\(\int \frac {(e+f x)^3 (A+B x+C x^2)}{\sqrt {1-d x} \sqrt {1+d x}} \, dx\) [38]

Optimal result

Integrand size = 37, antiderivative size = 332 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\left (15 B e-\frac {3 C e^2}{f}+20 A f+\frac {16 C f}{d^2}\right ) (e+f x)^2 \sqrt {1-d^2 x^2}}{60 d^2}-\frac {\left (5 B-\frac {C e}{f}\right ) (e+f x)^3 \sqrt {1-d^2 x^2}}{20 d^2}-\frac {C (e+f x)^4 \sqrt {1-d^2 x^2}}{5 d^2 f}+\frac {\left (4 \left (C \left (3 d^4 e^4-52 d^2 e^2 f^2-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )+d^2 f \left (6 C d^2 e^3-30 B d^2 e^2 f-71 C e f^2-100 A d^2 e f^2-45 B f^3\right ) x\right ) \sqrt {1-d^2 x^2}}{120 d^6 f}+\frac {\left (4 C d^2 e^3+8 A d^4 e^3+12 B d^2 e^2 f+9 C e f^2+12 A d^2 e f^2+3 B f^3\right ) \arcsin (d x)}{8 d^5} \] Output:

-1/60*(15*B*e-3*C*e^2/f+20*A*f+16*C*f/d^2)*(f*x+e)^2*(-d^2*x^2+1)^(1/2)/d^ 
2-1/20*(5*B-C*e/f)*(f*x+e)^3*(-d^2*x^2+1)^(1/2)/d^2-1/5*C*(f*x+e)^4*(-d^2* 
x^2+1)^(1/2)/d^2/f+1/120*(4*C*(3*d^4*e^4-52*d^2*e^2*f^2-16*f^4)-20*d^2*f*( 
4*A*f*(4*d^2*e^2+f^2)+3*B*(d^2*e^3+4*e*f^2))+d^2*f*(-100*A*d^2*e*f^2-30*B* 
d^2*e^2*f+6*C*d^2*e^3-45*B*f^3-71*C*e*f^2)*x)*(-d^2*x^2+1)^(1/2)/d^6/f+1/8 
*(8*A*d^4*e^3+12*A*d^2*e*f^2+12*B*d^2*e^2*f+4*C*d^2*e^3+3*B*f^3+9*C*e*f^2) 
*arcsin(d*x)/d^5
 

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.78 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-\sqrt {1-d^2 x^2} \left (20 A d^2 f \left (4 f^2+d^2 \left (18 e^2+9 e f x+2 f^2 x^2\right )\right )+15 B \left (d^2 f^2 (16 e+3 f x)+2 d^4 \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )\right )+C \left (64 f^3+d^2 f \left (240 e^2+135 e f x+32 f^2 x^2\right )+6 d^4 x \left (10 e^3+20 e^2 f x+15 e f^2 x^2+4 f^3 x^3\right )\right )\right )+30 d \left (4 C d^2 e^3+8 A d^4 e^3+12 B d^2 e^2 f+9 C e f^2+12 A d^2 e f^2+3 B f^3\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{120 d^6} \] Input:

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

Output:

