\(\int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx\) [373]

Optimal result

Integrand size = 22, antiderivative size = 178 \[ \int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx=\frac {3 (b c-a d) (5 b c-a d) \sqrt {c+d x}}{4 a^3 c \sqrt {a+b x}}+\frac {(5 b c-a d) (c+d x)^{3/2}}{4 a^2 c x \sqrt {a+b x}}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}-\frac {3 (b c-a d) (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} \sqrt {c}} \] Output:

3/4*(-a*d+b*c)*(-a*d+5*b*c)*(d*x+c)^(1/2)/a^3/c/(b*x+a)^(1/2)+1/4*(-a*d+5* 
b*c)*(d*x+c)^(3/2)/a^2/c/x/(b*x+a)^(1/2)-1/2*(d*x+c)^(5/2)/a/c/x^2/(b*x+a) 
^(1/2)-3/4*(-a*d+b*c)*(-a*d+5*b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/( 
d*x+c)^(1/2))/a^(7/2)/c^(1/2)
 

Mathematica [A] (verified)

Time = 10.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (15 b^2 c x^2+a b x (5 c-13 d x)-a^2 (2 c+5 d x)\right )}{4 a^3 x^2 \sqrt {a+b x}}-\frac {3 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} \sqrt {c}} \] Input:

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

Output:

(Sqrt[c + d*x]*(15*b^2*c*x^2 + a*b*x*(5*c - 13*d*x) - a^2*(2*c + 5*d*x)))/ 
(4*a^3*x^2*Sqrt[a + b*x]) - (3*(5*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*a^(7/2)*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {107, 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.

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

\(\Big \downarrow \) 107

\(\displaystyle -\frac {(5 b c-a d) \int \frac {(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}}dx}{4 a c}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(5 b c-a d) \left (-\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{x (a+b x)^{3/2}}dx}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a c}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(5 b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {c \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a c}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {(5 b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 c \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{a}+\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a c}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(5 b c-a d) \left (-\frac {3 (b c-a d) \left (\frac {2 \sqrt {c+d x}}{a \sqrt {a+b x}}-\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}\right )}{2 a}-\frac {(c+d x)^{3/2}}{a x \sqrt {a+b x}}\right )}{4 a c}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}}\)

Input:

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

Output:

-1/2*(c + d*x)^(5/2)/(a*c*x^2*Sqrt[a + b*x]) - ((5*b*c - a*d)*(-((c + d*x) 
^(3/2)/(a*x*Sqrt[a + b*x])) - (3*(b*c - a*d)*((2*Sqrt[c + d*x])/(a*Sqrt[a 
+ b*x]) - (2*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x 
])])/a^(3/2)))/(2*a)))/(4*a*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 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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. \(463\) vs. \(2(146)=292\).

Time = 0.24 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.61

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \left (3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b \,d^{2} x^{3}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c d \,x^{3}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{3} c^{2} x^{3}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{3} d^{2} x^{2}-18 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b c d \,x^{2}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} x^{2}+26 a b d \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-30 b^{2} c \,x^{2} \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+10 a^{2} d x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-10 a b c x \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+4 a^{2} c \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\right )}{8 a^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {a c}\, \sqrt {b x +a}}\)	\(464\)

Input:

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

Output:

-1/8*(d*x+c)^(1/2)*(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^2*b*d^2*x^3-18*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*d*x^3+15*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^2*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)*a^3*d^2*x^2-18*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*x^2+15*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*x^2+26*a*b*d*x^2*(a*c) 
^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*b^2*c*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^ 
(1/2)+10*a^2*d*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*a*b*c*x*(a*c)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)+4*a^2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/a^3 
/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.66 \[ \int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{3} c^{2} - {\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{3} c^{2} - {\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}\right ] \] Input:

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

Output:

[1/16*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x^3 + (5*a*b^2*c^2 - 6*a^2 
*b*c*d + a^3*d^2)*x^2)*sqrt(a*c)*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^3*c^2 - (15*a*b^2*c^2 - 13*a^ 
2*b*c*d)*x^2 - 5*(a^2*b*c^2 - a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^ 
4*b*c*x^3 + a^5*c*x^2), 1/8*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x^3 
+ (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-a*c)*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^3*c^2 - (15*a*b^2*c^2 - 13*a^2*b*c 
*d)*x^2 - 5*(a^2*b*c^2 - a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b*c 
*x^3 + a^5*c*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(3/2)/x^3/(b*x+a)^(3/2),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. 1167 vs. \(2 (146) = 292\).

Time = 1.25 (sec) , antiderivative size = 1167, normalized size of antiderivative = 6.56 \[ \int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-3/4*(5*sqrt(b*d)*b^2*c^2*abs(b) - 6*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)* 
a^2*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)* 
a^3*b) + 4*(sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b 
*d)*a^2*d^2*abs(b))/((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)*a^3) + 1/2*(7*sqrt(b*d)*b^8*c^5*abs(b) - 33 
*sqrt(b*d)*a*b^7*c^4*d*abs(b) + 62*sqrt(b*d)*a^2*b^6*c^3*d^2*abs(b) - 58*s 
qrt(b*d)*a^3*b^5*c^2*d^3*abs(b) + 27*sqrt(b*d)*a^4*b^4*c*d^4*abs(b) - 5*sq 
rt(b*d)*a^5*b^3*d^5*abs(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt( 
b^2*c + (b*x + a)*b*d - a*b*d))^2*b^6*c^4*abs(b) + 32*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^5*c^3*d*abs(b) 
 + 14*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^4*c^2*d^2*abs(b) - 40*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s 
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^3*c*d^3*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^4*b^2* 
d^4*abs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^4*b^4*c^3*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*c^2*d*abs(b) + 11*sqrt(b*d)*(sq 
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c*d 
^2*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 953, normalized size of antiderivative = 5.35 \[ \int \frac {(c+d x)^{3/2}}{x^3 (a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(9*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqr 
t(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d* 
*3*x**2 - 39*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqrt(d)*sqrt(c)*s 
qrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x 
))*a**2*b*c*d**2*x**2 - 45*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log( - sqrt(2*sqr 
t(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**2*c**2*d*x**2 + 75*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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**3*c**3*x**2 + 9*sqrt(c)*sqrt(a)*sqrt(a + 
 b*x)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sq 
rt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**3*d**3*x**2 - 39*sqrt(c)*sqrt(a)*s 
qrt(a + b*x)*log(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqr 
t(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a**2*b*c*d**2*x**2 - 45*sqrt(c 
)*sqrt(a)*sqrt(a + b*x)*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**2*c**2*d*x**2 
+ 75*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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**3*c** 
3*x**2 - 9*sqrt(c)*sqrt(a)*sqrt(a + b*x)*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)*a**3*d**3* 
x**2 + 39*sqrt(c)*sqrt(a)*sqrt(a + b*x)*log(2*sqrt(d)*sqrt(b)*sqrt(c + ...