\(\int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx\) [43]

Optimal result

Integrand size = 37, antiderivative size = 163 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\frac {\left (C e^2-B e f+A f^2\right ) \sqrt {1-d^2 x^2}}{f \left (d^2 e^2-f^2\right ) (e+f x)}+\frac {C \arcsin (d x)}{d f^2}-\frac {\left (C d^2 e^3-2 C e f^2-A d^2 e f^2+B f^3\right ) \arctan \left (\frac {f+d^2 e x}{\sqrt {d^2 e^2-f^2} \sqrt {1-d^2 x^2}}\right )}{f^2 \left (d^2 e^2-f^2\right )^{3/2}} \] Output:

(A*f^2-B*e*f+C*e^2)*(-d^2*x^2+1)^(1/2)/f/(d^2*e^2-f^2)/(f*x+e)+C*arcsin(d* 
x)/d/f^2-(-A*d^2*e*f^2+C*d^2*e^3+B*f^3-2*C*e*f^2)*arctan((d^2*e*x+f)/(d^2* 
e^2-f^2)^(1/2)/(-d^2*x^2+1)^(1/2))/f^2/(d^2*e^2-f^2)^(3/2)
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\frac {\frac {f \left (C e^2+f (-B e+A f)\right ) \sqrt {1-d^2 x^2}}{(d e-f) (d e+f) (e+f x)}+\frac {2 C \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{d}+\frac {2 \sqrt {d^2 e^2-f^2} \left (C d^2 e^3-2 C e f^2-A d^2 e f^2+B f^3\right ) \arctan \left (\frac {\sqrt {d^2 e^2-f^2} x}{e+f x-e \sqrt {1-d^2 x^2}}\right )}{(-d e+f)^2 (d e+f)^2}}{f^2} \] Input:

Integrate[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^2),x]
 

Output:

((f*(C*e^2 + f*(-(B*e) + A*f))*Sqrt[1 - d^2*x^2])/((d*e - f)*(d*e + f)*(e 
+ f*x)) + (2*C*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/d + (2*Sqrt[d^2*e^2 
 - f^2]*(C*d^2*e^3 - 2*C*e*f^2 - A*d^2*e*f^2 + B*f^3)*ArcTan[(Sqrt[d^2*e^2 
 - f^2]*x)/(e + f*x - e*Sqrt[1 - d^2*x^2])])/((-(d*e) + f)^2*(d*e + f)^2)) 
/f^2
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {2112, 2182, 719, 223, 488, 217}

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.

\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {d x+1} (e+f x)^2} \, dx\)

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2182

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\arctan \left (\frac {d^2 e x+f}{\sqrt {1-d^2 x^2} \sqrt {d^2 e^2-f^2}}\right ) \left (A d^2 e-B f-\frac {C d^2 e^3}{f^2}+2 C e\right )}{\sqrt {d^2 e^2-f^2}}+\frac {C \arcsin (d x) (d e-f) (d e+f)}{d f^2}}{d^2 e^2-f^2}+\frac {\sqrt {1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}\)

Input:

Int[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^2),x]
 

Output:

((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(f*(d^2*e^2 - f^2)*(e + f*x)) 
+ ((C*(d*e - f)*(d*e + f)*ArcSin[d*x])/(d*f^2) + ((2*C*e + A*d^2*e - (C*d^ 
2*e^3)/f^2 - B*f)*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x 
^2])])/Sqrt[d^2*e^2 - f^2])/(d^2*e^2 - f^2)
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 719

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

rule 2112

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f 
_.)*(x_))^(p_.), x_Symbol] :> Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; F 
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] & 
& EqQ[m, n] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 2182

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, 
 d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* 
d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2))   Int[(d + e*x)^(m + 
1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b 
*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, 
 x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.16 (sec) , antiderivative size = 899, normalized size of antiderivative = 5.52

method	result	size
default	\(\frac {\left (-A \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d^{3} e \,f^{3} x +C \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d^{3} e^{3} f x -A \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d^{3} e^{2} f^{2}+C \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d^{3} e^{4}+C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2} e^{2} f^{2} x \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}+A \,\operatorname {csgn}\left (d \right ) d \,f^{4} \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}+B \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d \,f^{4} x -B \,\operatorname {csgn}\left (d \right ) d e \,f^{3} \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}-2 C \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d e \,f^{3} x +C \,\operatorname {csgn}\left (d \right ) d \,e^{2} f^{2} \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}+C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2} e^{3} f \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}+B \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d e \,f^{3}-2 C \,\operatorname {csgn}\left (d \right ) \ln \left (\frac {2 d^{2} x e +2 \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f +2 f}{f x +e}\right ) d \,e^{2} f^{2}-C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) f^{4} x \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}-C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) e \,f^{3} \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\right ) \operatorname {csgn}\left (d \right ) \sqrt {-x d +1}\, \sqrt {x d +1}}{\sqrt {-d^{2} x^{2}+1}\, \left (d e +f \right ) \left (d e -f \right ) \left (f x +e \right ) d \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f^{3}}\)	\(899\)

