\(\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx\) [380]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.93 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=-\frac {\sqrt {c+d x} \left (3 b^2 c^2 x+2 a^2 d^2 x+a b c (c-4 d x)\right )}{a^2 b x \sqrt {a+b x}}+\frac {c^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{5/2}}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{3/2}} \] Input:

Integrate[(c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {109, 27, 167, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x]
 

Output:

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

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 109

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

rule 167

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 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. \(501\) vs. \(2(131)=262\).

Time = 0.24 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.08

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \left (5 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{2} \sqrt {d b}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{2} \sqrt {d b}-2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b \,d^{3} x^{2} \sqrt {a c}+5 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} b \,c^{2} d x -3 \sqrt {d b}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{3} x -2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) \sqrt {a c}\, a^{3} d^{3} x +4 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a^{2} d^{2} x -8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b c d x +6 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, b^{2} c^{2} x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\, a b \,c^{2}\right )}{2 a^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x \sqrt {d b}\, \sqrt {a c}\, \sqrt {b x +a}\, b}\)	\(502\)

Input:

int((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(d*x+c)^(1/2)*(5*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^2*(d*b)^(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)*b^3*c^3*x^2*(d*b)^(1/2)-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^2*b*d^3*x^2* 
(a*c)^(1/2)+5*(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^2*b*c^2*d*x-3*(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^2*c^3*x-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^3*d^3*x+4* 
((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a^2*d^2*x-8*((b*x+a)*(d*x+ 
c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)*a*b*c*d*x+6*((b*x+a)*(d*x+c))^(1/2)*(d*b 
)^(1/2)*(a*c)^(1/2)*b^2*c^2*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^ 
(1/2)*a*b*c^2)/a^2/((b*x+a)*(d*x+c))^(1/2)/x/(d*b)^(1/2)/(a*c)^(1/2)/(b*x+ 
a)^(1/2)/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (131) = 262\).

Time = 0.95 (sec) , antiderivative size = 1234, normalized size of antiderivative = 7.57 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/4*(2*(a^2*b*d^2*x^2 + a^3*d^2*x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 
+ 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d 
*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - ((3*b^3*c^2 - 5*a*b^2*c*d)* 
x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 
 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)* 
sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(a*b*c^2 + (3* 
b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^2* 
x^2 + a^3*b*x), -1/4*(4*(a^2*b*d^2*x^2 + a^3*d^2*x)*sqrt(-d/b)*arctan(1/2* 
(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + 
a*c*d + (b*c*d + a*d^2)*x)) + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (3*a*b^2*c^ 
2 - 5*a^2*b*c*d)*x)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2* 
d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqr 
t(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(a*b*c^2 + (3*b^2*c^2 - 4*a*b*c 
*d + 2*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^2*x^2 + a^3*b*x), - 
1/2*(((3*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt( 
-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt( 
-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (a^2*b*d^2*x^2 + a^3*d^2* 
x)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2* 
d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + 
a*b*d^2)*x) + 2*(a*b*c^2 + (3*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2)*x)*sqrt(...
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{2} \left (a + b x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((d*x+c)**(5/2)/x**2/(b*x+a)**(3/2),x)
 

Output:

Integral((c + d*x)**(5/2)/(x**2*(a + b*x)**(3/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(5/2)/x^2/(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 [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),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. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^2\,{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:

int((c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x)
                                                                                    
                                                                                    
 

Output:

int((c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 850, normalized size of antiderivative = 5.21 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x)
 

Output:

(5*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**2*b* 
*2*c*d**2*x + 12*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**3*c**2*d*x - 9*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))*b**4*c**3*x + 5*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) + sqr 
t(b)*sqrt(c + d*x))*a**2*b**2*c*d**2*x + 12*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**3*c**2*d*x - 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)*s 
qrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**4*c**3*x - 5*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)*sq 
rt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*a**2*b**2*c*d**2*x - 12*sqrt(c)*sqrt(a)*s 
qrt(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*b**3*c**2*d*x + 9*sqrt(c)*sqrt(a)*sq 
rt(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)*b**4*c**3*x + 4*sqrt(d)*sqrt(b)*sqrt(a 
+ b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b...