\(\int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx\) [78]

Optimal result

Integrand size = 29, antiderivative size = 164 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=-\frac {(3 C d e+5 c C f-4 B d f) \sqrt {c+d x} \sqrt {e+f x}}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{4 d^{5/2} f^{5/2}} \] Output:

-1/4*(-4*B*d*f+5*C*c*f+3*C*d*e)*(d*x+c)^(1/2)*(f*x+e)^(1/2)/d^2/f^2+1/2*C* 
(d*x+c)^(3/2)*(f*x+e)^(1/2)/d^2/f+1/4*(C*(3*c^2*f^2+2*c*d*e*f+3*d^2*e^2)+4 
*d*f*(2*A*d*f-B*(c*f+d*e)))*arctanh(f^(1/2)*(d*x+c)^(1/2)/d^(1/2)/(f*x+e)^ 
(1/2))/d^(5/2)/f^(5/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {\sqrt {c+d x} \sqrt {e+f x} (4 B d f+C (-3 d e-3 c f+2 d f x))}{4 d^2 f^2}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \sqrt {c+d x}}\right )}{4 d^{5/2} f^{5/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*Sqrt[e + f*x]*(4*B*d*f + C*(-3*d*e - 3*c*f + 2*d*f*x)))/(4* 
d^2*f^2) + ((C*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^2) + 4*d*f*(2*A*d*f - B*(d 
*e + c*f)))*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/(Sqrt[f]*Sqrt[c + d*x])])/(4*d 
^(5/2)*f^(5/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1194, 27, 90, 66, 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.

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

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\int -\frac {C f c^2+3 C d e c-4 A d^2 f+d (3 C d e+5 c C f-4 B d f) x}{2 \sqrt {c+d x} \sqrt {e+f x}}dx}{2 d^2 f}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}-\frac {\int \frac {C f c^2+3 C d e c-4 A d^2 f+d (3 C d e+5 c C f-4 B d f) x}{\sqrt {c+d x} \sqrt {e+f x}}dx}{4 d^2 f}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} (-4 B d f+5 c C f+3 C d e)}{f}-\frac {\left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{2 f}}{4 d^2 f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} (-4 B d f+5 c C f+3 C d e)}{f}-\frac {\left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right ) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{f}}{4 d^2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}-\frac {\frac {\sqrt {c+d x} \sqrt {e+f x} (-4 B d f+5 c C f+3 C d e)}{f}-\frac {\text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{\sqrt {d} f^{3/2}}}{4 d^2 f}\)

Input:

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

Output:

(C*(c + d*x)^(3/2)*Sqrt[e + f*x])/(2*d^2*f) - (((3*C*d*e + 5*c*C*f - 4*B*d 
*f)*Sqrt[c + d*x]*Sqrt[e + f*x])/f - ((C*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^ 
2) + 4*d*f*(2*A*d*f - B*(d*e + c*f)))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqr 
t[d]*Sqrt[e + f*x])])/(Sqrt[d]*f^(3/2)))/(4*d^2*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 90

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

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 1194

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(138)=276\).

Time = 0.68 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.59

method	result	size
default	\(\frac {\left (8 A \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{2} f^{2}-4 B \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c d \,f^{2}-4 B \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{2} e f +4 C \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}\, d f x +3 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c^{2} f^{2}+2 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) c d e f +3 C \ln \left (\frac {2 d f x +2 \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, \sqrt {d f}+c f +d e}{2 \sqrt {d f}}\right ) d^{2} e^{2}+8 B \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d f -6 C \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, c f -6 C \sqrt {d f}\, \sqrt {\left (f x +e \right ) \left (x d +c \right )}\, d e \right ) \sqrt {x d +c}\, \sqrt {f x +e}}{8 \sqrt {d f}\, f^{2} d^{2} \sqrt {\left (f x +e \right ) \left (x d +c \right )}}\)	\(425\)

Input:

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

Output:

1/8*(8*A*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))*d^2*f^2-4*B*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))*c*d*f^2-4*B*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))*d^2*e*f+4*C*((f*x+e)*(d*x+c))^(1/2) 
*(d*f)^(1/2)*d*f*x+3*C*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))*c^2*f^2+2*C*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))*c*d*e*f+3*C*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))*d^2*e^2+8*B*(d*f)^(1/ 
2)*((f*x+e)*(d*x+c))^(1/2)*d*f-6*C*(d*f)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*c*f 
-6*C*(d*f)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*d*e)*(d*x+c)^(1/2)*(f*x+e)^(1/2)/ 
(d*f)^(1/2)/f^2/d^2/((f*x+e)*(d*x+c))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.32 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\left [\frac {{\left (3 \, C d^{2} e^{2} + 2 \, {\left (C c d - 2 \, B d^{2}\right )} e f + {\left (3 \, C c^{2} - 4 \, B c d + 8 \, A d^{2}\right )} f^{2}\right )} \sqrt {d f} \log \left (8 \, d^{2} f^{2} x^{2} + d^{2} e^{2} + 6 \, c d e f + c^{2} f^{2} + 4 \, {\left (2 \, d f x + d e + c f\right )} \sqrt {d f} \sqrt {d x + c} \sqrt {f x + e} + 8 \, {\left (d^{2} e f + c d f^{2}\right )} x\right ) + 4 \, {\left (2 \, C d^{2} f^{2} x - 3 \, C d^{2} e f - {\left (3 \, C c d - 4 \, B d^{2}\right )} f^{2}\right )} \sqrt {d x + c} \sqrt {f x + e}}{16 \, d^{3} f^{3}}, -\frac {{\left (3 \, C d^{2} e^{2} + 2 \, {\left (C c d - 2 \, B d^{2}\right )} e f + {\left (3 \, C c^{2} - 4 \, B c d + 8 \, A d^{2}\right )} f^{2}\right )} \sqrt {-d f} \arctan \left (\frac {{\left (2 \, d f x + d e + c f\right )} \sqrt {-d f} \sqrt {d x + c} \sqrt {f x + e}}{2 \, {\left (d^{2} f^{2} x^{2} + c d e f + {\left (d^{2} e f + c d f^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, C d^{2} f^{2} x - 3 \, C d^{2} e f - {\left (3 \, C c d - 4 \, B d^{2}\right )} f^{2}\right )} \sqrt {d x + c} \sqrt {f x + e}}{8 \, d^{3} f^{3}}\right ] \] Input:

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

Output:

[1/16*((3*C*d^2*e^2 + 2*(C*c*d - 2*B*d^2)*e*f + (3*C*c^2 - 4*B*c*d + 8*A*d 
^2)*f^2)*sqrt(d*f)*log(8*d^2*f^2*x^2 + d^2*e^2 + 6*c*d*e*f + c^2*f^2 + 4*( 
2*d*f*x + d*e + c*f)*sqrt(d*f)*sqrt(d*x + c)*sqrt(f*x + e) + 8*(d^2*e*f + 
c*d*f^2)*x) + 4*(2*C*d^2*f^2*x - 3*C*d^2*e*f - (3*C*c*d - 4*B*d^2)*f^2)*sq 
rt(d*x + c)*sqrt(f*x + e))/(d^3*f^3), -1/8*((3*C*d^2*e^2 + 2*(C*c*d - 2*B* 
d^2)*e*f + (3*C*c^2 - 4*B*c*d + 8*A*d^2)*f^2)*sqrt(-d*f)*arctan(1/2*(2*d*f 
*x + d*e + c*f)*sqrt(-d*f)*sqrt(d*x + c)*sqrt(f*x + e)/(d^2*f^2*x^2 + c*d* 
e*f + (d^2*e*f + c*d*f^2)*x)) - 2*(2*C*d^2*f^2*x - 3*C*d^2*e*f - (3*C*c*d 
- 4*B*d^2)*f^2)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^3)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((C*x^2+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(c*f-d*e>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {{\left (\sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \sqrt {d x + c} {\left (\frac {2 \, {\left (d x + c\right )} C}{d^{3} f} - \frac {3 \, C d^{6} e f + 5 \, C c d^{5} f^{2} - 4 \, B d^{6} f^{2}}{d^{8} f^{3}}\right )} - \frac {{\left (3 \, C d^{2} e^{2} + 2 \, C c d e f - 4 \, B d^{2} e f + 3 \, C c^{2} f^{2} - 4 \, B c d f^{2} + 8 \, A d^{2} f^{2}\right )} \log \left ({\left | -\sqrt {d f} \sqrt {d x + c} + \sqrt {d^{2} e + {\left (d x + c\right )} d f - c d f} \right |}\right )}{\sqrt {d f} d^{2} f^{2}}\right )} d}{4 \, {\left | d \right |}} \] Input:

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

Output:

1/4*(sqrt(d^2*e + (d*x + c)*d*f - c*d*f)*sqrt(d*x + c)*(2*(d*x + c)*C/(d^3 
*f) - (3*C*d^6*e*f + 5*C*c*d^5*f^2 - 4*B*d^6*f^2)/(d^8*f^3)) - (3*C*d^2*e^ 
2 + 2*C*c*d*e*f - 4*B*d^2*e*f + 3*C*c^2*f^2 - 4*B*c*d*f^2 + 8*A*d^2*f^2)*l 
og(abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt(d^2*e + (d*x + c)*d*f - c*d*f)))/(s 
qrt(d*f)*d^2*f^2))*d/abs(d)
 

Mupad [B] (verification not implemented)

Time = 18.25 (sec) , antiderivative size = 833, normalized size of antiderivative = 5.08 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx =\text {Too large to display} \] Input:

