\(\int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx\) [19]

Optimal result

Integrand size = 25, antiderivative size = 101 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\frac {2 f \sqrt {a+b x}}{d}+\frac {2 \sqrt {b c-a d} (d e-c f) \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{c d^{3/2}}-\frac {2 \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{c} \]

[Out]

-2*e*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)/c+2*(-c*f+d*e)*arctan(d^(1/2)*(b*x+a)^(1/2)/(-a*d+b*c)^(1/2))*(-a*
d+b*c)^(1/2)/c/d^(3/2)+2*f*(b*x+a)^(1/2)/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {159, 162, 65, 214, 211} \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\frac {2 \sqrt {b c-a d} (d e-c f) \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{c d^{3/2}}-\frac {2 \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{c}+\frac {2 f \sqrt {a+b x}}{d} \]

[In]

Int[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)),x]

[Out]

(2*f*Sqrt[a + b*x])/d + (2*Sqrt[b*c - a*d]*(d*e - c*f)*ArcTan[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(c*d^(
3/2)) - (2*Sqrt[a]*e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/c

Rule 65

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 159

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 162

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 f \sqrt {a+b x}}{d}+\frac {2 \int \frac {\frac {a d e}{2}+\frac {1}{2} (b d e-b c f+a d f) x}{x \sqrt {a+b x} (c+d x)} \, dx}{d} \\ & = \frac {2 f \sqrt {a+b x}}{d}+\frac {(a e) \int \frac {1}{x \sqrt {a+b x}} \, dx}{c}+\frac {((b c-a d) (d e-c f)) \int \frac {1}{\sqrt {a+b x} (c+d x)} \, dx}{c d} \\ & = \frac {2 f \sqrt {a+b x}}{d}+\frac {(2 a e) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b c}+\frac {(2 (b c-a d) (d e-c f)) \text {Subst}\left (\int \frac {1}{c-\frac {a d}{b}+\frac {d x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b c d} \\ & = \frac {2 f \sqrt {a+b x}}{d}+\frac {2 \sqrt {b c-a d} (d e-c f) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{c d^{3/2}}-\frac {2 \sqrt {a} e \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\frac {2 f \sqrt {a+b x}}{d}-\frac {2 \sqrt {b c-a d} (-d e+c f) \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{c d^{3/2}}-\frac {2 \sqrt {a} e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{c} \]

[In]

Integrate[(Sqrt[a + b*x]*(e + f*x))/(x*(c + d*x)),x]

[Out]

(2*f*Sqrt[a + b*x])/d - (2*Sqrt[b*c - a*d]*(-(d*e) + c*f)*ArcTan[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(c*
d^(3/2)) - (2*Sqrt[a]*e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/c

Maple [A] (verified)

Time = 5.38 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {2 f \sqrt {b x +a}}{d}-\frac {2 e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}}{c}-\frac {2 \left (a c d f -a e \,d^{2}-c^{2} b f +b c d e \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {\left (a d -b c \right ) d}}\right )}{d c \sqrt {\left (a d -b c \right ) d}}\)	\(103\)
default	\(\frac {2 f \sqrt {b x +a}}{d}-\frac {2 e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}}{c}-\frac {2 \left (a c d f -a e \,d^{2}-c^{2} b f +b c d e \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {\left (a d -b c \right ) d}}\right )}{d c \sqrt {\left (a d -b c \right ) d}}\)	\(103\)
pseudoelliptic	\(\frac {-2 \left (c f -d e \right ) \left (a d -b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {\left (a d -b c \right ) d}}\right )+2 \left (-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\, d e +\sqrt {b x +a}\, c f \right ) \sqrt {\left (a d -b c \right ) d}}{d c \sqrt {\left (a d -b c \right ) d}}\)	\(105\)

[In]

int((f*x+e)*(b*x+a)^(1/2)/x/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

2*f*(b*x+a)^(1/2)/d-2*e*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)/c-2/d*(a*c*d*f-a*d^2*e-b*c^2*f+b*c*d*e)/c/((a*d
-b*c)*d)^(1/2)*arctanh(d*(b*x+a)^(1/2)/((a*d-b*c)*d)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 450, normalized size of antiderivative = 4.46 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\left [\frac {\sqrt {a} d e \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a} c f - {\left (d e - c f\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {b d x - b c + 2 \, a d - 2 \, \sqrt {b x + a} d \sqrt {-\frac {b c - a d}{d}}}{d x + c}\right )}{c d}, \frac {2 \, \sqrt {-a} d e \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + 2 \, \sqrt {b x + a} c f - {\left (d e - c f\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {b d x - b c + 2 \, a d - 2 \, \sqrt {b x + a} d \sqrt {-\frac {b c - a d}{d}}}{d x + c}\right )}{c d}, \frac {\sqrt {a} d e \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a} c f - 2 \, {\left (d e - c f\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {\sqrt {b x + a} d \sqrt {\frac {b c - a d}{d}}}{b c - a d}\right )}{c d}, \frac {2 \, {\left (\sqrt {-a} d e \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \sqrt {b x + a} c f - {\left (d e - c f\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {\sqrt {b x + a} d \sqrt {\frac {b c - a d}{d}}}{b c - a d}\right )\right )}}{c d}\right ] \]

