\(\int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx\) [176]

Optimal result

Integrand size = 22, antiderivative size = 165 \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\frac {(3 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b}+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}-2 \sqrt {a} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2} \sqrt {d}} \] Output:

1/4*(a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b+1/2*(b*x+a)^(1/2)*(d*x+c)^(3 
/2)-2*a^(1/2)*c^(3/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2)) 
+1/4*(-a^2*d^2+6*a*b*c*d+3*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/ 
(d*x+c)^(1/2))/b^(3/2)/d^(1/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c+a d+2 b d x)}{b}-8 \sqrt {a} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {112, 27, 171, 27, 175, 66, 104, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 112

\(\displaystyle \frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}-\frac {1}{2} \int -\frac {\sqrt {c+d x} (4 a c+(3 b c+a d) x)}{2 x \sqrt {a+b x}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\sqrt {c+d x} (4 a c+(3 b c+a d) x)}{x \sqrt {a+b x}}dx+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {8 a b c^2+\left (3 b^2 c^2+6 a b d c-a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {8 a b c^2+\left (3 b^2 c^2+6 a b d c-a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{4} \left (\frac {\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+8 a b c^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {1}{4} \left (\frac {2 \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+8 a b c^2 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{4} \left (\frac {2 \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+16 a b c^2 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} \left (\frac {\frac {2 \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}}-16 \sqrt {a} b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{b}\right )+\frac {1}{2} \sqrt {a+b x} (c+d x)^{3/2}\)

Input:

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

Output:

(Sqrt[a + b*x]*(c + d*x)^(3/2))/2 + (((3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + 
 d*x])/b + (-16*Sqrt[a]*b*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a] 
*Sqrt[c + d*x])] + (2*(3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*S 
qrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*Sqrt[d]))/(2*b))/4
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 66

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

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 112

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

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[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] /; Fre 
eQ[{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 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(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 221

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(127)=254\).

Time = 0.22 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.01

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {x d +c}\, \left (\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a^{2} d^{2}-6 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a b c d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, b^{2} c^{2}+8 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b \,c^{2}-4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b d x -2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a d -10 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b c \right )}{8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b \sqrt {d b}\, \sqrt {a c}}\)	\(332\)

Input:

int((b*x+a)^(1/2)*(d*x+c)^(3/2)/x,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 985, normalized size of antiderivative = 5.97 \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(8*sqrt(a*c)*b^2*c*d*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) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^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*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) + 4*( 
2*b^2*d^2*x + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d), 1 
/8*(4*sqrt(a*c)*b^2*c*d*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) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(-b*d)* 
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)) + 2*(2*b^2*d^2*x + 5*b^2*c* 
d + a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d), 1/16*(16*sqrt(-a*c)*b^2 
*c*d*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)) - (3*b^2*c^2 + 6*a*b 
*c*d - a^2*d^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*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) + 4*(2*b^2*d^2*x + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a)*s 
qrt(d*x + c))/(b^2*d), 1/8*(8*sqrt(-a*c)*b^2*c*d*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)) - (3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*sqrt(-b*d)*...
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}{x}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(3/2)/x,x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(3/2)/x,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}}{x} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.08 \[ \int \frac {\sqrt {a+b x} (c+d x)^{3/2}}{x} \, dx=\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, a b \,d^{2}+5 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} c d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} d^{2} x +4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{2} c d +4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{2} c d -4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {b}\, \sqrt {d x +c}\, \sqrt {b x +a}+2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+2 b d x \right ) b^{2} c d -\sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} d^{2}+6 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a b c d +3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{2} c^{2}}{4 b^{2} d} \] Input:

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

Output:

(sqrt(c + d*x)*sqrt(a + b*x)*a*b*d**2 + 5*sqrt(c + d*x)*sqrt(a + b*x)*b**2 
*c*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*b**2*d**2*x + 4*sqrt(c)*sqrt(a)*log( 
- sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b 
*x) + sqrt(b)*sqrt(c + d*x))*b**2*c*d + 4*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt( 
d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)* 
sqrt(c + d*x))*b**2*c*d - 4*sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt(b)*sqrt(c + 
 d*x)*sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*b**2*c* 
d - sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*a**2*d**2 + 6*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + 
sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*c*d + 3*sqrt(d)*sqrt(b)*log((s 
qrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**2)/ 
(4*b**2*d)