\(\int \frac {\sqrt {c+d x} (A+B x+C x^2)}{(a+b x) \sqrt {e+f x}} \, dx\) [72]

Optimal result

Integrand size = 36, antiderivative size = 292 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{(a+b x) \sqrt {e+f x}} \, dx=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {\left (2 d f (4 A b d f-a C (3 d e+c f))+\frac {(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 B d f))}{b}\right ) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{4 b^2 d^{3/2} f^{5/2}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b e-a f} \sqrt {c+d x}}{\sqrt {b c-a d} \sqrt {e+f x}}\right )}{b^3 \sqrt {b e-a f}} \] Output:

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

Mathematica [A] (verified)

Time = 12.04 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{(a+b x) \sqrt {e+f x}} \, dx=\frac {\frac {8 \left (A b^2+a (-b B+a C)\right ) \sqrt {d e-c f} \sqrt {\frac {d (e+f x)}{d e-c f}} \text {arcsinh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )}{\sqrt {f} \sqrt {e+f x}}+\frac {4 b (b C e-b B f+a C f) \sqrt {e+f x} \left (-\sqrt {f} \sqrt {d e-c f} (c+d x) \sqrt {\frac {d (e+f x)}{d e-c f}}+(d e-c f) \sqrt {c+d x} \text {arcsinh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )\right )}{f^{5/2} \sqrt {d e-c f} \sqrt {c+d x} \sqrt {\frac {d (e+f x)}{d e-c f}}}+\frac {b^2 C \sqrt {e+f x} \left (\sqrt {f} \sqrt {c+d x} (c f+d (e+2 f x))-\frac {(d e-c f)^{3/2} \text {arcsinh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )}{\sqrt {\frac {d (e+f x)}{d e-c f}}}\right )}{d f^{5/2}}-\frac {8 \left (A b^2+a (-b B+a C)\right ) \sqrt {-b c+a d} \text {arctanh}\left (\frac {\sqrt {-b e+a f} \sqrt {c+d x}}{\sqrt {-b c+a d} \sqrt {e+f x}}\right )}{\sqrt {-b e+a f}}}{4 b^3} \] Input:

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

Output:

((8*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[d*e - c*f]*Sqrt[(d*(e + f*x))/(d*e - c 
*f)]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d*e - c*f]])/(Sqrt[f]*Sqrt[e + f 
*x]) + (4*b*(b*C*e - b*B*f + a*C*f)*Sqrt[e + f*x]*(-(Sqrt[f]*Sqrt[d*e - c* 
f]*(c + d*x)*Sqrt[(d*(e + f*x))/(d*e - c*f)]) + (d*e - c*f)*Sqrt[c + d*x]* 
ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d*e - c*f]]))/(f^(5/2)*Sqrt[d*e - c*f 
]*Sqrt[c + d*x]*Sqrt[(d*(e + f*x))/(d*e - c*f)]) + (b^2*C*Sqrt[e + f*x]*(S 
qrt[f]*Sqrt[c + d*x]*(c*f + d*(e + 2*f*x)) - ((d*e - c*f)^(3/2)*ArcSinh[(S 
qrt[f]*Sqrt[c + d*x])/Sqrt[d*e - c*f]])/Sqrt[(d*(e + f*x))/(d*e - c*f)]))/ 
(d*f^(5/2)) - (8*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[-(b*c) + a*d]*ArcTanh[(Sq 
rt[-(b*e) + a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/Sqrt[ 
-(b*e) + a*f])/(4*b^3)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2118, 27, 171, 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.

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

\(\Big \downarrow \) 2118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

\(\displaystyle \frac {\frac {\frac {8 d f^2 (b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{b}+\frac {(2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e))) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}}dx}{b}}{2 b f}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{b f}}{4 b d f}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\frac {8 d f^2 (b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}}dx}{b}+\frac {2 (2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e))) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}}{2 b f}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{b f}}{4 b d f}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\frac {\frac {16 d f^2 (b c-a d) \left (A b^2-a (b B-a C)\right ) \int \frac {1}{-b c+a d+\frac {(b e-a f) (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}+\frac {2 (2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e))) \int \frac {1}{d-\frac {f (c+d x)}{e+f x}}d\frac {\sqrt {c+d x}}{\sqrt {e+f x}}}{b}}{2 b f}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{b f}}{4 b d f}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e)))}{b \sqrt {d} \sqrt {f}}-\frac {16 d f^2 \sqrt {b c-a d} \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right )}{b \sqrt {b e-a f}}}{2 b f}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{b f}}{4 b d f}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}\)

Input:

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

Output:

(C*(c + d*x)^(3/2)*Sqrt[e + f*x])/(2*b*d*f) + (-(((4*a*C*d*f + b*(3*C*d*e 
+ c*C*f - 4*B*d*f))*Sqrt[c + d*x]*Sqrt[e + f*x])/(b*f)) + ((2*(2*b*d*f*(4* 
A*b*d*f - a*C*(3*d*e + c*f)) + (b*d*e - b*c*f + 2*a*d*f)*(4*a*C*d*f + b*(3 
*C*d*e + c*C*f - 4*B*d*f)))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[ 
e + f*x])])/(b*Sqrt[d]*Sqrt[f]) - (16*(A*b^2 - a*(b*B - a*C))*d*Sqrt[b*c - 
 a*d]*f^2*ArcTanh[(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e 
+ f*x])])/(b*Sqrt[b*e - a*f]))/(2*b*f))/(4*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 171

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

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. \(1821\) vs. \(2(254)=508\).

Time = 0.92 (sec) , antiderivative size = 1822, normalized size of antiderivative = 6.24

method	result	size
default	\(\text {Expression too large to display}\)	\(1822\)

Input:

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

Output:

1/8*(8*A*((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^3*d^2*f^2+8*A*l 
n((-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))*(d*f)^(1/2)*a* 
b^2*d^2*f^2-8*A*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 
))*(d*f)^(1/2)*b^3*c*d*f^2-8*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))*a*b^2*d^2*f^2+4*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^3 
*c*d*f^2-4*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^3*d^2*e*f-8 
*B*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))*(d*f)^(1/2 
)*a^2*b*d^2*f^2+8*B*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))*(d*f)^(1/2)*a*b^2*c*d*f^2+4*C*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^ 
2)^(1/2)*(d*f)^(1/2)*((f*x+e)*(d*x+c))^(1/2)*b^3*d*f*x+8*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^2*b*d^2*f^2-4*C*((a^2*d*f-a*b*c*f-a*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(1/2)*(C*x^2+B*x+A)/(b*x+a)/(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. \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{(a+b x) \sqrt {e+f x}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*x+c)^(1/2)*(C*x^2+B*x+A)/(b*x+a)/(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 [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

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