\(\int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx\) [79]

Optimal result

Integrand size = 36, antiderivative size = 188 \[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}-\frac {(2 a C d f+b (C d e+c C f-2 B d f)) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{b^2 d^{3/2} f^{3/2}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {b e-a f} \sqrt {c+d x}}{\sqrt {b c-a d} \sqrt {e+f x}}\right )}{b^2 \sqrt {b c-a d} \sqrt {b e-a f}} \] Output:

C*(d*x+c)^(1/2)*(f*x+e)^(1/2)/b/d/f-(2*a*C*d*f+b*(-2*B*d*f+C*c*f+C*d*e))*a 
rctanh(f^(1/2)*(d*x+c)^(1/2)/d^(1/2)/(f*x+e)^(1/2))/b^2/d^(3/2)/f^(3/2)-2* 
(A*b^2-a*(B*b-C*a))*arctanh((-a*f+b*e)^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2 
)/(f*x+e)^(1/2))/b^2/(-a*d+b*c)^(1/2)/(-a*f+b*e)^(1/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {\frac {b C \sqrt {c+d x} \sqrt {e+f x}}{d f}+\frac {2 \left (A b^2+a (-b B+a C)\right ) \arctan \left (\frac {\sqrt {b c-a d} \sqrt {e+f x}}{\sqrt {-b e+a f} \sqrt {c+d x}}\right )}{\sqrt {b c-a d} \sqrt {-b e+a f}}-\frac {(2 a C d f+b (C d e+c C f-2 B d f)) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \sqrt {c+d x}}\right )}{d^{3/2} f^{3/2}}}{b^2} \] Input:

Integrate[(A + B*x + C*x^2)/((a + b*x)*Sqrt[c + d*x]*Sqrt[e + f*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2118, 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.

\(\displaystyle \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx\)

\(\Big \downarrow \) 2118

\(\displaystyle \frac {\int \frac {b (2 A b d f-a C (d e+c f)-(2 a C d f+b (C d e+c C f-2 B d f)) x)}{2 (a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{b^2 d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {2 A b d f-a C (d e+c f)-(2 a C d f+b (C d e+c C f-2 B d f)) x}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{2 b d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {\frac {2 d f \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{b}-\frac {(2 a C d f+b (-2 B d f+c C f+C d e)) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{b}}{2 b d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {2 d f \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{b}-\frac {2 (2 a C d f+b (-2 B d f+c C f+C d e)) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}}{2 b d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\frac {4 d f \left (A b^2-a (b B-a C)\right ) \int \frac {1}{-b c+a d+\frac {(b e-a f) (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}-\frac {2 (2 a C d f+b (-2 B d f+c C f+C d e)) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}}{2 b d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {4 d f \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right )}{b \sqrt {b c-a d} \sqrt {b e-a f}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (2 a C d f+b (-2 B d f+c C f+C d e))}{b \sqrt {d} \sqrt {f}}}{2 b d f}+\frac {C \sqrt {c+d x} \sqrt {e+f x}}{b d f}\)

Input:

Int[(A + B*x + C*x^2)/((a + b*x)*Sqrt[c + d*x]*Sqrt[e + f*x]),x]
 

Output:

(C*Sqrt[c + d*x]*Sqrt[e + f*x])/(b*d*f) + ((-2*(2*a*C*d*f + b*(C*d*e + c*C 
*f - 2*B*d*f))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/( 
b*Sqrt[d]*Sqrt[f]) - (4*(A*b^2 - a*(b*B - a*C))*d*f*ArcTanh[(Sqrt[b*e - a* 
f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e + f*x])])/(b*Sqrt[b*c - a*d]*Sqr 
t[b*e - a*f]))/(2*b*d*f)
 

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 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]
 

rule 2118

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f 
_.)*(x_))^(p_.), x_Symbol] :> With[{q = Expon[Px, x], k = Coeff[Px, x, Expo 
n[Px, x]]}, Simp[k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 
1)/(d*f*b^(q - 1)*(m + n + p + q + 1))), x] + Simp[1/(d*f*b^q*(m + n + p + 
q + 1))   Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + 
n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a + b*x)^(q 
- 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + 
 c*f*(p + 1))) + b*(a*d*f*(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m 
 + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] && PolyQ[Px, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(745\) vs. \(2(160)=320\).

Time = 1.06 (sec) , antiderivative size = 746, normalized size of antiderivative = 3.97

method	result	size
default	\(-\frac {\left (2 A \sqrt {d f}\, \ln \left (\frac {-2 a d f x +b c f x +b d e x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, b -a c f -a d e +2 b c e}{b x +a}\right ) b^{2} d f -2 B \sqrt {d f}\, \ln \left (\frac {-2 a d f x +b c f x +b d e x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, b -a c f -a d e +2 b c e}{b x +a}\right ) a b d f -2 B \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) b^{2} d f +2 C \sqrt {d f}\, \ln \left (\frac {-2 a d f x +b c f x +b d e x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, b -a c f -a d e +2 b c e}{b x +a}\right ) a^{2} d f +2 C \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) a b d f +C \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) b^{2} c f +C \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) b^{2} d e -2 C \sqrt {d f}\, \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, b^{2}\right ) \sqrt {x d +c}\, \sqrt {f x +e}}{2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, b^{3} \sqrt {d f}\, \sqrt {\frac {a^{2} d f -a b c f -a b d e +c e \,b^{2}}{b^{2}}}\, d f}\)	\(746\)

Input:

int((C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(1/2)/(f*x+e)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

