3.2237 \(\int \frac{(A+B x) (d+e x)^{3/2}}{\sqrt{a+b x}} \, dx\)

Optimal. Leaf size=193 \[ -\frac{(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{7/2} e^{3/2}}-\frac{\sqrt{a+b x} (d+e x)^{3/2} (5 a B e-6 A b e+b B d)}{12 b^2 e}-\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (5 a B e-6 A b e+b B d)}{8 b^3 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e} \]

[Out]

-((b*d - a*e)*(b*B*d - 6*A*b*e + 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^3*e) - ((b*B*d - 6*A*b*e + 5*a*B*e
)*Sqrt[a + b*x]*(d + e*x)^(3/2))/(12*b^2*e) + (B*Sqrt[a + b*x]*(d + e*x)^(5/2))/(3*b*e) - ((b*d - a*e)^2*(b*B*
d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))

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Rubi [A]  time = 0.148768, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {80, 50, 63, 217, 206} \[ -\frac{(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{7/2} e^{3/2}}-\frac{\sqrt{a+b x} (d+e x)^{3/2} (5 a B e-6 A b e+b B d)}{12 b^2 e}-\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e) (5 a B e-6 A b e+b B d)}{8 b^3 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*d - a*e)*(b*B*d - 6*A*b*e + 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^3*e) - ((b*B*d - 6*A*b*e + 5*a*B*e
)*Sqrt[a + b*x]*(d + e*x)^(3/2))/(12*b^2*e) + (B*Sqrt[a + b*x]*(d + e*x)^(5/2))/(3*b*e) - ((b*d - a*e)^2*(b*B*
d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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])

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^{3/2}}{\sqrt{a+b x}} \, dx &=\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}+\frac{\left (3 A b e-B \left (\frac{b d}{2}+\frac{5 a e}{2}\right )\right ) \int \frac{(d+e x)^{3/2}}{\sqrt{a+b x}} \, dx}{3 b e}\\ &=-\frac{(b B d-6 A b e+5 a B e) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}-\frac{((b d-a e) (b B d-6 A b e+5 a B e)) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x}} \, dx}{8 b^2 e}\\ &=-\frac{(b d-a e) (b B d-6 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{8 b^3 e}-\frac{(b B d-6 A b e+5 a B e) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}-\frac{\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{16 b^3 e}\\ &=-\frac{(b d-a e) (b B d-6 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{8 b^3 e}-\frac{(b B d-6 A b e+5 a B e) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}-\frac{\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{8 b^4 e}\\ &=-\frac{(b d-a e) (b B d-6 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{8 b^3 e}-\frac{(b B d-6 A b e+5 a B e) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}-\frac{\left ((b d-a e)^2 (b B d-6 A b e+5 a B e)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{8 b^4 e}\\ &=-\frac{(b d-a e) (b B d-6 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{8 b^3 e}-\frac{(b B d-6 A b e+5 a B e) \sqrt{a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac{B \sqrt{a+b x} (d+e x)^{5/2}}{3 b e}-\frac{(b d-a e)^2 (b B d-6 A b e+5 a B e) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{8 b^{7/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.572491, size = 182, normalized size = 0.94 \[ \frac{\sqrt{d+e x} \left (\sqrt{e} \sqrt{a+b x} \left (15 a^2 B e^2-2 a b e (9 A e+11 B d+5 B e x)+b^2 \left (6 A e (5 d+2 e x)+B \left (3 d^2+14 d e x+8 e^2 x^2\right )\right )\right )-\frac{3 (b d-a e)^{3/2} (5 a B e-6 A b e+b B d) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{\frac{b (d+e x)}{b d-a e}}}\right )}{24 b^3 e^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(Sqrt[e]*Sqrt[a + b*x]*(15*a^2*B*e^2 - 2*a*b*e*(11*B*d + 9*A*e + 5*B*e*x) + b^2*(6*A*e*(5*d + 2
*e*x) + B*(3*d^2 + 14*d*e*x + 8*e^2*x^2))) - (3*(b*d - a*e)^(3/2)*(b*B*d - 6*A*b*e + 5*a*B*e)*ArcSinh[(Sqrt[e]
*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/Sqrt[(b*(d + e*x))/(b*d - a*e)]))/(24*b^3*e^(3/2))