[In]

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

[Out]

[(sqrt(a)*d*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*sqrt(b*x + a)*c*f - (d*e - c*f)*sqrt(-(b*c - a*
d)/d)*log((b*d*x - b*c + 2*a*d - 2*sqrt(b*x + a)*d*sqrt(-(b*c - a*d)/d))/(d*x + c)))/(c*d), (2*sqrt(-a)*d*e*ar
ctan(sqrt(b*x + a)*sqrt(-a)/a) + 2*sqrt(b*x + a)*c*f - (d*e - c*f)*sqrt(-(b*c - a*d)/d)*log((b*d*x - b*c + 2*a
*d - 2*sqrt(b*x + a)*d*sqrt(-(b*c - a*d)/d))/(d*x + c)))/(c*d), (sqrt(a)*d*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a
) + 2*a)/x) + 2*sqrt(b*x + a)*c*f - 2*(d*e - c*f)*sqrt((b*c - a*d)/d)*arctan(-sqrt(b*x + a)*d*sqrt((b*c - a*d)
/d)/(b*c - a*d)))/(c*d), 2*(sqrt(-a)*d*e*arctan(sqrt(b*x + a)*sqrt(-a)/a) + sqrt(b*x + a)*c*f - (d*e - c*f)*sq
rt((b*c - a*d)/d)*arctan(-sqrt(b*x + a)*d*sqrt((b*c - a*d)/d)/(b*c - a*d)))/(c*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (90) = 180\).

Time = 12.59 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\begin {cases} \frac {2 a e \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{c \sqrt {- a}} + \frac {2 f \sqrt {a + b x}}{d} + \frac {2 \left (a d - b c\right ) \left (c f - d e\right ) \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- \frac {a d - b c}{d}}} \right )}}{c d^{2} \sqrt {- \frac {a d - b c}{d}}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\left (- f + \frac {d e}{2 c}\right ) \left (\frac {2 c \left (\begin {cases} - \frac {\frac {1}{x} + \frac {d}{2 c}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (2 c \left (\frac {1}{x} + \frac {d}{2 c}\right ) - d \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{d} - \frac {2 c \left (\begin {cases} \frac {\frac {1}{x} + \frac {d}{2 c}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (2 c \left (\frac {1}{x} + \frac {d}{2 c}\right ) + d \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{d}\right ) - \frac {e \log {\left (\frac {c}{x^{2}} + \frac {d}{x} \right )}}{2 c}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a*e*atan(sqrt(a + b*x)/sqrt(-a))/(c*sqrt(-a)) + 2*f*sqrt(a + b*x)/d + 2*(a*d - b*c)*(c*f - d*e)*a
tan(sqrt(a + b*x)/sqrt(-(a*d - b*c)/d))/(c*d**2*sqrt(-(a*d - b*c)/d)), Ne(b, 0)), (sqrt(a)*((-f + d*e/(2*c))*(
2*c*Piecewise((-(1/x + d/(2*c))/d, Eq(c, 0)), (log(2*c*(1/x + d/(2*c)) - d)/(2*c), True))/d - 2*c*Piecewise(((
1/x + d/(2*c))/d, Eq(c, 0)), (log(2*c*(1/x + d/(2*c)) + d)/(2*c), True))/d) - e*log(c/x**2 + d/x)/(2*c)), True
))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\frac {2 \, a e \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} c} + \frac {2 \, \sqrt {b x + a} f}{d} + \frac {2 \, {\left (b c d e - a d^{2} e - b c^{2} f + a c d f\right )} \arctan \left (\frac {\sqrt {b x + a} d}{\sqrt {b c d - a d^{2}}}\right )}{\sqrt {b c d - a d^{2}} c d} \]

[In]

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

[Out]

2*a*e*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*c) + 2*sqrt(b*x + a)*f/d + 2*(b*c*d*e - a*d^2*e - b*c^2*f + a*c
*d*f)*arctan(sqrt(b*x + a)*d/sqrt(b*c*d - a*d^2))/(sqrt(b*c*d - a*d^2)*c*d)

Mupad [B] (verification not implemented)

