\(\int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx\) [220]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 a c^2 x}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 a c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 a x^3}+\frac {(b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{3/2} c^{5/2}} \] Output:

1/8*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c^2/x+1/12*(-a*d+b*c)*(b*x+ 
a)^(3/2)*(d*x+c)^(1/2)/a/c/x^2-1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/a/x^3+1/8*( 
-a*d+b*c)^3*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(3/2)/c 
^(5/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\frac {(-b c+a d)^3 \left (\frac {\sqrt {a} \sqrt {c} (a+b x)^{5/2} \sqrt {c+d x} \left (-3 c^2-\frac {8 a c (c+d x)}{a+b x}+\frac {3 a^2 (c+d x)^2}{(a+b x)^2}\right )}{(-b c x+a d x)^3}-3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )\right )}{24 a^{3/2} c^{5/2}} \] Input:

Integrate[((a + b*x)^(3/2)*Sqrt[c + d*x])/x^4,x]
 

Output:

((-(b*c) + a*d)^3*((Sqrt[a]*Sqrt[c]*(a + b*x)^(5/2)*Sqrt[c + d*x]*(-3*c^2 
- (8*a*c*(c + d*x))/(a + b*x) + (3*a^2*(c + d*x)^2)/(a + b*x)^2))/(-(b*c*x 
) + a*d*x)^3 - 3*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])]) 
)/(24*a^(3/2)*c^(5/2))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {105, 105, 105, 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.

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x)^(3/2)*Sqrt[c + d*x])/x^4,x]
 

Output:

-1/3*((a + b*x)^(3/2)*(c + d*x)^(3/2))/(c*x^3) + ((b*c - a*d)*(-1/2*(Sqrt[ 
a + b*x]*(c + d*x)^(3/2))/(c*x^2) + ((b*c - a*d)*(-((Sqrt[a + b*x]*Sqrt[c 
+ d*x])/(a*x)) + ((b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqr 
t[c + d*x])])/(a^(3/2)*Sqrt[c])))/(4*c)))/(2*c)
 

Defintions of rubi rules used

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 105

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

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. \(407\) vs. \(2(131)=262\).

Time = 0.22 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.50

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {x d +c}\, \left (3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{3}-9 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{3}+9 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{3}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{3}-6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+16 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}+6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}+4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, a^{2} c d x +28 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {a c}\, a b \,c^{2} x +16 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} c^{2} \sqrt {a c}\right )}{48 a \,c^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{3} \sqrt {a c}}\)	\(408\)

Input:

int((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c^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)*a^3*d^3*x^3-9*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^2*b*c*d^2*x^3+9*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^2*c^2*d*x^3-3*ln((a*d*x+b 
*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*b^3*c^3*x^3-6*((b*x+a 
)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2+16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^ 
(1/2)*a*b*c*d*x^2+6*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2+4*((b* 
x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c*d*x+28*((b*x+a)*(d*x+c))^(1/2)*(a*c) 
^(1/2)*a*b*c^2*x+16*((b*x+a)*(d*x+c))^(1/2)*a^2*c^2*(a*c)^(1/2))/((b*x+a)* 
(d*x+c))^(1/2)/x^3/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.67 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\left [-\frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a c} x^{3} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (8 \, a^{3} c^{3} + {\left (3 \, a b^{2} c^{3} + 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c^{3} + a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, a^{2} c^{3} x^{3}}, -\frac {3 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-a c} x^{3} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (8 \, a^{3} c^{3} + {\left (3 \, a b^{2} c^{3} + 8 \, a^{2} b c^{2} d - 3 \, a^{3} c d^{2}\right )} x^{2} + 2 \, {\left (7 \, a^{2} b c^{3} + a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, a^{2} c^{3} x^{3}}\right ] \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^4,x, algorithm="fricas")
 

Output:

[-1/96*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a*c)*x^ 
3*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*(8*a^3*c^3 + (3*a*b^2*c^3 + 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2* 
(7*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^3*x^3), - 
1/48*(3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-a*c)*x^3 
*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*(8*a^3*c^3 + (3*a*b^ 
2*c^3 + 8*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(7*a^2*b*c^3 + a^3*c^2*d)*x)* 
sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^3*x^3)]
 

Sympy [F]

\[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}} \sqrt {c + d x}}{x^{4}}\, dx \] Input:

integrate((b*x+a)**(3/2)*(d*x+c)**(1/2)/x**4,x)
 

Output:

Integral((a + b*x)**(3/2)*sqrt(c + d*x)/x**4, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^4,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2158 vs. \(2 (131) = 262\).

Time = 1.61 (sec) , antiderivative size = 2158, normalized size of antiderivative = 13.24 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^4,x, algorithm="giac")
 

Output:

1/24*b^3*(3*(sqrt(b*d)*b^3*c^3*abs(b) - 3*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 3 
*sqrt(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)*a^3*d^3*abs(b))*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)*a*b^4*c^2) - 2*(3*sqrt(b*d)*b^13 
*c^8*abs(b) - 10*sqrt(b*d)*a*b^12*c^7*d*abs(b) - 6*sqrt(b*d)*a^2*b^11*c^6* 
d^2*abs(b) + 78*sqrt(b*d)*a^3*b^10*c^5*d^3*abs(b) - 160*sqrt(b*d)*a^4*b^9* 
c^4*d^4*abs(b) + 162*sqrt(b*d)*a^5*b^8*c^3*d^5*abs(b) - 90*sqrt(b*d)*a^6*b 
^7*c^2*d^6*abs(b) + 26*sqrt(b*d)*a^7*b^6*c*d^7*abs(b) - 3*sqrt(b*d)*a^8*b^ 
5*d^8*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))^2*b^11*c^7*abs(b) + 33*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^10*c^6*d*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^2*b 
^9*c^5*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))^2*a^3*b^8*c^4*d^3*abs(b) - 93*sqrt(b*d)*(sqrt(b*d)*s 
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^7*c^3*d^4*abs( 
b) + 147*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - 
 a*b*d))^2*a^5*b^6*c^2*d^5*abs(b) - 81*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) 
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^5*c*d^6*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))^2*a^7*b 
^4*d^7*abs(b) + 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x}}{x^4} \,d x \] Input:

int(((a + b*x)^(3/2)*(c + d*x)^(1/2))/x^4,x)
 

Output:

int(((a + b*x)^(3/2)*(c + d*x)^(1/2))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 950, normalized size of antiderivative = 5.83 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^4} \, dx =\text {Too large to display} \] Input:

int((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^4,x)
 

Output:

( - 16*sqrt(c + d*x)*sqrt(a + b*x)*a**4*c**3*d - 4*sqrt(c + d*x)*sqrt(a + 
b*x)*a**4*c**2*d**2*x + 6*sqrt(c + d*x)*sqrt(a + b*x)*a**4*c*d**3*x**2 - 1 
6*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c**4 - 32*sqrt(c + d*x)*sqrt(a + b*x) 
*a**3*b*c**3*d*x - 10*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*c**2*d**2*x**2 - 
28*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*c**4*x - 22*sqrt(c + d*x)*sqrt(a 
+ b*x)*a**2*b**2*c**3*d*x**2 - 6*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**4*x 
**2 + 3*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))*a**4*d**4*x**3 - 
 6*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))*a**3*b*c*d**3*x**3 + 
6*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))*a*b**3*c**3*d*x**3 - 3 
*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**4*c**4*x**3 + 3*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))*a**4*d**4*x**3 - 6*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))*a**3*b*c*d**3*x**3 + 6*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))*a*b**3*c**3*d*x**3 - 3*sqrt(c)*sqrt(a)*l...