\(\int \frac {\sqrt {1-d x} \sqrt {1+d x} (A+B x+C x^2)}{e+f x} \, dx\) [35]

Optimal result

Integrand size = 37, antiderivative size = 205 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{e+f x} \, dx=\frac {\left (2 \left (C e^2-B e f+A f^2\right )-f (C e-B f) x\right ) \sqrt {1-d^2 x^2}}{2 f^3}-\frac {C \left (1-d^2 x^2\right )^{3/2}}{3 d^2 f}+\frac {\left (2 A d^2 e+\frac {(C e-B f) \left (2 d^2 e^2-f^2\right )}{f^2}\right ) \arcsin (d x)}{2 d f^2}-\frac {\sqrt {d^2 e^2-f^2} \left (C e^2-B e f+A f^2\right ) \arctan \left (\frac {f+d^2 e x}{\sqrt {d^2 e^2-f^2} \sqrt {1-d^2 x^2}}\right )}{f^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{e+f x} \, dx=\frac {\frac {f \sqrt {1-d^2 x^2} \left (-2 C f^2+3 d^2 f (-2 B e+2 A f+B f x)+C d^2 \left (6 e^2-3 e f x+2 f^2 x^2\right )\right )}{d^2}+\frac {6 \left (2 C d^2 e^3-2 B d^2 e^2 f-C e f^2+2 A d^2 e f^2+B f^3\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{d}+12 \sqrt {d^2 e^2-f^2} \left (C e^2+f (-B e+A f)\right ) \arctan \left (\frac {\sqrt {d^2 e^2-f^2} x}{e+f x-e \sqrt {1-d^2 x^2}}\right )}{6 f^4} \] Input:

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

Output:

((f*Sqrt[1 - d^2*x^2]*(-2*C*f^2 + 3*d^2*f*(-2*B*e + 2*A*f + B*f*x) + C*d^2 
*(6*e^2 - 3*e*f*x + 2*f^2*x^2)))/d^2 + (6*(2*C*d^2*e^3 - 2*B*d^2*e^2*f - C 
*e*f^2 + 2*A*d^2*e*f^2 + B*f^3)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/d 
+ 12*Sqrt[d^2*e^2 - f^2]*(C*e^2 + f*(-(B*e) + A*f))*ArcTan[(Sqrt[d^2*e^2 - 
 f^2]*x)/(e + f*x - e*Sqrt[1 - d^2*x^2])])/(6*f^4)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {2112, 2185, 27, 682, 25, 27, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 682

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 682

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p 
+ 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p 
+ 2))), x] + Simp[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)))   Int[(d + e*x) 
^m*(a + c*x^2)^(p - 1)*Simp[f*a*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f* 
d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))*x, x], x 
], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  ! 
RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(422\) vs. \(2(188)=376\).

