\(\int \sqrt {1-d x} \sqrt {1+d x} (e+f x) (A+B x+C x^2) \, dx\) [33]

Optimal result

Integrand size = 35, antiderivative size = 164 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx=\frac {\left (C e+4 A d^2 e+B f\right ) x \sqrt {1-d^2 x^2}}{8 d^2}-\frac {C (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^2 f}-\frac {\left (4 \left (\left (2 C+5 A d^2\right ) f-\frac {d^2 e (3 C e-5 B f)}{f}\right )-3 d^2 (3 C e-5 B f) x\right ) \left (1-d^2 x^2\right )^{3/2}}{60 d^4}+\frac {\left (C e+4 A d^2 e+B f\right ) \arcsin (d x)}{8 d^3} \] Output:

1/8*(4*A*d^2*e+B*f+C*e)*x*(-d^2*x^2+1)^(1/2)/d^2-1/5*C*(f*x+e)^2*(-d^2*x^2 
+1)^(3/2)/d^2/f-1/60*(4*(5*A*d^2+2*C)*f-4*d^2*e*(-5*B*f+3*C*e)/f-3*d^2*(-5 
*B*f+3*C*e)*x)*(-d^2*x^2+1)^(3/2)/d^4+1/8*(4*A*d^2*e+B*f+C*e)*arcsin(d*x)/ 
d^3
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.97 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {1-d^2 x^2} \left (60 A d^4 e x+40 A d^2 f \left (-1+d^2 x^2\right )+15 C d^2 e x \left (-1+2 d^2 x^2\right )+5 B d^2 \left (-8 e-3 f x+8 d^2 e x^2+6 d^2 f x^3\right )+8 C f \left (-2-d^2 x^2+3 d^4 x^4\right )\right )+30 d \left (C e+4 A d^2 e+B f\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{120 d^4} \] Input:

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

Output:

(Sqrt[1 - d^2*x^2]*(60*A*d^4*e*x + 40*A*d^2*f*(-1 + d^2*x^2) + 15*C*d^2*e* 
x*(-1 + 2*d^2*x^2) + 5*B*d^2*(-8*e - 3*f*x + 8*d^2*e*x^2 + 6*d^2*f*x^3) + 
8*C*f*(-2 - d^2*x^2 + 3*d^4*x^4)) + 30*d*(C*e + 4*A*d^2*e + B*f)*ArcTan[(d 
*x)/(-1 + Sqrt[1 - d^2*x^2])])/(120*d^4)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2112, 2185, 25, 27, 676, 211, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

-1/5*(C*(e + f*x)^2*(1 - d^2*x^2)^(3/2))/(d^2*f) + (-1/3*((5*f*(B*e + A*f) 
 - C*(3*e^2 - (2*f^2)/d^2))*(1 - d^2*x^2)^(3/2)) + (f*(3*C*e - 5*B*f)*x*(1 
 - d^2*x^2)^(3/2))/4 + (5*f*(C*e + 4*A*d^2*e + B*f)*((x*Sqrt[1 - d^2*x^2]) 
/2 + ArcSin[d*x]/(2*d)))/4)/(5*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 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 = 0.44 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.41

method	result	size
risch	\(-\frac {\left (24 C f \,x^{4} d^{4}+30 B \,d^{4} f \,x^{3}+30 C \,d^{4} e \,x^{3}+40 A \,d^{4} f \,x^{2}+40 B \,d^{4} e \,x^{2}+60 A \,d^{4} e x -8 C \,d^{2} f \,x^{2}-15 B \,d^{2} f x -15 C \,d^{2} e x -40 A \,d^{2} f -40 B \,d^{2} e -16 C f \right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{120 d^{4} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (4 A \,d^{2} e +B f +C e \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{8 d^{2} \sqrt {d^{2}}\, \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(231\)
default	\(\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \left (24 C \,\operatorname {csgn}\left (d \right ) d^{4} f \,x^{4} \sqrt {-d^{2} x^{2}+1}+30 B \,\operatorname {csgn}\left (d \right ) d^{4} f \,x^{3} \sqrt {-d^{2} x^{2}+1}+30 C \,\operatorname {csgn}\left (d \right ) d^{4} e \,x^{3} \sqrt {-d^{2} x^{2}+1}+40 A \,\operatorname {csgn}\left (d \right ) d^{4} f \,x^{2} \sqrt {-d^{2} x^{2}+1}+40 B \,\operatorname {csgn}\left (d \right ) d^{4} e \,x^{2} \sqrt {-d^{2} x^{2}+1}+60 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e x -8 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f \,x^{2}-15 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f x -15 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} e x -40 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f +60 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{3} e -40 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} e +15 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d f -16 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) f +15 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d e \right ) \operatorname {csgn}\left (d \right )}{120 d^{4} \sqrt {-d^{2} x^{2}+1}}\)	\(377\)

Input:

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

Output:

-1/120*(24*C*d^4*f*x^4+30*B*d^4*f*x^3+30*C*d^4*e*x^3+40*A*d^4*f*x^2+40*B*d 
^4*e*x^2+60*A*d^4*e*x-8*C*d^2*f*x^2-15*B*d^2*f*x-15*C*d^2*e*x-40*A*d^2*f-4 
0*B*d^2*e-16*C*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/d^2*(4*A*d^2*e+B*f+C*e)/(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 = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.04 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx=\frac {{\left (24 \, C d^{4} f x^{4} - 40 \, B d^{2} e + 30 \, {\left (C d^{4} e + B d^{4} f\right )} x^{3} + 8 \, {\left (5 \, B d^{4} e + {\left (5 \, A d^{4} - C d^{2}\right )} f\right )} x^{2} - 8 \, {\left (5 \, A d^{2} + 2 \, C\right )} f - 15 \, {\left (B d^{2} f - {\left (4 \, A d^{4} - C d^{2}\right )} e\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 30 \, {\left (B d f + {\left (4 \, A d^{3} + C d\right )} e\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{120 \, d^{4}} \] Input:

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

Output:

1/120*((24*C*d^4*f*x^4 - 40*B*d^2*e + 30*(C*d^4*e + B*d^4*f)*x^3 + 8*(5*B* 
d^4*e + (5*A*d^4 - C*d^2)*f)*x^2 - 8*(5*A*d^2 + 2*C)*f - 15*(B*d^2*f - (4* 
A*d^4 - C*d^2)*e)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 30*(B*d*f + (4*A*d^3 + 
 C*d)*e)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^4
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.06 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx=\frac {1}{2} \, \sqrt {-d^{2} x^{2} + 1} A e x - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f x^{2}}{5 \, d^{2}} + \frac {A e \arcsin \left (d x\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} B e}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} A f}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (C e + B f\right )} x}{4 \, d^{2}} + \frac {\sqrt {-d^{2} x^{2} + 1} {\left (C e + B f\right )} x}{8 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f}{15 \, d^{4}} + \frac {{\left (C e + B f\right )} \arcsin \left (d x\right )}{8 \, d^{3}} \] Input:

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

Output:

1/2*sqrt(-d^2*x^2 + 1)*A*e*x - 1/5*(-d^2*x^2 + 1)^(3/2)*C*f*x^2/d^2 + 1/2* 
A*e*arcsin(d*x)/d - 1/3*(-d^2*x^2 + 1)^(3/2)*B*e/d^2 - 1/3*(-d^2*x^2 + 1)^ 
(3/2)*A*f/d^2 - 1/4*(-d^2*x^2 + 1)^(3/2)*(C*e + B*f)*x/d^2 + 1/8*sqrt(-d^2 
*x^2 + 1)*(C*e + B*f)*x/d^2 - 2/15*(-d^2*x^2 + 1)^(3/2)*C*f/d^4 + 1/8*(C*e 
 + B*f)*arcsin(d*x)/d^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 631 vs. \(2 (148) = 296\).

Time = 0.20 (sec) , antiderivative size = 631, normalized size of antiderivative = 3.85 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/120*(60*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*s 
qrt(d*x + 1)))*A*d^3*e + 120*(sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*arcsin(1/2* 
sqrt(2)*sqrt(d*x + 1)))*A*d^3*e + 20*(((2*d*x - 5)*(d*x + 1) + 9)*sqrt(d*x 
 + 1)*sqrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B*d^2*e + 60*( 
sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1 
)))*B*d^2*e + 20*(((2*d*x - 5)*(d*x + 1) + 9)*sqrt(d*x + 1)*sqrt(-d*x + 1) 
 + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*A*d^2*f + 60*(sqrt(d*x + 1)*(d*x - 
 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*A*d^2*f + 5*(((2 
*(3*d*x - 10)*(d*x + 1) + 43)*(d*x + 1) - 39)*sqrt(d*x + 1)*sqrt(-d*x + 1) 
 - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*C*d*e + 20*(((2*d*x - 5)*(d*x + 1 
) + 9)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))) 
*C*d*e + 5*(((2*(3*d*x - 10)*(d*x + 1) + 43)*(d*x + 1) - 39)*sqrt(d*x + 1) 
*sqrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B*d*f + 20*(((2*d* 
x - 5)*(d*x + 1) + 9)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)* 
sqrt(d*x + 1)))*B*d*f + (((2*(3*(4*d*x - 17)*(d*x + 1) + 133)*(d*x + 1) - 
295)*(d*x + 1) + 195)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 90*arcsin(1/2*sqrt(2) 
*sqrt(d*x + 1)))*C*f + 5*(((2*(3*d*x - 10)*(d*x + 1) + 43)*(d*x + 1) - 39) 
*sqrt(d*x + 1)*sqrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*C*f) 
/d^4
 

Mupad [B] (verification not implemented)

Time = 10.33 (sec) , antiderivative size = 736, normalized size of antiderivative = 4.49 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

((B*f*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1)^(1/2) - 1)) - (35*B*f*((1 - d*x 
)^(1/2) - 1)^3)/(2*((d*x + 1)^(1/2) - 1)^3) + (273*B*f*((1 - d*x)^(1/2) - 
1)^5)/(2*((d*x + 1)^(1/2) - 1)^5) - (715*B*f*((1 - d*x)^(1/2) - 1)^7)/(2*( 
(d*x + 1)^(1/2) - 1)^7) + (715*B*f*((1 - d*x)^(1/2) - 1)^9)/(2*((d*x + 1)^ 
(1/2) - 1)^9) - (273*B*f*((1 - d*x)^(1/2) - 1)^11)/(2*((d*x + 1)^(1/2) - 1 
)^11) + (35*B*f*((1 - d*x)^(1/2) - 1)^13)/(2*((d*x + 1)^(1/2) - 1)^13) - ( 
B*f*((1 - d*x)^(1/2) - 1)^15)/(2*((d*x + 1)^(1/2) - 1)^15))/(d^3*(((1 - d* 
x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^8) - (1 - d*x)^(1/2)*((2*C*f* 
(d*x + 1)^(1/2))/(15*d^4) - (C*f*x^4*(d*x + 1)^(1/2))/5 + (C*f*x^2*(d*x + 
1)^(1/2))/(15*d^2)) + ((C*e*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1)^(1/2) - 1 
)) - (35*C*e*((1 - d*x)^(1/2) - 1)^3)/(2*((d*x + 1)^(1/2) - 1)^3) + (273*C 
*e*((1 - d*x)^(1/2) - 1)^5)/(2*((d*x + 1)^(1/2) - 1)^5) - (715*C*e*((1 - d 
*x)^(1/2) - 1)^7)/(2*((d*x + 1)^(1/2) - 1)^7) + (715*C*e*((1 - d*x)^(1/2) 
- 1)^9)/(2*((d*x + 1)^(1/2) - 1)^9) - (273*C*e*((1 - d*x)^(1/2) - 1)^11)/( 
2*((d*x + 1)^(1/2) - 1)^11) + (35*C*e*((1 - d*x)^(1/2) - 1)^13)/(2*((d*x + 
 1)^(1/2) - 1)^13) - (C*e*((1 - d*x)^(1/2) - 1)^15)/(2*((d*x + 1)^(1/2) - 
1)^15))/(d^3*(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^8) - (B 
*f*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/(2*d^3) - (C*e*atan( 
((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/(2*d^3) + (A*e*x*(1 - d*x)^( 
1/2)*(d*x + 1)^(1/2))/2 - (A*d^(1/2)*e*log((-d)^(1/2)*(1 - d*x)^(1/2)*(...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.95 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x) \left (A+B x+C x^2\right ) \, dx=\frac {-120 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{3} e -30 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) b d f -30 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c d e +60 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{4} e x +40 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{4} f \,x^{2}-40 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{2} f +40 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} e \,x^{2}+30 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} f \,x^{3}-40 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{2} e -15 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{2} f x +30 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} e \,x^{3}+24 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} f \,x^{4}-15 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{2} e x -8 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{2} f \,x^{2}-16 \sqrt {d x +1}\, \sqrt {-d x +1}\, c f}{120 d^{4}} \] Input:

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

Output:

( - 120*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**3*e - 30*asin(sqrt( - d*x + 1) 
/sqrt(2))*b*d*f - 30*asin(sqrt( - d*x + 1)/sqrt(2))*c*d*e + 60*sqrt(d*x + 
1)*sqrt( - d*x + 1)*a*d**4*e*x + 40*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**4* 
f*x**2 - 40*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**2*f + 40*sqrt(d*x + 1)*sqr 
t( - d*x + 1)*b*d**4*e*x**2 + 30*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**4*f*x 
**3 - 40*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**2*e - 15*sqrt(d*x + 1)*sqrt( 
- d*x + 1)*b*d**2*f*x + 30*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**4*e*x**3 + 
24*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**4*f*x**4 - 15*sqrt(d*x + 1)*sqrt( - 
 d*x + 1)*c*d**2*e*x - 8*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**2*f*x**2 - 16 
*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*f)/(120*d**4)