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

Output:

(((2*B*c*f + 2*B*d*e)*((c + d*x)^(1/2) - c^(1/2)))/(f^3*((e + f*x)^(1/2) - 
 e^(1/2))) + ((2*B*c*f + 2*B*d*e)*((c + d*x)^(1/2) - c^(1/2))^3)/(d*f^2*(( 
e + f*x)^(1/2) - e^(1/2))^3) - (8*B*c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c^( 
1/2))^2)/(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)) - ((((c + d*x)^(1/2) - c^(1/2))* 
((3*C*d^3*e^2)/2 + (3*C*c^2*d*f^2)/2 + C*c*d^2*e*f))/(f^6*((e + f*x)^(1/2) 
 - e^(1/2))) - (((c + d*x)^(1/2) - c^(1/2))^3*((11*C*c^2*f^2)/2 + (11*C*d^ 
2*e^2)/2 + 25*C*c*d*e*f))/(f^5*((e + f*x)^(1/2) - e^(1/2))^3) + (((c + d*x 
)^(1/2) - c^(1/2))^7*((3*C*c^2*f^2)/2 + (3*C*d^2*e^2)/2 + C*c*d*e*f))/(d^2 
*f^3*((e + f*x)^(1/2) - e^(1/2))^7) - (((c + d*x)^(1/2) - c^(1/2))^5*((11* 
C*c^2*f^2)/2 + (11*C*d^2*e^2)/2 + 25*C*c*d*e*f))/(d*f^4*((e + f*x)^(1/2) - 
 e^(1/2))^5) + (c^(1/2)*e^(1/2)*(32*C*c*f + 32*C*d*e)*((c + d*x)^(1/2) - c 
^(1/2))^4)/(f^4*((e + f*x)^(1/2) - e^(1/2))^4))/(((c + d*x)^(1/2) - c^(1/2 
))^8/((e + f*x)^(1/2) - e^(1/2))^8 + d^4/f^4 - (4*d*((c + d*x)^(1/2) - c^( 
1/2))^6)/(f*((e + f*x)^(1/2) - e^(1/2))^6) - (4*d^3*((c + d*x)^(1/2) - c^( 
1/2))^2)/(f^3*((e + f*x)^(1/2) - e^(1/2))^2) + (6*d^2*((c + d*x)^(1/2) - c 
^(1/2))^4)/(f^2*((e + f*x)^(1/2) - e^(1/2))^4)) - (4*A*atan((d*((e + f*x)^ 
(1/2) - e^(1/2)))/((-d*f)^(1/2)*((c + d*x)^(1/2) - c^(1/2)))))/(-d*f)^(1/2 
) - (2*B*atanh((f^(1/2)*((c + d*x)^(1/2) - c^(1/2)))/(d^(1/2)*((e + f*x...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.19 \[ \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx=\frac {4 \sqrt {f x +e}\, \sqrt {d x +c}\, b \,d^{2} f^{2}-3 \sqrt {f x +e}\, \sqrt {d x +c}\, c^{2} d \,f^{2}-3 \sqrt {f x +e}\, \sqrt {d x +c}\, c \,d^{2} e f +2 \sqrt {f x +e}\, \sqrt {d x +c}\, c \,d^{2} f^{2} x +8 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) a \,d^{2} f^{2}-4 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) b c d \,f^{2}-4 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) b \,d^{2} e f +3 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c^{3} f^{2}+2 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c^{2} d e f +3 \sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {f}\, \sqrt {d x +c}+\sqrt {d}\, \sqrt {f x +e}}{\sqrt {c f -d e}}\right ) c \,d^{2} e^{2}}{4 d^{3} f^{3}} \] Input:

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

Output:

(4*sqrt(e + f*x)*sqrt(c + d*x)*b*d**2*f**2 - 3*sqrt(e + f*x)*sqrt(c + d*x) 
*c**2*d*f**2 - 3*sqrt(e + f*x)*sqrt(c + d*x)*c*d**2*e*f + 2*sqrt(e + f*x)* 
sqrt(c + d*x)*c*d**2*f**2*x + 8*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) 
 + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*a*d**2*f**2 - 4*sqrt(f)*sqrt(d) 
*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*b*c* 
d*f**2 - 4*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f 
*x))/sqrt(c*f - d*e))*b*d**2*e*f + 3*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + 
 d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*c**3*f**2 + 2*sqrt(f)*sqrt 
(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e + f*x))/sqrt(c*f - d*e))*c 
**2*d*e*f + 3*sqrt(f)*sqrt(d)*log((sqrt(f)*sqrt(c + d*x) + sqrt(d)*sqrt(e 
+ f*x))/sqrt(c*f - d*e))*c*d**2*e**2)/(4*d**3*f**3)