d)*a*b^8*c^4*d*abs(b) + 26*sqrt(b*d)*a^2*b^7*c^3*d^2*abs(b) - 14*sqrt(
b*d)*a^3*b^6*c^2*d^3*abs(b) + sqrt(b*d)*a^4*b^5*c*d^4*abs(b) + sqrt(b*d)*a^5*b^4*d^5*abs(b) - 15*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^4*abs(b) + 16*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^6*c^3*d*abs(b) + 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^2*b^5*c^2*d^2*abs(b) - 8*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^4*c*d^3*abs(b) - 3*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^3*d^4*abs(b) + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4*b^5*c^3*abs(b) + 13*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^4*c^2*d*abs(b) + 17*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^3*c*d
^2*abs(b) + 3*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^2*d^3*abs(b) -
5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^3*c^2*abs(b) - 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*b^2*c*d*abs(b) - 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*d^2*abs(b))/((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
))/b