\(\int \frac {(e+f x)^2 (A+B x+C x^2)}{\sqrt {1-d x} \sqrt {1+d x}} \, dx\) [39]

Optimal result

Integrand size = 37, antiderivative size = 228 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\left (4 B-\frac {C e}{f}\right ) (e+f x)^2 \sqrt {1-d^2 x^2}}{12 d^2}-\frac {C (e+f x)^3 \sqrt {1-d^2 x^2}}{4 d^2 f}+\frac {\left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f \left (3 \left (3 C+4 A d^2\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right ) \sqrt {1-d^2 x^2}}{24 d^4 f}+\frac {\left (C \left (4 d^2 e^2+3 f^2\right )+4 d^2 \left (2 B e f+A \left (2 d^2 e^2+f^2\right )\right )\right ) \arcsin (d x)}{8 d^5} \] Output:

-1/12*(4*B-C*e/f)*(f*x+e)^2*(-d^2*x^2+1)^(1/2)/d^2-1/4*C*(f*x+e)^3*(-d^2*x 
^2+1)^(1/2)/d^2/f+1/24*(4*C*(d^2*e^3-8*e*f^2)-16*f*(3*A*d^2*e*f+B*(d^2*e^2 
+f^2))-f*(3*(4*A*d^2+3*C)*f^2-2*d^2*e*(-4*B*f+C*e))*x)*(-d^2*x^2+1)^(1/2)/ 
d^4/f+1/8*(C*(4*d^2*e^2+3*f^2)+4*d^2*(2*B*e*f+A*(2*d^2*e^2+f^2)))*arcsin(d 
*x)/d^5
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.78 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-d \sqrt {1-d^2 x^2} \left (12 A d^2 f (4 e+f x)+C \left (12 d^2 e^2 x+16 e f \left (2+d^2 x^2\right )+3 f^2 x \left (3+2 d^2 x^2\right )\right )+8 B \left (2 f^2+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right )\right )+6 \left (C \left (4 d^2 e^2+3 f^2\right )+4 d^2 \left (2 B e f+A \left (2 d^2 e^2+f^2\right )\right )\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{24 d^5} \] Input:

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

Output:

(-(d*Sqrt[1 - d^2*x^2]*(12*A*d^2*f*(4*e + f*x) + C*(12*d^2*e^2*x + 16*e*f* 
(2 + d^2*x^2) + 3*f^2*x*(3 + 2*d^2*x^2)) + 8*B*(2*f^2 + d^2*(3*e^2 + 3*e*f 
*x + f^2*x^2)))) + 6*(C*(4*d^2*e^2 + 3*f^2) + 4*d^2*(2*B*e*f + A*(2*d^2*e^ 
2 + f^2)))*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/(24*d^5)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {2112, 2185, 25, 27, 687, 25, 27, 676, 223}

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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)-\frac {\int -\frac {d^2 (e+f x) \left (f \left (12 A e d^2+7 C e+8 B f\right )+\left (3 \left (4 A d^2+3 C\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{3 d^2}}{4 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {d^2 (e+f x) \left (f \left (12 A e d^2+7 C e+8 B f\right )+\left (3 \left (4 A d^2+3 C\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{3 d^2}+\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{4 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(e+f x) \left (f \left (12 A e d^2+7 C e+8 B f\right )+\left (3 \left (4 A d^2+3 C\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx+\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{4 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 f \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right ) \int \frac {1}{\sqrt {1-d^2 x^2}}dx}{2 d^2}+\frac {1}{2} f x \sqrt {1-d^2 x^2} \left (-12 A f^2-8 B e f-\frac {9 C f^2}{d^2}+2 C e^2\right )+\frac {2 \sqrt {1-d^2 x^2} \left (-12 A d^2 e f^2-4 B d^2 e^2 f-4 B f^3+C d^2 e^3-8 C e f^2\right )}{d^2}\right )+\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{4 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 f \arcsin (d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )}{2 d^3}+\frac {1}{2} f x \sqrt {1-d^2 x^2} \left (-12 A f^2-8 B e f-\frac {9 C f^2}{d^2}+2 C e^2\right )+\frac {2 \sqrt {1-d^2 x^2} \left (-12 A d^2 e f^2-4 B d^2 e^2 f-4 B f^3+C d^2 e^3-8 C e f^2\right )}{d^2}\right )+\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{4 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f}\)

Input:

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

Output:

-1/4*(C*(e + f*x)^3*Sqrt[1 - d^2*x^2])/(d^2*f) + (((C*e - 4*B*f)*(e + f*x) 
^2*Sqrt[1 - d^2*x^2])/3 + ((2*(C*d^2*e^3 - 4*B*d^2*e^2*f - 8*C*e*f^2 - 12* 
A*d^2*e*f^2 - 4*B*f^3)*Sqrt[1 - d^2*x^2])/d^2 + (f*(2*C*e^2 - 8*B*e*f - 12 
*A*f^2 - (9*C*f^2)/d^2)*x*Sqrt[1 - d^2*x^2])/2 + (3*f*(C*(4*d^2*e^2 + 3*f^ 
2) + 4*d^2*(2*B*e*f + A*(2*d^2*e^2 + f^2)))*ArcSin[d*x])/(2*d^3))/3)/(4*d^ 
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 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 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 687

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

Time = 0.74 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {\left (6 C \,d^{2} f^{2} x^{3}+8 B \,d^{2} f^{2} x^{2}+16 C \,d^{2} e f \,x^{2}+12 A \,d^{2} f^{2} x +24 B \,d^{2} e f x +12 C \,d^{2} e^{2} x +48 A \,d^{2} e f +24 B \,d^{2} e^{2}+9 C \,f^{2} x +16 B \,f^{2}+32 C e f \right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{24 d^{4} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (8 A \,d^{4} e^{2}+4 A \,d^{2} f^{2}+8 B \,d^{2} e f +4 C \,d^{2} e^{2}+3 C \,f^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{8 d^{4} \sqrt {d^{2}}\, \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(256\)
default	\(-\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \left (6 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} f^{2} x^{3}+8 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} f^{2} x^{2}+16 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} e f \,x^{2}+12 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} f^{2} x +24 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} e f x +12 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} e^{2} x +48 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} e f -24 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{4} e^{2}+24 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} e^{2}+9 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d \,f^{2} x -12 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2} f^{2}+16 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d \,f^{2}-24 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2} e f +32 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d e f -12 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2} e^{2}-9 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) f^{2}\right ) \operatorname {csgn}\left (d \right )}{24 d^{5} \sqrt {-d^{2} x^{2}+1}}\)	\(423\)

Input:

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

Output:

1/24*(6*C*d^2*f^2*x^3+8*B*d^2*f^2*x^2+16*C*d^2*e*f*x^2+12*A*d^2*f^2*x+24*B 
*d^2*e*f*x+12*C*d^2*e^2*x+48*A*d^2*e*f+24*B*d^2*e^2+9*C*f^2*x+16*B*f^2+32* 
C*e*f)*(d*x+1)^(1/2)*(d*x-1)/d^4/(-(d*x+1)*(d*x-1))^(1/2)*((-d*x+1)*(d*x+1 
))^(1/2)/(-d*x+1)^(1/2)+1/8*(8*A*d^4*e^2+4*A*d^2*f^2+8*B*d^2*e*f+4*C*d^2*e 
^2+3*C*f^2)/d^4/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+1)^(1/2))*((-d* 
x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (6 \, C d^{3} f^{2} x^{3} + 24 \, B d^{3} e^{2} + 16 \, B d f^{2} + 16 \, {\left (3 \, A d^{3} + 2 \, C d\right )} e f + 8 \, {\left (2 \, C d^{3} e f + B d^{3} f^{2}\right )} x^{2} + 3 \, {\left (4 \, C d^{3} e^{2} + 8 \, B d^{3} e f + {\left (4 \, A d^{3} + 3 \, C d\right )} f^{2}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (8 \, B d^{2} e f + 4 \, {\left (2 \, A d^{4} + C d^{2}\right )} e^{2} + {\left (4 \, A d^{2} + 3 \, C\right )} f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{5}} \] Input:

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

Output:

-1/24*((6*C*d^3*f^2*x^3 + 24*B*d^3*e^2 + 16*B*d*f^2 + 16*(3*A*d^3 + 2*C*d) 
*e*f + 8*(2*C*d^3*e*f + B*d^3*f^2)*x^2 + 3*(4*C*d^3*e^2 + 8*B*d^3*e*f + (4 
*A*d^3 + 3*C*d)*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*(8*B*d^2*e*f + 4* 
(2*A*d^4 + C*d^2)*e^2 + (4*A*d^2 + 3*C)*f^2)*arctan((sqrt(d*x + 1)*sqrt(-d 
*x + 1) - 1)/(d*x)))/d^5
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.01 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\sqrt {-d^{2} x^{2} + 1} C f^{2} x^{3}}{4 \, d^{2}} + \frac {A e^{2} \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} B e^{2}}{d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} A e f}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (2 \, C e f + B f^{2}\right )} x^{2}}{3 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (C e^{2} + 2 \, B e f + A f^{2}\right )} x}{2 \, d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} C f^{2} x}{8 \, d^{4}} + \frac {{\left (C e^{2} + 2 \, B e f + A f^{2}\right )} \arcsin \left (d x\right )}{2 \, d^{3}} + \frac {3 \, C f^{2} \arcsin \left (d x\right )}{8 \, d^{5}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} {\left (2 \, C e f + B f^{2}\right )}}{3 \, d^{4}} \] Input:

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

Output:

-1/4*sqrt(-d^2*x^2 + 1)*C*f^2*x^3/d^2 + A*e^2*arcsin(d*x)/d - sqrt(-d^2*x^ 
2 + 1)*B*e^2/d^2 - 2*sqrt(-d^2*x^2 + 1)*A*e*f/d^2 - 1/3*sqrt(-d^2*x^2 + 1) 
*(2*C*e*f + B*f^2)*x^2/d^2 - 1/2*sqrt(-d^2*x^2 + 1)*(C*e^2 + 2*B*e*f + A*f 
^2)*x/d^2 - 3/8*sqrt(-d^2*x^2 + 1)*C*f^2*x/d^4 + 1/2*(C*e^2 + 2*B*e*f + A* 
f^2)*arcsin(d*x)/d^3 + 3/8*C*f^2*arcsin(d*x)/d^5 - 2/3*sqrt(-d^2*x^2 + 1)* 
(2*C*e*f + B*f^2)/d^4
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.02 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (24 \, B d^{3} e^{2} + 48 \, A d^{3} e f - 12 \, C d^{2} e^{2} - 24 \, B d^{2} e f - 12 \, A d^{2} f^{2} + 48 \, C d e f + 24 \, B d f^{2} - 15 \, C f^{2} + {\left (12 \, C d^{2} e^{2} + 24 \, B d^{2} e f + 12 \, A d^{2} f^{2} - 32 \, C d e f - 16 \, B d f^{2} + 27 \, C f^{2} + 2 \, {\left (8 \, C d e f + 3 \, {\left (d x + 1\right )} C f^{2} + 4 \, B d f^{2} - 9 \, C f^{2}\right )} {\left (d x + 1\right )}\right )} {\left (d x + 1\right )}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 6 \, {\left (8 \, A d^{4} e^{2} + 4 \, C d^{2} e^{2} + 8 \, B d^{2} e f + 4 \, A d^{2} f^{2} + 3 \, C f^{2}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{24 \, d^{5}} \] Input:

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

Output:

-1/24*((24*B*d^3*e^2 + 48*A*d^3*e*f - 12*C*d^2*e^2 - 24*B*d^2*e*f - 12*A*d 
^2*f^2 + 48*C*d*e*f + 24*B*d*f^2 - 15*C*f^2 + (12*C*d^2*e^2 + 24*B*d^2*e*f 
 + 12*A*d^2*f^2 - 32*C*d*e*f - 16*B*d*f^2 + 27*C*f^2 + 2*(8*C*d*e*f + 3*(d 
*x + 1)*C*f^2 + 4*B*d*f^2 - 9*C*f^2)*(d*x + 1))*(d*x + 1))*sqrt(d*x + 1)*s 
qrt(-d*x + 1) - 6*(8*A*d^4*e^2 + 4*C*d^2*e^2 + 8*B*d^2*e*f + 4*A*d^2*f^2 + 
 3*C*f^2)*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))/d^5
 

Mupad [B] (verification not implemented)

Time = 26.44 (sec) , antiderivative size = 1732, normalized size of antiderivative = 7.60 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((14*A*f^2*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 - (2*A*f^2*( 
(1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1) - (14*A*f^2*((1 - d*x)^(1/2) - 
 1)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*A*f^2*((1 - d*x)^(1/2) - 1)^7)/((d*x + 
 1)^(1/2) - 1)^7 + (16*A*d*e*f*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 
 1)^2 + (32*A*d*e*f*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (16 
*A*d*e*f*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6)/(d^3 + (4*d^3*( 
(1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (6*d^3*((1 - d*x)^(1/2) 
- 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (4*d^3*((1 - d*x)^(1/2) - 1)^6)/((d*x + 
1)^(1/2) - 1)^6 + (d^3*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8) - 
 ((((1 - d*x)^(1/2) - 1)^4*(64*B*f^2 + 32*B*d^2*e^2))/((d*x + 1)^(1/2) - 1 
)^4 + (((1 - d*x)^(1/2) - 1)^8*(64*B*f^2 + 32*B*d^2*e^2))/((d*x + 1)^(1/2) 
 - 1)^8 - (((1 - d*x)^(1/2) - 1)^6*((128*B*f^2)/3 - 48*B*d^2*e^2))/((d*x + 
 1)^(1/2) - 1)^6 + (8*B*d^2*e^2*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) 
- 1)^2 + (8*B*d^2*e^2*((1 - d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + 
 (20*B*d*e*f*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 + (24*B*d*e* 
f*((1 - d*x)^(1/2) - 1)^5)/((d*x + 1)^(1/2) - 1)^5 - (24*B*d*e*f*((1 - d*x 
)^(1/2) - 1)^7)/((d*x + 1)^(1/2) - 1)^7 - (20*B*d*e*f*((1 - d*x)^(1/2) - 1 
)^9)/((d*x + 1)^(1/2) - 1)^9 + (4*B*d*e*f*((1 - d*x)^(1/2) - 1)^11)/((d*x 
+ 1)^(1/2) - 1)^11 - (4*B*d*e*f*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 
1))/(d^4 + (6*d^4*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (1...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.57 \[ \int \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-48 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{4} e^{2}-24 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{2} f^{2}-48 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) b \,d^{2} e f -24 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c \,d^{2} e^{2}-18 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c \,f^{2}-48 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{3} e f -12 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{3} f^{2} x -24 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{3} e^{2}-24 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{3} e f x -8 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{3} f^{2} x^{2}-16 \sqrt {d x +1}\, \sqrt {-d x +1}\, b d \,f^{2}-12 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{3} e^{2} x -16 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{3} e f \,x^{2}-6 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{3} f^{2} x^{3}-32 \sqrt {d x +1}\, \sqrt {-d x +1}\, c d e f -9 \sqrt {d x +1}\, \sqrt {-d x +1}\, c d \,f^{2} x}{24 d^{5}} \] Input:

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

Output:

( - 48*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**4*e**2 - 24*asin(sqrt( - d*x + 
1)/sqrt(2))*a*d**2*f**2 - 48*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**2*e*f - 2 
4*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**2*e**2 - 18*asin(sqrt( - d*x + 1)/sq 
rt(2))*c*f**2 - 48*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**3*e*f - 12*sqrt(d*x 
 + 1)*sqrt( - d*x + 1)*a*d**3*f**2*x - 24*sqrt(d*x + 1)*sqrt( - d*x + 1)*b 
*d**3*e**2 - 24*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**3*e*f*x - 8*sqrt(d*x + 
 1)*sqrt( - d*x + 1)*b*d**3*f**2*x**2 - 16*sqrt(d*x + 1)*sqrt( - d*x + 1)* 
b*d*f**2 - 12*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**3*e**2*x - 16*sqrt(d*x + 
 1)*sqrt( - d*x + 1)*c*d**3*e*f*x**2 - 6*sqrt(d*x + 1)*sqrt( - d*x + 1)*c* 
d**3*f**2*x**3 - 32*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d*e*f - 9*sqrt(d*x + 
1)*sqrt( - d*x + 1)*c*d*f**2*x)/(24*d**5)