-1/2*(2*A*(d*f)^(1/2)*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+2*((f*x+e)*(d*x+c))^( 
1/2)*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*b-a*c*f-a*d*e+2*b*c*e)/ 
(b*x+a))*b^2*d*f-2*B*(d*f)^(1/2)*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+2*((f*x+e) 
*(d*x+c))^(1/2)*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*b-a*c*f-a*d* 
e+2*b*c*e)/(b*x+a))*a*b*d*f-2*B*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1 
/2)*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^( 
1/2))*b^2*d*f+2*C*(d*f)^(1/2)*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+2*((f*x+e)*(d 
*x+c))^(1/2)*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*b-a*c*f-a*d*e+2 
*b*c*e)/(b*x+a))*a^2*d*f+2*C*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2) 
*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2 
))*a*b*d*f+C*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*ln(1/2*(2*d*f*x 
+2*((f*x+e)*(d*x+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*b^2*c*f+C*((a 
^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*ln(1/2*(2*d*f*x+2*((f*x+e)*(d*x 
+c))^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*b^2*d*e-2*C*(d*f)^(1/2)*((a^2 
*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*b^2)*(d*x 
+c)^(1/2)*(f*x+e)^(1/2)/((f*x+e)*(d*x+c))^(1/2)/b^3/(d*f)^(1/2)/((a^2*d*f- 
a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)/d/f
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\int \frac {A + B x + C x^{2}}{\left (a + b x\right ) \sqrt {c + d x} \sqrt {e + f x}}\, dx \] Input:

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

Output:

Integral((A + B*x + C*x**2)/((a + b*x)*sqrt(c + d*x)*sqrt(e + f*x)), x)
 

Maxima [F(-2)]

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

integrate((C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(1/2)/(f*x+e)^(1/2),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(((-(2*a*d*f)/b^2)>0)', see `assu 
me?` for m
 

Giac [F(-2)]

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

integrate((C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(1/2)/(f*x+e)^(1/2),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 [B] (verification not implemented)

Time = 159.40 (sec) , antiderivative size = 96118, normalized size of antiderivative = 511.27 \[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\text {Too large to display} \] Input:

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

Output:

(((2*C*c*f + 2*C*d*e)*((c + d*x)^(1/2) - c^(1/2)))/(b*f^3*((e + f*x)^(1/2) 
 - e^(1/2))) + ((2*C*c*f + 2*C*d*e)*((c + d*x)^(1/2) - c^(1/2))^3)/(b*d*f^ 
2*((e + f*x)^(1/2) - e^(1/2))^3) - (8*C*c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - 
 c^(1/2))^2)/(b*f^2*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d*x)^(1/2) - c^ 
(1/2))^4/((e + f*x)^(1/2) - e^(1/2))^4 + d^2/f^2 - (2*d*((c + d*x)^(1/2) - 
 c^(1/2))^2)/(f*((e + f*x)^(1/2) - e^(1/2))^2)) + (A*atan((c^(1/2)*d*e^(1/ 
2)*(b^2*c*e + a^2*d*f - a*b*c*f - a*b*d*e)^(1/2)*2i - (c*f*((c + d*x)^(1/2 
) - c^(1/2))*(b^2*c*e + a^2*d*f - a*b*c*f - a*b*d*e)^(1/2)*2i)/((e + f*x)^ 
(1/2) - e^(1/2)) - (d*e*((c + d*x)^(1/2) - c^(1/2))*(b^2*c*e + a^2*d*f - a 
*b*c*f - a*b*d*e)^(1/2)*2i)/((e + f*x)^(1/2) - e^(1/2)) + (c^(1/2)*e^(1/2) 
*f*((c + d*x)^(1/2) - c^(1/2))^2*(b^2*c*e + a^2*d*f - a*b*c*f - a*b*d*e)^( 
1/2)*2i)/((e + f*x)^(1/2) - e^(1/2))^2)/(a*d^2*e + a*c*d*f - 2*b*c*d*e + ( 
a*c*f^2*((c + d*x)^(1/2) - c^(1/2))^2)/((e + f*x)^(1/2) - e^(1/2))^2 + (a* 
d*e*f*((c + d*x)^(1/2) - c^(1/2))^2)/((e + f*x)^(1/2) - e^(1/2))^2 - (2*b* 
c*e*f*((c + d*x)^(1/2) - c^(1/2))^2)/((e + f*x)^(1/2) - e^(1/2))^2 + (2*b* 
c^(1/2)*d*e^(3/2)*((c + d*x)^(1/2) - c^(1/2)))/((e + f*x)^(1/2) - e^(1/2)) 
 + (2*b*c^(3/2)*e^(1/2)*f*((c + d*x)^(1/2) - c^(1/2)))/((e + f*x)^(1/2) - 
e^(1/2)) - (4*a*c^(1/2)*d*e^(1/2)*f*((c + d*x)^(1/2) - c^(1/2)))/((e + f*x 
)^(1/2) - e^(1/2))))*2i)/((a*d - b*c)*(a*f - b*e))^(1/2) - (B*atan(((B*((6 
5536*((c + d*x)^(1/2) - c^(1/2))*(32*B^4*a^2*b^2*d^8*e^6 + 6*B^4*a^4*c^...
 

Reduce [F]

\[ \int \frac {A+B x+C x^2}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {2 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right )+\left (\int \frac {x^{2}}{\sqrt {f x +e}\, \sqrt {d x +c}\, a +\sqrt {f x +e}\, \sqrt {d x +c}\, b x}d x \right ) c d f}{d f} \] Input:

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

Output:

(2*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqr 
t(c*f - d*e)) + int(x**2/(sqrt(e + f*x)*sqrt(c + d*x)*a + sqrt(e + f*x)*sq 
rt(c + d*x)*b*x),x)*c*d*f)/(d*f)