Input:

int((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^2,x,method=_RETURNV 
ERBOSE)
 

Output:

(-A*csgn(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+ 
f)/(f*x+e))*d^3*e*f^3*x+C*csgn(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2* 
e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^3*e^3*f*x-A*csgn(d)*ln(2*(d^2*x*e+(-d^ 
2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^3*e^2*f^2+C*csgn 
(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+ 
e))*d^3*e^4+C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*e^2*f^2*x*(-(d^2* 
e^2-f^2)/f^2)^(1/2)+A*csgn(d)*d*f^4*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2 
)^(1/2)+B*csgn(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1 
/2)*f+f)/(f*x+e))*d*f^4*x-B*csgn(d)*d*e*f^3*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2- 
f^2)/f^2)^(1/2)-2*C*csgn(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^ 
2)/f^2)^(1/2)*f+f)/(f*x+e))*d*e*f^3*x+C*csgn(d)*d*e^2*f^2*(-d^2*x^2+1)^(1/ 
2)*(-(d^2*e^2-f^2)/f^2)^(1/2)+C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2 
*e^3*f*(-(d^2*e^2-f^2)/f^2)^(1/2)+B*csgn(d)*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/ 
2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d*e*f^3-2*C*csgn(d)*ln(2*(d^2* 
x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d*e^2*f^2- 
C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*f^4*x*(-(d^2*e^2-f^2)/f^2)^(1/2)- 
C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*e*f^3*(-(d^2*e^2-f^2)/f^2)^(1/2)) 
*csgn(d)*(-d*x+1)^(1/2)*(d*x+1)^(1/2)/(-d^2*x^2+1)^(1/2)/(d*e+f)/(d*e-f)/( 
f*x+e)/d/(-(d^2*e^2-f^2)/f^2)^(1/2)/f^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (155) = 310\).

Time = 14.65 (sec) , antiderivative size = 1025, normalized size of antiderivative = 6.29 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[(C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e^ 
3*f^3 + (C*d^3*e^5 + B*d*e^2*f^3 - (A*d^3 + 2*C*d)*e^3*f^2 + (C*d^3*e^4*f 
+ B*d*e*f^4 - (A*d^3 + 2*C*d)*e^2*f^3)*x)*sqrt(-d^2*e^2 + f^2)*log((d^2*e* 
f*x + f^2 - sqrt(-d^2*e^2 + f^2)*(d^2*e*x + f) - (sqrt(-d^2*e^2 + f^2)*sqr 
t(-d*x + 1)*f + (d^2*e^2 - f^2)*sqrt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e)) 
+ (C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e 
^3*f^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) + (C*d^3*e^4*f^2 - B*d^3*e^3*f^3 + B* 
d*e*f^5 - A*d*f^6 + (A*d^3 - C*d)*e^2*f^4)*x - 2*(C*d^4*e^6 - 2*C*d^2*e^4* 
f^2 + C*e^2*f^4 + (C*d^4*e^5*f - 2*C*d^2*e^3*f^3 + C*e*f^5)*x)*arctan((sqr 
t(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^5*e^6*f^2 - 2*d^3*e^4*f^4 + d*e^ 
2*f^6 + (d^5*e^5*f^3 - 2*d^3*e^3*f^5 + d*e*f^7)*x), (C*d^3*e^5*f - B*d^3*e 
^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e^3*f^3 - 2*(C*d^3*e^5 + 
B*d*e^2*f^3 - (A*d^3 + 2*C*d)*e^3*f^2 + (C*d^3*e^4*f + B*d*e*f^4 - (A*d^3 
+ 2*C*d)*e^2*f^3)*x)*sqrt(d^2*e^2 - f^2)*arctan(-(sqrt(d^2*e^2 - f^2)*sqrt 
(d*x + 1)*sqrt(-d*x + 1)*e - sqrt(d^2*e^2 - f^2)*(f*x + e))/((d^2*e^2 - f^ 
2)*x)) + (C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - 
 C*d)*e^3*f^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) + (C*d^3*e^4*f^2 - B*d^3*e^3*f 
^3 + B*d*e*f^5 - A*d*f^6 + (A*d^3 - C*d)*e^2*f^4)*x - 2*(C*d^4*e^6 - 2*C*d 
^2*e^4*f^2 + C*e^2*f^4 + (C*d^4*e^5*f - 2*C*d^2*e^3*f^3 + C*e*f^5)*x)*arct 
an((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^5*e^6*f^2 - 2*d^3*e^4*...
 