Time = 3.29 (sec) , antiderivative size = 2355, normalized size of antiderivative = 23.32 \[ \int \frac {\sqrt {a+b x} (e+f x)}{x (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(2*f*(a + b*x)^(1/2))/d - (a^(1/2)*e*atan(((a^(1/2)*e*((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b
^4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d
^2*e*f - 2*a^2*b^2*c*d^3*e*f))/d + (a^(1/2)*e*((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/d + (8*a^(1/2)*e*(b^3
*c^3*d^3 - 2*a*b^2*c^2*d^4)*(a + b*x)^(1/2))/(c*d)))/c)*1i)/c + (a^(1/2)*e*((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 +
2*a^2*b^2*d^4*e^2 + b^4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*
d*f^2 + 4*a*b^3*c^2*d^2*e*f - 2*a^2*b^2*c*d^3*e*f))/d - (a^(1/2)*e*((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/
d - (8*a^(1/2)*e*(b^3*c^3*d^3 - 2*a*b^2*c^2*d^4)*(a + b*x)^(1/2))/(c*d)))/c)*1i)/c)/((16*(a^2*b^3*d^3*e^3 - a*
b^4*c*d^2*e^3 - a*b^4*c^3*e*f^2 + a^3*b^2*d^3*e^2*f - 3*a^2*b^3*c*d^2*e^2*f + 2*a^2*b^3*c^2*d*e*f^2 - a^3*b^2*
c*d^2*e*f^2 + 2*a*b^4*c^2*d*e^2*f))/d - (a^(1/2)*e*((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b^4*
c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d^2*
e*f - 2*a^2*b^2*c*d^3*e*f))/d + (a^(1/2)*e*((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/d + (8*a^(1/2)*e*(b^3*c^
3*d^3 - 2*a*b^2*c^2*d^4)*(a + b*x)^(1/2))/(c*d)))/c))/c + (a^(1/2)*e*((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 + 2*a^2*
b^2*d^4*e^2 + b^4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*d*f^2
+ 4*a*b^3*c^2*d^2*e*f - 2*a^2*b^2*c*d^3*e*f))/d - (a^(1/2)*e*((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/d - (8
*a^(1/2)*e*(b^3*c^3*d^3 - 2*a*b^2*c^2*d^4)*(a + b*x)^(1/2))/(c*d)))/c))/c))*2i)/c - (atan(((((8*(a + b*x)^(1/2
)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b^4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e
^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d^2*e*f - 2*a^2*b^2*c*d^3*e*f))/d + (((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d
^3*f))/d + (8*(b^3*c^3*d^3 - 2*a*b^2*c^2*d^4)*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*(a + b*x)^(1/2))/(c*d^4))*(c
*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d^3))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*1i)/(c*d^3) + (((8*(a + b*x)^(
1/2)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b^4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^
3*e^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d^2*e*f - 2*a^2*b^2*c*d^3*e*f))/d - (((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^
2*d^3*f))/d - (8*(b^3*c^3*d^3 - 2*a*b^2*c^2*d^4)*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*(a + b*x)^(1/2))/(c*d^4))
*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d^3))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*1i)/(c*d^3))/((16*(a^2*b^3*
d^3*e^3 - a*b^4*c*d^2*e^3 - a*b^4*c^3*e*f^2 + a^3*b^2*d^3*e^2*f - 3*a^2*b^3*c*d^2*e^2*f + 2*a^2*b^3*c^2*d*e*f^
2 - a^3*b^2*c*d^2*e*f^2 + 2*a*b^4*c^2*d*e^2*f))/d - (((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b^
4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d^
2*e*f - 2*a^2*b^2*c*d^3*e*f))/d + (((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/d + (8*(b^3*c^3*d^3 - 2*a*b^2*c^
2*d^4)*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*(a + b*x)^(1/2))/(c*d^4))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d
^3))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d^3) + (((8*(a + b*x)^(1/2)*(b^4*c^4*f^2 + 2*a^2*b^2*d^4*e^2 + b^
4*c^2*d^2*e^2 - 2*b^4*c^3*d*e*f + a^2*b^2*c^2*d^2*f^2 - 2*a*b^3*c*d^3*e^2 - 2*a*b^3*c^3*d*f^2 + 4*a*b^3*c^2*d^
2*e*f - 2*a^2*b^2*c*d^3*e*f))/d - (((8*(a*b^3*c^3*d^2*f - a^2*b^2*c^2*d^3*f))/d - (8*(b^3*c^3*d^3 - 2*a*b^2*c^
2*d^4)*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*(a + b*x)^(1/2))/(c*d^4))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d
^3))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2))/(c*d^3)))*(c*f - d*e)*(d^3*(a*d - b*c))^(1/2)*2i)/(c*d^3)