Time = 0.86 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.06

method	result	size
risch	\(-\frac {\left (2 C \,d^{2} f^{2} x^{2}+3 B \,d^{2} f^{2} x -3 C \,d^{2} e f x +6 A \,d^{2} f^{2}-6 B \,d^{2} e f +6 C \,d^{2} e^{2}-2 C \,f^{2}\right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{6 d^{2} f^{3} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (\frac {\left (2 A \,d^{2} e \,f^{2}-2 B \,d^{2} e^{2} f +2 C \,d^{2} e^{3}+B \,f^{3}-C e \,f^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{f \sqrt {d^{2}}}+\frac {2 \left (A \,d^{2} e^{2} f^{2}-e^{3} B \,d^{2} f +e^{4} d^{2} C -A \,f^{4}+B e \,f^{3}-C \,e^{2} f^{2}\right ) \ln \left (\frac {-\frac {2 \left (d^{2} e^{2}-f^{2}\right )}{f^{2}}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}+2 \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, \sqrt {-d^{2} \left (x +\frac {e}{f}\right )^{2}+\frac {2 d^{2} e \left (x +\frac {e}{f}\right )}{f}-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}{x +\frac {e}{f}}\right )}{f^{2} \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{2 f^{3} \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(423\)
default	\(\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \left (2 C \,\operatorname {csgn}\left (d \right ) d^{2} f^{4} x^{2} \sqrt {-d^{2} x^{2}+1}\, \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}+6 A \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 ) \operatorname {csgn}\left (d \right ) d^{4} e^{2} f^{2}-6 B \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 ) \operatorname {csgn}\left (d \right ) d^{4} e^{3} f +3 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{2} f^{4} x +6 C \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 ) \operatorname {csgn}\left (d \right ) d^{4} e^{4}-3 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{2} e \,f^{3} x +6 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{2} f^{4}+6 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{3} e \,f^{3}-6 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{2} e \,f^{3}-6 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{3} e^{2} f^{2}+6 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{2} e^{2} f^{2}+6 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d^{3} e^{3} f -6 A \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 ) \operatorname {csgn}\left (d \right ) d^{2} f^{4}+6 B \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 ) \operatorname {csgn}\left (d \right ) d^{2} e \,f^{3}-6 C \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 ) \operatorname {csgn}\left (d \right ) d^{2} e^{2} f^{2}+3 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d \,f^{4}-2 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, f^{4}-3 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}\, d e \,f^{3}\right ) \operatorname {csgn}\left (d \right )}{6 d^{2} \sqrt {-d^{2} x^{2}+1}\, f^{5} \sqrt {-\frac {d^{2} e^{2}-f^{2}}{f^{2}}}}\)	\(990\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 591, normalized size of antiderivative = 2.88 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{e+f x} \, dx=\left [\frac {6 \, {\left (C d^{2} e^{2} - B d^{2} e f + A d^{2} f^{2}\right )} \sqrt {-d^{2} e^{2} + f^{2}} \log \left (\frac {d^{2} e f x + f^{2} - \sqrt {-d^{2} e^{2} + f^{2}} {\left (d^{2} e x + f\right )} - {\left (\sqrt {-d^{2} e^{2} + f^{2}} \sqrt {-d x + 1} f + {\left (d^{2} e^{2} - f^{2}\right )} \sqrt {-d x + 1}\right )} \sqrt {d x + 1}}{f x + e}\right ) + {\left (2 \, C d^{2} f^{3} x^{2} + 6 \, C d^{2} e^{2} f - 6 \, B d^{2} e f^{2} + 2 \, {\left (3 \, A d^{2} - C\right )} f^{3} - 3 \, {\left (C d^{2} e f^{2} - B d^{2} f^{3}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 6 \, {\left (2 \, C d^{3} e^{3} - 2 \, B d^{3} e^{2} f + B d f^{3} + {\left (2 \, A d^{3} - C d\right )} e f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{6 \, d^{2} f^{4}}, -\frac {12 \, {\left (C d^{2} e^{2} - B d^{2} e f + A d^{2} f^{2}\right )} \sqrt {d^{2} e^{2} - f^{2}} \arctan \left (-\frac {\sqrt {d^{2} e^{2} - f^{2}} \sqrt {d x + 1} \sqrt {-d x + 1} e - \sqrt {d^{2} e^{2} - f^{2}} {\left (f x + e\right )}}{{\left (d^{2} e^{2} - f^{2}\right )} x}\right ) - {\left (2 \, C d^{2} f^{3} x^{2} + 6 \, C d^{2} e^{2} f - 6 \, B d^{2} e f^{2} + 2 \, {\left (3 \, A d^{2} - C\right )} f^{3} - 3 \, {\left (C d^{2} e f^{2} - B d^{2} f^{3}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (2 \, C d^{3} e^{3} - 2 \, B d^{3} e^{2} f + B d f^{3} + {\left (2 \, A d^{3} - C d\right )} e f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{6 \, d^{2} f^{4}}\right ] \] Input:

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

Output:

[1/6*(6*(C*d^2*e^2 - B*d^2*e*f + A*d^2*f^2)*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)*s 
qrt(-d*x + 1)*f + (d^2*e^2 - f^2)*sqrt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e) 
) + (2*C*d^2*f^3*x^2 + 6*C*d^2*e^2*f - 6*B*d^2*e*f^2 + 2*(3*A*d^2 - C)*f^3 
 - 3*(C*d^2*e*f^2 - B*d^2*f^3)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(2*C*d^ 
3*e^3 - 2*B*d^3*e^2*f + B*d*f^3 + (2*A*d^3 - C*d)*e*f^2)*arctan((sqrt(d*x 
+ 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^2*f^4), -1/6*(12*(C*d^2*e^2 - B*d^2*e* 
f + A*d^2*f^2)*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)) 
 - (2*C*d^2*f^3*x^2 + 6*C*d^2*e^2*f - 6*B*d^2*e*f^2 + 2*(3*A*d^2 - C)*f^3 
- 3*(C*d^2*e*f^2 - B*d^2*f^3)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*(2*C*d^3 
*e^3 - 2*B*d^3*e^2*f + B*d*f^3 + (2*A*d^3 - C*d)*e*f^2)*arctan((sqrt(d*x + 
 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^2*f^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Giac [F(-2)]