Sympy [F]

\[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\int \frac {A + B x + C x^{2}}{\left (e + f x\right )^{2} \sqrt {- d x + 1} \sqrt {d x + 1}}\, dx \] Input:

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

Output:

Integral((A + B*x + C*x**2)/((e + f*x)**2*sqrt(-d*x + 1)*sqrt(d*x + 1)), x 
)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^2,x, algorith 
m="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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 34.20 (sec) , antiderivative size = 10198, normalized size of antiderivative = 62.56 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(A*d^5*e^5*atan(((f + d*e)^(3/2)*(f - d*e)^(3/2)*1i - (((1 - d*x)^(1/2) - 
1)^2*(f + d*e)^(3/2)*(f - d*e)^(3/2)*1i)/((d*x + 1)^(1/2) - 1)^2)/(f^3 - d 
^2*e^2*f - (f^3*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (2*d^3* 
e^3*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1) + (2*d*e*f^2*((1 - d*x)^( 
1/2) - 1))/((d*x + 1)^(1/2) - 1) + (d^2*e^2*f*((1 - d*x)^(1/2) - 1)^2)/((d 
*x + 1)^(1/2) - 1)^2))*2i - A*d^3*e^3*f^2*atan(((f + d*e)^(3/2)*(f - d*e)^ 
(3/2)*1i - (((1 - d*x)^(1/2) - 1)^2*(f + d*e)^(3/2)*(f - d*e)^(3/2)*1i)/(( 
d*x + 1)^(1/2) - 1)^2)/(f^3 - d^2*e^2*f - (f^3*((1 - d*x)^(1/2) - 1)^2)/(( 
d*x + 1)^(1/2) - 1)^2 - (2*d^3*e^3*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) 
 - 1) + (2*d*e*f^2*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1) + (d^2*e^2 
*f*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2))*2i + (4*A*f^2*((1 - 
d*x)^(1/2) - 1)*(f + d*e)^(3/2)*(f - d*e)^(3/2))/((d*x + 1)^(1/2) - 1) + ( 
A*d^5*e^5*atan(((f + d*e)^(3/2)*(f - d*e)^(3/2)*1i - (((1 - d*x)^(1/2) - 1 
)^2*(f + d*e)^(3/2)*(f - d*e)^(3/2)*1i)/((d*x + 1)^(1/2) - 1)^2)/(f^3 - d^ 
2*e^2*f - (f^3*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (2*d^3*e 
^3*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1) + (2*d*e*f^2*((1 - d*x)^(1 
/2) - 1))/((d*x + 1)^(1/2) - 1) + (d^2*e^2*f*((1 - d*x)^(1/2) - 1)^2)/((d* 
x + 1)^(1/2) - 1)^2))*((1 - d*x)^(1/2) - 1)^2*4i)/((d*x + 1)^(1/2) - 1)^2 
+ (A*d^5*e^5*atan(((f + d*e)^(3/2)*(f - d*e)^(3/2)*1i - (((1 - d*x)^(1/2) 
- 1)^2*(f + d*e)^(3/2)*(f - d*e)^(3/2)*1i)/((d*x + 1)^(1/2) - 1)^2)/(f^...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1375, normalized size of antiderivative = 8.44 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**4*e**5 - 2*asin(sqrt( - d*x + 1) 
/sqrt(2))*c*d**4*e**4*f*x + 4*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**2*e**3*f 
**2 + 4*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**2*e**2*f**3*x - 2*asin(sqrt( - 
 d*x + 1)/sqrt(2))*c*e*f**4 - 2*asin(sqrt( - d*x + 1)/sqrt(2))*c*f**5*x - 
2*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1 
)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*a*d**3*e**2*f**2 - 2*sqrt( 
d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt( 
2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*a*d**3*e*f**3*x + 2*sqrt(d*e + f) 
*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - 
 sqrt(f)*sqrt(2))/sqrt(d*e - f))*b*d*e*f**3 + 2*sqrt(d*e + f)*sqrt(d*e - f 
)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt 
(2))/sqrt(d*e - f))*b*d*f**4*x + 2*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt( 
d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e 
 - f))*c*d**3*e**4 + 2*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan 
(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*c*d** 
3*e**3*f*x - 4*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sq 
rt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*c*d*e**2*f**2 
 - 4*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x 
+ 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*c*d*e*f**3*x - 2*sqrt(d 
*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqr...