\(\int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 (A+B x+C x^2) \, dx\) [31]

Optimal result

Integrand size = 37, antiderivative size = 404 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\frac {\left (2 C d^2 e^3+8 A d^4 e^3+6 B d^2 e^2 f+3 C e f^2+6 A d^2 e f^2+B f^3\right ) x \sqrt {1-d^2 x^2}}{16 d^4}-\frac {\left (7 B e-\frac {3 C e^2}{f}+14 A f+\frac {8 C f}{d^2}\right ) (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{70 d^2}-\frac {\left (7 B-\frac {3 C e}{f}\right ) (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{42 d^2}-\frac {C (e+f x)^4 \left (1-d^2 x^2\right )^{3/2}}{7 d^2 f}+\frac {\left (8 \left (C \left (3 d^4 e^4-30 d^2 e^2 f^2-8 f^4\right )-7 d^2 f \left (2 A f \left (6 d^2 e^2+f^2\right )+B \left (d^2 e^3+6 e f^2\right )\right )\right )+3 d^2 f \left (6 C d^2 e^3-14 B d^2 e^2 f-41 C e f^2-98 A d^2 e f^2-35 B f^3\right ) x\right ) \left (1-d^2 x^2\right )^{3/2}}{840 d^6 f}+\frac {\left (2 C d^2 e^3+8 A d^4 e^3+6 B d^2 e^2 f+3 C e f^2+6 A d^2 e f^2+B f^3\right ) \arcsin (d x)}{16 d^5} \] Output:

1/16*(8*A*d^4*e^3+6*A*d^2*e*f^2+6*B*d^2*e^2*f+2*C*d^2*e^3+B*f^3+3*C*e*f^2) 
*x*(-d^2*x^2+1)^(1/2)/d^4-1/70*(7*B*e-3*C*e^2/f+14*A*f+8*C*f/d^2)*(f*x+e)^ 
2*(-d^2*x^2+1)^(3/2)/d^2-1/42*(7*B-3*C*e/f)*(f*x+e)^3*(-d^2*x^2+1)^(3/2)/d 
^2-1/7*C*(f*x+e)^4*(-d^2*x^2+1)^(3/2)/d^2/f+1/840*(8*C*(3*d^4*e^4-30*d^2*e 
^2*f^2-8*f^4)-56*d^2*f*(2*A*f*(6*d^2*e^2+f^2)+B*(d^2*e^3+6*e*f^2))+3*d^2*f 
*(-98*A*d^2*e*f^2-14*B*d^2*e^2*f+6*C*d^2*e^3-35*B*f^3-41*C*e*f^2)*x)*(-d^2 
*x^2+1)^(3/2)/d^6/f+1/16*(8*A*d^4*e^3+6*A*d^2*e*f^2+6*B*d^2*e^2*f+2*C*d^2* 
e^3+B*f^3+3*C*e*f^2)*arcsin(d*x)/d^5
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 373, normalized size of antiderivative = 0.92 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {1-d^2 x^2} \left (14 A d^2 \left (-16 f^3-d^2 f \left (120 e^2+45 e f x+8 f^2 x^2\right )+6 d^4 x \left (10 e^3+20 e^2 f x+15 e f^2 x^2+4 f^3 x^3\right )\right )+7 B \left (-3 d^2 f^2 (32 e+5 f x)-2 d^4 \left (40 e^3+45 e^2 f x+24 e f^2 x^2+5 f^3 x^3\right )+4 d^6 x^2 \left (20 e^3+45 e^2 f x+36 e f^2 x^2+10 f^3 x^3\right )\right )+C \left (-128 f^3-d^2 f \left (672 e^2+315 e f x+64 f^2 x^2\right )-6 d^4 x \left (35 e^3+56 e^2 f x+35 e f^2 x^2+8 f^3 x^3\right )+12 d^6 x^3 \left (35 e^3+84 e^2 f x+70 e f^2 x^2+20 f^3 x^3\right )\right )\right )+210 d \left (2 C d^2 e^3+8 A d^4 e^3+6 B d^2 e^2 f+3 C e f^2+6 A d^2 e f^2+B f^3\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{1680 d^6} \] Input:

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

Output:

(Sqrt[1 - d^2*x^2]*(14*A*d^2*(-16*f^3 - d^2*f*(120*e^2 + 45*e*f*x + 8*f^2* 
x^2) + 6*d^4*x*(10*e^3 + 20*e^2*f*x + 15*e*f^2*x^2 + 4*f^3*x^3)) + 7*B*(-3 
*d^2*f^2*(32*e + 5*f*x) - 2*d^4*(40*e^3 + 45*e^2*f*x + 24*e*f^2*x^2 + 5*f^ 
3*x^3) + 4*d^6*x^2*(20*e^3 + 45*e^2*f*x + 36*e*f^2*x^2 + 10*f^3*x^3)) + C* 
(-128*f^3 - d^2*f*(672*e^2 + 315*e*f*x + 64*f^2*x^2) - 6*d^4*x*(35*e^3 + 5 
6*e^2*f*x + 35*e*f^2*x^2 + 8*f^3*x^3) + 12*d^6*x^3*(35*e^3 + 84*e^2*f*x + 
70*e*f^2*x^2 + 20*f^3*x^3))) + 210*d*(2*C*d^2*e^3 + 8*A*d^4*e^3 + 6*B*d^2* 
e^2*f + 3*C*e*f^2 + 6*A*d^2*e*f^2 + B*f^3)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2 
*x^2])])/(1680*d^6)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {2112, 2185, 25, 27, 687, 27, 687, 25, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)-\frac {\int -3 d^2 (e+f x)^2 \left (f \left (14 A e d^2+5 C e+7 B f\right )+\left (7 d^2 f (B e+2 A f)-C \left (3 d^2 e^2-8 f^2\right )\right ) x\right ) \sqrt {1-d^2 x^2}dx}{6 d^2}}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int (e+f x)^2 \left (f \left (14 A e d^2+5 C e+7 B f\right )+\left (7 d^2 f (B e+2 A f)-C \left (3 d^2 e^2-8 f^2\right )\right ) x\right ) \sqrt {1-d^2 x^2}dx+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{2} \left (-\frac {\int -\left ((e+f x) \left (f \left (70 A e^2 d^4+19 C e^2 d^2+28 A f^2 d^2+49 B e f d^2+16 C f^2\right )-d^2 \left (6 C d^2 e^3-14 B d^2 f e^2-98 A d^2 f^2 e-41 C f^2 e-35 B f^3\right ) x\right ) \sqrt {1-d^2 x^2}\right )dx}{5 d^2}-\frac {1}{5} \left (1-d^2 x^2\right )^{3/2} (e+f x)^2 \left (7 f (2 A f+B e)-C \left (3 e^2-\frac {8 f^2}{d^2}\right )\right )\right )+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{2} \left (\frac {\int (e+f x) \left (f \left (70 A e^2 d^4+19 C e^2 d^2+28 A f^2 d^2+49 B e f d^2+16 C f^2\right )-d^2 \left (6 C d^2 e^3-14 B d^2 f e^2-98 A d^2 f^2 e-41 C f^2 e-35 B f^3\right ) x\right ) \sqrt {1-d^2 x^2}dx}{5 d^2}-\frac {1}{5} \left (1-d^2 x^2\right )^{3/2} (e+f x)^2 \left (7 f (2 A f+B e)-C \left (3 e^2-\frac {8 f^2}{d^2}\right )\right )\right )+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{2} \left (\frac {\frac {35}{4} f \left (8 A d^4 e^3+6 A d^2 e f^2+6 B d^2 e^2 f+B f^3+2 C d^2 e^3+3 C e f^2\right ) \int \sqrt {1-d^2 x^2}dx+\frac {1}{4} f x \left (1-d^2 x^2\right )^{3/2} \left (-98 A d^2 e f^2-14 B d^2 e^2 f-35 B f^3+6 C d^2 e^3-41 C e f^2\right )+\frac {2 \left (1-d^2 x^2\right )^{3/2} \left (C \left (3 d^4 e^4-30 d^2 e^2 f^2-8 f^4\right )-7 d^2 f \left (2 A f \left (6 d^2 e^2+f^2\right )+B \left (d^2 e^3+6 e f^2\right )\right )\right )}{3 d^2}}{5 d^2}-\frac {1}{5} \left (1-d^2 x^2\right )^{3/2} (e+f x)^2 \left (7 f (2 A f+B e)-C \left (3 e^2-\frac {8 f^2}{d^2}\right )\right )\right )+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{2} \left (\frac {\frac {35}{4} f \left (8 A d^4 e^3+6 A d^2 e f^2+6 B d^2 e^2 f+B f^3+2 C d^2 e^3+3 C e f^2\right ) \left (\frac {1}{2} \int \frac {1}{\sqrt {1-d^2 x^2}}dx+\frac {1}{2} x \sqrt {1-d^2 x^2}\right )+\frac {1}{4} f x \left (1-d^2 x^2\right )^{3/2} \left (-98 A d^2 e f^2-14 B d^2 e^2 f-35 B f^3+6 C d^2 e^3-41 C e f^2\right )+\frac {2 \left (1-d^2 x^2\right )^{3/2} \left (C \left (3 d^4 e^4-30 d^2 e^2 f^2-8 f^4\right )-7 d^2 f \left (2 A f \left (6 d^2 e^2+f^2\right )+B \left (d^2 e^3+6 e f^2\right )\right )\right )}{3 d^2}}{5 d^2}-\frac {1}{5} \left (1-d^2 x^2\right )^{3/2} (e+f x)^2 \left (7 f (2 A f+B e)-C \left (3 e^2-\frac {8 f^2}{d^2}\right )\right )\right )+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {1}{2} \left (\frac {\frac {35}{4} f \left (\frac {\arcsin (d x)}{2 d}+\frac {1}{2} x \sqrt {1-d^2 x^2}\right ) \left (8 A d^4 e^3+6 A d^2 e f^2+6 B d^2 e^2 f+B f^3+2 C d^2 e^3+3 C e f^2\right )+\frac {1}{4} f x \left (1-d^2 x^2\right )^{3/2} \left (-98 A d^2 e f^2-14 B d^2 e^2 f-35 B f^3+6 C d^2 e^3-41 C e f^2\right )+\frac {2 \left (1-d^2 x^2\right )^{3/2} \left (C \left (3 d^4 e^4-30 d^2 e^2 f^2-8 f^4\right )-7 d^2 f \left (2 A f \left (6 d^2 e^2+f^2\right )+B \left (d^2 e^3+6 e f^2\right )\right )\right )}{3 d^2}}{5 d^2}-\frac {1}{5} \left (1-d^2 x^2\right )^{3/2} (e+f x)^2 \left (7 f (2 A f+B e)-C \left (3 e^2-\frac {8 f^2}{d^2}\right )\right )\right )+\frac {1}{6} \left (1-d^2 x^2\right )^{3/2} (e+f x)^3 (3 C e-7 B f)}{7 d^2 f}-\frac {C \left (1-d^2 x^2\right )^{3/2} (e+f x)^4}{7 d^2 f}\)

Input:

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

Output:

-1/7*(C*(e + f*x)^4*(1 - d^2*x^2)^(3/2))/(d^2*f) + (((3*C*e - 7*B*f)*(e + 
f*x)^3*(1 - d^2*x^2)^(3/2))/6 + (-1/5*((7*f*(B*e + 2*A*f) - C*(3*e^2 - (8* 
f^2)/d^2))*(e + f*x)^2*(1 - d^2*x^2)^(3/2)) + ((2*(C*(3*d^4*e^4 - 30*d^2*e 
^2*f^2 - 8*f^4) - 7*d^2*f*(2*A*f*(6*d^2*e^2 + f^2) + B*(d^2*e^3 + 6*e*f^2) 
))*(1 - d^2*x^2)^(3/2))/(3*d^2) + (f*(6*C*d^2*e^3 - 14*B*d^2*e^2*f - 41*C* 
e*f^2 - 98*A*d^2*e*f^2 - 35*B*f^3)*x*(1 - d^2*x^2)^(3/2))/4 + (35*f*(2*C*d 
^2*e^3 + 8*A*d^4*e^3 + 6*B*d^2*e^2*f + 3*C*e*f^2 + 6*A*d^2*e*f^2 + B*f^3)* 
((x*Sqrt[1 - d^2*x^2])/2 + ArcSin[d*x]/(2*d)))/4)/(5*d^2))/2)/(7*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 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 = 0.53 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.26

method	result	size
risch	\(-\frac {\left (240 C \,f^{3} x^{6} d^{6}+280 B \,d^{6} f^{3} x^{5}+840 C \,d^{6} e \,f^{2} x^{5}+336 A \,d^{6} f^{3} x^{4}+1008 B \,d^{6} e \,f^{2} x^{4}+1008 C \,d^{6} e^{2} f \,x^{4}+1260 A \,d^{6} e \,f^{2} x^{3}+1260 B \,d^{6} e^{2} f \,x^{3}+420 C \,d^{6} e^{3} x^{3}+1680 A \,d^{6} e^{2} f \,x^{2}+560 B \,d^{6} e^{3} x^{2}-48 C \,d^{4} f^{3} x^{4}+840 A \,d^{6} e^{3} x -70 B \,d^{4} f^{3} x^{3}-210 C \,d^{4} e \,f^{2} x^{3}-112 A \,d^{4} f^{3} x^{2}-336 B \,d^{4} e \,f^{2} x^{2}-336 C \,d^{4} e^{2} f \,x^{2}-630 A \,d^{4} e \,f^{2} x -630 B \,d^{4} e^{2} f x -210 C \,d^{4} e^{3} x -1680 A \,d^{4} e^{2} f -560 B \,d^{4} e^{3}-64 C \,d^{2} f^{3} x^{2}-105 B \,d^{2} f^{3} x -315 C \,d^{2} e \,f^{2} x -224 A \,d^{2} f^{3}-672 B \,d^{2} e \,f^{2}-672 C \,d^{2} e^{2} f -128 C \,f^{3}\right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{1680 d^{6} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (8 A \,d^{4} e^{3}+6 A \,d^{2} e \,f^{2}+6 B \,d^{2} e^{2} f +2 C \,d^{2} e^{3}+B \,f^{3}+3 C e \,f^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{16 d^{4} \sqrt {d^{2}}\, \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(508\)
default	\(\text {Expression too large to display}\)	\(959\)

Input:

int((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(f*x+e)^3*(C*x^2+B*x+A),x,method=_RETURNV 
ERBOSE)
 

Output:

-1/1680*(240*C*d^6*f^3*x^6+280*B*d^6*f^3*x^5+840*C*d^6*e*f^2*x^5+336*A*d^6 
*f^3*x^4+1008*B*d^6*e*f^2*x^4+1008*C*d^6*e^2*f*x^4+1260*A*d^6*e*f^2*x^3+12 
60*B*d^6*e^2*f*x^3+420*C*d^6*e^3*x^3+1680*A*d^6*e^2*f*x^2+560*B*d^6*e^3*x^ 
2-48*C*d^4*f^3*x^4+840*A*d^6*e^3*x-70*B*d^4*f^3*x^3-210*C*d^4*e*f^2*x^3-11 
2*A*d^4*f^3*x^2-336*B*d^4*e*f^2*x^2-336*C*d^4*e^2*f*x^2-630*A*d^4*e*f^2*x- 
630*B*d^4*e^2*f*x-210*C*d^4*e^3*x-1680*A*d^4*e^2*f-560*B*d^4*e^3-64*C*d^2* 
f^3*x^2-105*B*d^2*f^3*x-315*C*d^2*e*f^2*x-224*A*d^2*f^3-672*B*d^2*e*f^2-67 
2*C*d^2*e^2*f-128*C*f^3)*(d*x+1)^(1/2)*(d*x-1)/d^6/(-(d*x+1)*(d*x-1))^(1/2 
)*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)+1/16/d^4*(8*A*d^4*e^3+6*A*d^2*e* 
f^2+6*B*d^2*e^2*f+2*C*d^2*e^3+B*f^3+3*C*e*f^2)/(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 = 406, normalized size of antiderivative = 1.00 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\frac {{\left (240 \, C d^{6} f^{3} x^{6} - 560 \, B d^{4} e^{3} - 672 \, B d^{2} e f^{2} + 280 \, {\left (3 \, C d^{6} e f^{2} + B d^{6} f^{3}\right )} x^{5} + 48 \, {\left (21 \, C d^{6} e^{2} f + 21 \, B d^{6} e f^{2} + {\left (7 \, A d^{6} - C d^{4}\right )} f^{3}\right )} x^{4} - 336 \, {\left (5 \, A d^{4} + 2 \, C d^{2}\right )} e^{2} f - 32 \, {\left (7 \, A d^{2} + 4 \, C\right )} f^{3} + 70 \, {\left (6 \, C d^{6} e^{3} + 18 \, B d^{6} e^{2} f - B d^{4} f^{3} + 3 \, {\left (6 \, A d^{6} - C d^{4}\right )} e f^{2}\right )} x^{3} + 16 \, {\left (35 \, B d^{6} e^{3} - 21 \, B d^{4} e f^{2} + 21 \, {\left (5 \, A d^{6} - C d^{4}\right )} e^{2} f - {\left (7 \, A d^{4} + 4 \, C d^{2}\right )} f^{3}\right )} x^{2} - 105 \, {\left (6 \, B d^{4} e^{2} f + B d^{2} f^{3} - 2 \, {\left (4 \, A d^{6} - C d^{4}\right )} e^{3} + 3 \, {\left (2 \, A d^{4} + C d^{2}\right )} e f^{2}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 210 \, {\left (6 \, B d^{3} e^{2} f + B d f^{3} + 2 \, {\left (4 \, A d^{5} + C d^{3}\right )} e^{3} + 3 \, {\left (2 \, A d^{3} + C d\right )} e f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{1680 \, d^{6}} \] Input:

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(f*x+e)^3*(C*x^2+B*x+A),x, algorith 
m="fricas")
 

Output:

1/1680*((240*C*d^6*f^3*x^6 - 560*B*d^4*e^3 - 672*B*d^2*e*f^2 + 280*(3*C*d^ 
6*e*f^2 + B*d^6*f^3)*x^5 + 48*(21*C*d^6*e^2*f + 21*B*d^6*e*f^2 + (7*A*d^6 
- C*d^4)*f^3)*x^4 - 336*(5*A*d^4 + 2*C*d^2)*e^2*f - 32*(7*A*d^2 + 4*C)*f^3 
 + 70*(6*C*d^6*e^3 + 18*B*d^6*e^2*f - B*d^4*f^3 + 3*(6*A*d^6 - C*d^4)*e*f^ 
2)*x^3 + 16*(35*B*d^6*e^3 - 21*B*d^4*e*f^2 + 21*(5*A*d^6 - C*d^4)*e^2*f - 
(7*A*d^4 + 4*C*d^2)*f^3)*x^2 - 105*(6*B*d^4*e^2*f + B*d^2*f^3 - 2*(4*A*d^6 
 - C*d^4)*e^3 + 3*(2*A*d^4 + C*d^2)*e*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) 
 - 210*(6*B*d^3*e^2*f + B*d*f^3 + 2*(4*A*d^5 + C*d^3)*e^3 + 3*(2*A*d^3 + C 
*d)*e*f^2)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^6
 

Sympy [F]

\[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\int \left (e + f x\right )^{3} \sqrt {- d x + 1} \sqrt {d x + 1} \left (A + B x + C x^{2}\right )\, dx \] Input:

integrate((-d*x+1)**(1/2)*(d*x+1)**(1/2)*(f*x+e)**3*(C*x**2+B*x+A),x)
 

Output:

Integral((e + f*x)**3*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 = 444, normalized size of antiderivative = 1.10 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=-\frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f^{3} x^{4}}{7 \, d^{2}} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + 1} A e^{3} x + \frac {A e^{3} \arcsin \left (d x\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} B e^{3}}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} A e^{2} f}{d^{2}} - \frac {4 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f^{3} x^{2}}{35 \, d^{4}} - \frac {{\left (3 \, C e f^{2} + B f^{3}\right )} {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}}{6 \, d^{2}} - \frac {{\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{5 \, d^{2}} - \frac {{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, d^{2}} + \frac {{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \sqrt {-d^{2} x^{2} + 1} x}{8 \, d^{2}} - \frac {8 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f^{3}}{105 \, d^{6}} - \frac {{\left (3 \, C e f^{2} + B f^{3}\right )} {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{8 \, d^{4}} + \frac {{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \arcsin \left (d x\right )}{8 \, d^{3}} - \frac {2 \, {\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{15 \, d^{4}} + \frac {{\left (3 \, C e f^{2} + B f^{3}\right )} \sqrt {-d^{2} x^{2} + 1} x}{16 \, d^{4}} + \frac {{\left (3 \, C e f^{2} + B f^{3}\right )} \arcsin \left (d x\right )}{16 \, d^{5}} \] Input:

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(f*x+e)^3*(C*x^2+B*x+A),x, algorith 
m="maxima")
 

Output:

-1/7*(-d^2*x^2 + 1)^(3/2)*C*f^3*x^4/d^2 + 1/2*sqrt(-d^2*x^2 + 1)*A*e^3*x + 
 1/2*A*e^3*arcsin(d*x)/d - 1/3*(-d^2*x^2 + 1)^(3/2)*B*e^3/d^2 - (-d^2*x^2 
+ 1)^(3/2)*A*e^2*f/d^2 - 4/35*(-d^2*x^2 + 1)^(3/2)*C*f^3*x^2/d^4 - 1/6*(3* 
C*e*f^2 + B*f^3)*(-d^2*x^2 + 1)^(3/2)*x^3/d^2 - 1/5*(3*C*e^2*f + 3*B*e*f^2 
 + A*f^3)*(-d^2*x^2 + 1)^(3/2)*x^2/d^2 - 1/4*(C*e^3 + 3*B*e^2*f + 3*A*e*f^ 
2)*(-d^2*x^2 + 1)^(3/2)*x/d^2 + 1/8*(C*e^3 + 3*B*e^2*f + 3*A*e*f^2)*sqrt(- 
d^2*x^2 + 1)*x/d^2 - 8/105*(-d^2*x^2 + 1)^(3/2)*C*f^3/d^6 - 1/8*(3*C*e*f^2 
 + B*f^3)*(-d^2*x^2 + 1)^(3/2)*x/d^4 + 1/8*(C*e^3 + 3*B*e^2*f + 3*A*e*f^2) 
*arcsin(d*x)/d^3 - 2/15*(3*C*e^2*f + 3*B*e*f^2 + A*f^3)*(-d^2*x^2 + 1)^(3/ 
2)/d^4 + 1/16*(3*C*e*f^2 + B*f^3)*sqrt(-d^2*x^2 + 1)*x/d^4 + 1/16*(3*C*e*f 
^2 + B*f^3)*arcsin(d*x)/d^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1539 vs. \(2 (380) = 760\).

Time = 0.30 (sec) , antiderivative size = 1539, normalized size of antiderivative = 3.81 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\text {Too large to display} \] Input:

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(f*x+e)^3*(C*x^2+B*x+A),x, algorith 
m="giac")
 

Output:

1/1680*(840*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2) 
*sqrt(d*x + 1)))*A*d^5*e^3 + 1680*(sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*arcsin 
(1/2*sqrt(2)*sqrt(d*x + 1)))*A*d^5*e^3 + 280*(((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^4* 
e^3 + 840*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*s 
qrt(d*x + 1)))*B*d^4*e^3 + 840*(((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^4*e^2*f + 2520*( 
sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1 
)))*A*d^4*e^2*f + 70*(((2*(3*d*x - 10)*(d*x + 1) + 43)*(d*x + 1) - 39)*sqr 
t(d*x + 1)*sqrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*C*d^3*e^ 
3 + 280*(((2*d*x - 5)*(d*x + 1) + 9)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*arcs 
in(1/2*sqrt(2)*sqrt(d*x + 1)))*C*d^3*e^3 + 210*(((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^3*e^2*f + 840*(((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^3*e^2*f + 2 
10*(((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)))*A*d^3*e*f^2 + 840*(((2*d*x 
 - 5)*(d*x + 1) + 9)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)*s 
qrt(d*x + 1)))*A*d^3*e*f^2 + 42*(((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(...
 

Mupad [B] (verification not implemented)

Time = 43.84 (sec) , antiderivative size = 3993, normalized size of antiderivative = 9.88 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

- ((((2048*C*f^3)/3 - 640*C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1) 
^(1/2) - 1)^6 + (((2048*C*f^3)/3 - 640*C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^ 
22)/((d*x + 1)^(1/2) - 1)^22 - (((20480*C*f^3)/3 - 448*C*d^2*e^2*f)*((1 - 
d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 - (((20480*C*f^3)/3 - 448*C*d^2 
*e^2*f)*((1 - d*x)^(1/2) - 1)^20)/((d*x + 1)^(1/2) - 1)^20 + (((458752*C*f 
^3)/15 + (27136*C*d^2*e^2*f)/5)*((1 - d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) 
 - 1)^10 + (((458752*C*f^3)/15 + (27136*C*d^2*e^2*f)/5)*((1 - d*x)^(1/2) - 
 1)^18)/((d*x + 1)^(1/2) - 1)^18 - (((1011712*C*f^3)/15 - (13184*C*d^2*e^2 
*f)/5)*((1 - d*x)^(1/2) - 1)^12)/((d*x + 1)^(1/2) - 1)^12 - (((1011712*C*f 
^3)/15 - (13184*C*d^2*e^2*f)/5)*((1 - d*x)^(1/2) - 1)^16)/((d*x + 1)^(1/2) 
 - 1)^16 + (((9293824*C*f^3)/105 - (15104*C*d^2*e^2*f)/5)*((1 - d*x)^(1/2) 
 - 1)^14)/((d*x + 1)^(1/2) - 1)^14 + (((1 - d*x)^(1/2) - 1)^3*((29*C*d^3*e 
^3)/2 - (41*C*d*e*f^2)/4))/((d*x + 1)^(1/2) - 1)^3 - (((1 - d*x)^(1/2) - 1 
)^25*((29*C*d^3*e^3)/2 - (41*C*d*e*f^2)/4))/((d*x + 1)^(1/2) - 1)^25 - ((( 
1 - d*x)^(1/2) - 1)^5*(39*C*d^3*e^3 - (1099*C*d*e*f^2)/2))/((d*x + 1)^(1/2 
) - 1)^5 + (((1 - d*x)^(1/2) - 1)^23*(39*C*d^3*e^3 - (1099*C*d*e*f^2)/2))/ 
((d*x + 1)^(1/2) - 1)^23 - (((1 - d*x)^(1/2) - 1)^7*(209*C*d^3*e^3 + (8755 
*C*d*e*f^2)/2))/((d*x + 1)^(1/2) - 1)^7 + (((1 - d*x)^(1/2) - 1)^21*(209*C 
*d^3*e^3 + (8755*C*d*e*f^2)/2))/((d*x + 1)^(1/2) - 1)^21 + (((1 - d*x)^(1/ 
2) - 1)^11*((1767*C*d^3*e^3)/2 - (8267*C*d*e*f^2)/4))/((d*x + 1)^(1/2) ...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 869, normalized size of antiderivative = 2.15 \[ \int \sqrt {1-d x} \sqrt {1+d x} (e+f x)^3 \left (A+B x+C x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

( - 1680*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**5*e**3 - 1260*asin(sqrt( - d* 
x + 1)/sqrt(2))*a*d**3*e*f**2 - 1260*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**3 
*e**2*f - 210*asin(sqrt( - d*x + 1)/sqrt(2))*b*d*f**3 - 420*asin(sqrt( - d 
*x + 1)/sqrt(2))*c*d**3*e**3 - 630*asin(sqrt( - d*x + 1)/sqrt(2))*c*d*e*f* 
*2 + 840*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**6*e**3*x + 1680*sqrt(d*x + 1) 
*sqrt( - d*x + 1)*a*d**6*e**2*f*x**2 + 1260*sqrt(d*x + 1)*sqrt( - d*x + 1) 
*a*d**6*e*f**2*x**3 + 336*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**6*f**3*x**4 
- 1680*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**4*e**2*f - 630*sqrt(d*x + 1)*sq 
rt( - d*x + 1)*a*d**4*e*f**2*x - 112*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**4 
*f**3*x**2 - 224*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**2*f**3 + 560*sqrt(d*x 
 + 1)*sqrt( - d*x + 1)*b*d**6*e**3*x**2 + 1260*sqrt(d*x + 1)*sqrt( - d*x + 
 1)*b*d**6*e**2*f*x**3 + 1008*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**6*e*f**2 
*x**4 + 280*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**6*f**3*x**5 - 560*sqrt(d*x 
 + 1)*sqrt( - d*x + 1)*b*d**4*e**3 - 630*sqrt(d*x + 1)*sqrt( - d*x + 1)*b* 
d**4*e**2*f*x - 336*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**4*e*f**2*x**2 - 70 
*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**4*f**3*x**3 - 672*sqrt(d*x + 1)*sqrt( 
 - d*x + 1)*b*d**2*e*f**2 - 105*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**2*f**3 
*x + 420*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**6*e**3*x**3 + 1008*sqrt(d*x + 
 1)*sqrt( - d*x + 1)*c*d**6*e**2*f*x**4 + 840*sqrt(d*x + 1)*sqrt( - d*x + 
1)*c*d**6*e*f**2*x**5 + 240*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**6*f**3*...