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

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(C*x^2+B*x+A)/(f*x+e),x, algorithm= 
"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 = 32.89 (sec) , antiderivative size = 13115, normalized size of antiderivative = 63.98 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{e+f x} \, dx=\text {Too large to display} \] Input:

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

Output:

(f*(8*A*(d*x + 1)^(1/2) - 8*A + 8*A*(1 - d*x)^(1/2) + 2*A*d^2*x^2 - 8*A*(1 
 - d*x)^(1/2)*(d*x + 1)^(1/2) - 4*A*d*x*(d*x + 1)^(1/2) + 4*A*d*x*(1 - d*x 
)^(1/2)) + 24*A*d*e*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)) - 12 
*A*atanh((4*f^2*(d*x + 1)^(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) - 4*f^2*(1 
 - d*x)^(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) - 2*d*e*f*(f + d*e)^(1/2)*(f 
 - d*e)^(1/2) - d^2*e^2*(d*x + 1)^(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) + 
d^2*e^2*(1 - d*x)^(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) - 4*d*f^2*x*(f + d 
*e)^(1/2)*(f - d*e)^(1/2) + d^3*e^2*x*(f + d*e)^(1/2)*(f - d*e)^(1/2) + 2* 
d*e*f*(d*x + 1)^(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) + 2*d*e*f*(1 - d*x)^ 
(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2) - 2*d*e*f*(1 - d*x)^(1/2)*(d*x + 1)^ 
(1/2)*(f + d*e)^(1/2)*(f - d*e)^(1/2))/(4*f^3*(d*x + 1)^(1/2) - 4*f^3*(1 - 
 d*x)^(1/2) + d^3*e^3 - 4*d*f^3*x - d^3*e^3*(d*x + 1)^(1/2) - d^3*e^3*(1 - 
 d*x)^(1/2) - 2*d*e*f^2 - 3*d^2*e^2*f*(d*x + 1)^(1/2) + 3*d^2*e^2*f*(1 - d 
*x)^(1/2) + 3*d^3*e^2*f*x + d^3*e^3*(1 - d*x)^(1/2)*(d*x + 1)^(1/2) + 2*d* 
e*f^2*(d*x + 1)^(1/2) + 2*d*e*f^2*(1 - d*x)^(1/2) - 2*d*e*f^2*(1 - d*x)^(1 
/2)*(d*x + 1)^(1/2)))*(f + d*e)^(1/2)*(f - d*e)^(1/2) - 12*A*d*e*atan(((1 
- d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1))*(d*x + 1)^(1/2) - 12*A*d*e*atan(( 
(1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1))*(1 - d*x)^(1/2) - 4*A*d*e*atan 
(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1))*(d*x + 1)^(3/2) - 4*A*d*e*at 
an(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1))*(1 - d*x)^(3/2) + 6*A*a...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.23 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{e+f x} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 12*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**3*e*f**2 + 12*asin(sqrt( - d*x 
+ 1)/sqrt(2))*b*d**3*e**2*f - 6*asin(sqrt( - d*x + 1)/sqrt(2))*b*d*f**3 - 
12*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**3*e**3 + 6*asin(sqrt( - d*x + 1)/sq 
rt(2))*c*d*e*f**2 + 12*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** 
2*f**2 - 12*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**2*e*f + 12* 
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**2*e**2 + 12*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**2*f**2 - 12*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**2*e*f + 12*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))/sqr 
t(d*e - f))*c*d**2*e**2 + 6*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**2*f**3 - 6 
*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**2*e*f**2 + 3*sqrt(d*x + 1)*sqrt( - d* 
x + 1)*b*d**2*f**3*x + 6*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**2*e**2*f - 3* 
sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**2*e*f**2*x + 2*sqrt(d*x + 1)*sqrt( - d 
*x + 1)*c*d**2*f**3*x**2 - 2*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*f**3)/(6*d** 
2*f**4)