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Maple [B]  time = 0.019, size = 636, normalized size = 3.3 \begin{align*}{\frac{1}{48\,{b}^{3}e}\sqrt{ex+d}\sqrt{bx+a} \left ( 16\,B{x}^{2}{b}^{2}{e}^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+18\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}b{e}^{3}-36\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{2}d{e}^{2}+18\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{2}e+24\,A\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}x{b}^{2}{e}^{2}-15\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{3}{e}^{3}+27\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}bd{e}^{2}-9\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{2}{d}^{2}e-3\,B\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{3}-20\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}xab{e}^{2}+28\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}x{b}^{2}de-36\,A\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }ab{e}^{2}+60\,A\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{b}^{2}de+30\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}{a}^{2}{e}^{2}-44\,B\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }abde+6\,B\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x)

[Out]

1/48*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(16*B*x^2*b^2*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+18*A*ln(1/2*(2*b*x*e+2*
((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*e^3-36*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(
1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e^2+18*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+
a*e+b*d)/(b*e)^(1/2))*b^3*d^2*e+24*A*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*b^2*e^2-15*B*ln(1/2*(2*b*x*e+2*((b*
x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*e^3+27*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d*e^2-9*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d
)/(b*e)^(1/2))*a*b^2*d^2*e-3*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3
*d^3-20*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*a*b*e^2+28*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*x*b^2*d*e-36*
A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*e^2+60*A*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b^2*d*e+30*B*((b*x+a)*(
e*x+d))^(1/2)*(b*e)^(1/2)*a^2*e^2-44*B*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*d*e+6*B*(b*e)^(1/2)*((b*x+a)*(e
*x+d))^(1/2)*b^2*d^2)/b^3/e/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(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)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53768, size = 1215, normalized size = 6.3 \begin{align*} \left [-\frac{3 \,{\left (B b^{3} d^{3} + 3 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \,{\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} +{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \,{\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \,{\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \,{\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \,{\left (7 \, B b^{3} d e^{2} -{\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{96 \, b^{4} e^{2}}, \frac{3 \,{\left (B b^{3} d^{3} + 3 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \,{\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} +{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \,{\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \,{\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \,{\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \,{\left (7 \, B b^{3} d e^{2} -{\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{48 \, b^{4} e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a^3 - 6*A*a^2*b)*e
^3)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x +
a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(8*B*b^3*e^3*x^2 + 3*B*b^3*d^2*e - 2*(11*B*a*b^2 - 15*A*b^3)*
d*e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 2*(7*B*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*
x + d))/(b^4*e^2), 1/48*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a
^3 - 6*A*a^2*b)*e^3)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e
^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) + 2*(8*B*b^3*e^3*x^2 + 3*B*b^3*d^2*e - 2*(11*B*a*b^2 - 15*A*b^3)*d*
e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 2*(7*B*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x
+ d))/(b^4*e^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{\frac{3}{2}}}{\sqrt{a + b x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**(3/2)/sqrt(a + b*x), x)

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Giac [B]  time = 1.77904, size = 790, normalized size = 4.09 \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)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/48*(48*((b^2*d - a*b*e)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*
e)))/sqrt(b) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a))*A*d*abs(b)/b^2 - 2*(sqrt(b^2*d + (b*x + a)*b
*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)*e^(-4)/b^7) - 3*(b^7*d^2*
e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)*e^(-4)/b^7) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*e^(-
5/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(3/2))*B*abs(b)*e/b^2 -
(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e^(-2)/b^4 + (b*d*e - 5*a*e^2)*e^(-4)/b^4) + (
b^2*d^2 + 2*a*b*d*e - 3*a^2*e^2)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e
- a*b*e)))/b^(7/2))*B*d*abs(b)/b^3 - (sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e^(-2)/b^
4 + (b*d*e - 5*a*e^2)*e^(-4)/b^4) + (b^2*d^2 + 2*a*b*d*e - 3*a^2*e^2)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*
e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(7/2))*A*abs(b)*e/b^3)/b