(-(Sqrt[1 - d^2*x^2]*(20*A*d^2*f*(4*f^2 + d^2*(18*e^2 + 9*e*f*x + 2*f^2*x^ 
2)) + 15*B*(d^2*f^2*(16*e + 3*f*x) + 2*d^4*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^ 
2 + f^3*x^3)) + C*(64*f^3 + d^2*f*(240*e^2 + 135*e*f*x + 32*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)))) + 30*d*(4*C*d^2*e 
^3 + 8*A*d^4*e^3 + 12*B*d^2*e^2*f + 9*C*e*f^2 + 12*A*d^2*e*f^2 + 3*B*f^3)* 
ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/(120*d^6)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.08, 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, 25, 27, 687, 25, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)-\frac {\int -\frac {d^2 (e+f x)^2 \left (f \left (20 A e d^2+13 C e+15 B f\right )+\left (4 \left (5 A d^2+4 C\right ) f^2-3 d^2 e (C e-5 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{4 d^2}}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {d^2 (e+f x)^2 \left (f \left (20 A e d^2+13 C e+15 B f\right )+\left (4 \left (5 A d^2+4 C\right ) f^2-3 d^2 e (C e-5 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{4 d^2}+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \int \frac {(e+f x)^2 \left (f \left (20 A e d^2+13 C e+15 B f\right )+\left (4 \left (5 A d^2+4 C\right ) f^2-3 d^2 e (C e-5 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}}dx+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {\int -\frac {(e+f x) \left (f \left (60 A e^2 d^4+33 C e^2 d^2+40 A f^2 d^2+75 B e f d^2+32 C f^2\right )-d^2 \left (6 C d^2 e^3-30 B d^2 f e^2-100 A d^2 f^2 e-71 C f^2 e-45 B f^3\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{3 d^2}-\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 \left (5 f (4 A f+3 B e)-C \left (3 e^2-\frac {16 f^2}{d^2}\right )\right )\right )+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\int \frac {(e+f x) \left (f \left (60 A e^2 d^4+33 C e^2 d^2+40 A f^2 d^2+75 B e f d^2+32 C f^2\right )-d^2 \left (6 C d^2 e^3-30 B d^2 f e^2-100 A d^2 f^2 e-71 C f^2 e-45 B f^3\right ) x\right )}{\sqrt {1-d^2 x^2}}dx}{3 d^2}-\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 \left (5 f (4 A f+3 B e)-C \left (3 e^2-\frac {16 f^2}{d^2}\right )\right )\right )+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\frac {15}{2} f \left (8 A d^4 e^3+12 A d^2 e f^2+12 B d^2 e^2 f+3 B f^3+4 C d^2 e^3+9 C e f^2\right ) \int \frac {1}{\sqrt {1-d^2 x^2}}dx+\frac {1}{2} f x \sqrt {1-d^2 x^2} \left (-100 A d^2 e f^2-30 B d^2 e^2 f-45 B f^3+6 C d^2 e^3-71 C e f^2\right )+\frac {2 \sqrt {1-d^2 x^2} \left (C \left (3 d^4 e^4-52 d^2 e^2 f^2-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )}{d^2}}{3 d^2}-\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 \left (5 f (4 A f+3 B e)-C \left (3 e^2-\frac {16 f^2}{d^2}\right )\right )\right )+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\frac {15 f \arcsin (d x) \left (8 A d^4 e^3+12 A d^2 e f^2+12 B d^2 e^2 f+3 B f^3+4 C d^2 e^3+9 C e f^2\right )}{2 d}+\frac {1}{2} f x \sqrt {1-d^2 x^2} \left (-100 A d^2 e f^2-30 B d^2 e^2 f-45 B f^3+6 C d^2 e^3-71 C e f^2\right )+\frac {2 \sqrt {1-d^2 x^2} \left (C \left (3 d^4 e^4-52 d^2 e^2 f^2-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )}{d^2}}{3 d^2}-\frac {1}{3} \sqrt {1-d^2 x^2} (e+f x)^2 \left (5 f (4 A f+3 B e)-C \left (3 e^2-\frac {16 f^2}{d^2}\right )\right )\right )+\frac {1}{4} \sqrt {1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{5 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^4}{5 d^2 f}\)

Input:

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

Output:

-1/5*(C*(e + f*x)^4*Sqrt[1 - d^2*x^2])/(d^2*f) + (((C*e - 5*B*f)*(e + f*x) 
^3*Sqrt[1 - d^2*x^2])/4 + (-1/3*((5*f*(3*B*e + 4*A*f) - C*(3*e^2 - (16*f^2 
)/d^2))*(e + f*x)^2*Sqrt[1 - d^2*x^2]) + ((2*(C*(3*d^4*e^4 - 52*d^2*e^2*f^ 
2 - 16*f^4) - 5*d^2*f*(4*A*f*(4*d^2*e^2 + f^2) + 3*B*(d^2*e^3 + 4*e*f^2))) 
*Sqrt[1 - d^2*x^2])/d^2 + (f*(6*C*d^2*e^3 - 30*B*d^2*e^2*f - 71*C*e*f^2 - 
100*A*d^2*e*f^2 - 45*B*f^3)*x*Sqrt[1 - d^2*x^2])/2 + (15*f*(4*C*d^2*e^3 + 
8*A*d^4*e^3 + 12*B*d^2*e^2*f + 9*C*e*f^2 + 12*A*d^2*e*f^2 + 3*B*f^3)*ArcSi 
n[d*x])/(2*d))/(3*d^2))/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 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 = 0.79 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {\left (24 C \,d^{4} f^{3} x^{4}+30 B \,d^{4} f^{3} x^{3}+90 C \,d^{4} e \,f^{2} x^{3}+40 A \,d^{4} f^{3} x^{2}+120 B \,d^{4} e \,f^{2} x^{2}+120 C \,d^{4} e^{2} f \,x^{2}+180 A \,d^{4} e \,f^{2} x +180 B \,d^{4} e^{2} f x +60 C \,d^{4} e^{3} x +360 A \,d^{4} e^{2} f +120 B \,d^{4} e^{3}+32 C \,d^{2} f^{3} x^{2}+45 B \,d^{2} f^{3} x +135 C \,d^{2} e \,f^{2} x +80 A \,d^{2} f^{3}+240 B \,d^{2} e \,f^{2}+240 C \,d^{2} e^{2} f +64 C \,f^{3}\right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{120 d^{6} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (8 A \,d^{4} e^{3}+12 A \,d^{2} e \,f^{2}+12 B \,d^{2} e^{2} f +4 C \,d^{2} e^{3}+3 B \,f^{3}+9 C e \,f^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{8 d^{4} \sqrt {d^{2}}\, \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(361\)
default	\(-\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \left (24 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} f^{3} x^{4}+30 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} f^{3} x^{3}+90 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e \,f^{2} x^{3}+40 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} f^{3} x^{2}+120 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e \,f^{2} x^{2}+120 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e^{2} f \,x^{2}+180 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e \,f^{2} x +180 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e^{2} f x +60 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e^{3} x +360 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e^{2} f -120 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{5} e^{3}+120 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{4} e^{3}+32 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f^{3} x^{2}+45 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f^{3} x +135 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} e \,f^{2} x +80 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} f^{3}-180 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{3} e \,f^{2}+240 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} e \,f^{2}-180 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{3} e^{2} f +240 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{2} e^{2} f -60 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{3} e^{3}-45 B \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d \,f^{3}+64 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) f^{3}-135 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d e \,f^{2}\right ) \operatorname {csgn}\left (d \right )}{120 d^{6} \sqrt {-d^{2} x^{2}+1}}\)	\(643\)

Input:

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

Output:

1/120*(24*C*d^4*f^3*x^4+30*B*d^4*f^3*x^3+90*C*d^4*e*f^2*x^3+40*A*d^4*f^3*x 
^2+120*B*d^4*e*f^2*x^2+120*C*d^4*e^2*f*x^2+180*A*d^4*e*f^2*x+180*B*d^4*e^2 
*f*x+60*C*d^4*e^3*x+360*A*d^4*e^2*f+120*B*d^4*e^3+32*C*d^2*f^3*x^2+45*B*d^ 
2*f^3*x+135*C*d^2*e*f^2*x+80*A*d^2*f^3+240*B*d^2*e*f^2+240*C*d^2*e^2*f+64* 
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)+1/8*(8*A*d^4*e^3+12*A*d^2*e*f^2+12*B*d^2*e^2*f+4*C 
*d^2*e^3+3*B*f^3+9*C*e*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 = 0.08 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.86 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (24 \, C d^{4} f^{3} x^{4} + 120 \, B d^{4} e^{3} + 240 \, B d^{2} e f^{2} + 120 \, {\left (3 \, A d^{4} + 2 \, C d^{2}\right )} e^{2} f + 16 \, {\left (5 \, A d^{2} + 4 \, C\right )} f^{3} + 30 \, {\left (3 \, C d^{4} e f^{2} + B d^{4} f^{3}\right )} x^{3} + 8 \, {\left (15 \, C d^{4} e^{2} f + 15 \, B d^{4} e f^{2} + {\left (5 \, A d^{4} + 4 \, C d^{2}\right )} f^{3}\right )} x^{2} + 15 \, {\left (4 \, C d^{4} e^{3} + 12 \, B d^{4} e^{2} f + 3 \, B d^{2} f^{3} + 3 \, {\left (4 \, A d^{4} + 3 \, C d^{2}\right )} e f^{2}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 30 \, {\left (12 \, B d^{3} e^{2} f + 3 \, B d f^{3} + 4 \, {\left (2 \, A d^{5} + C d^{3}\right )} e^{3} + 3 \, {\left (4 \, A d^{3} + 3 \, C d\right )} e f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{120 \, d^{6}} \] Input:

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

Output:

-1/120*((24*C*d^4*f^3*x^4 + 120*B*d^4*e^3 + 240*B*d^2*e*f^2 + 120*(3*A*d^4 
 + 2*C*d^2)*e^2*f + 16*(5*A*d^2 + 4*C)*f^3 + 30*(3*C*d^4*e*f^2 + B*d^4*f^3 
)*x^3 + 8*(15*C*d^4*e^2*f + 15*B*d^4*e*f^2 + (5*A*d^4 + 4*C*d^2)*f^3)*x^2 
+ 15*(4*C*d^4*e^3 + 12*B*d^4*e^2*f + 3*B*d^2*f^3 + 3*(4*A*d^4 + 3*C*d^2)*e 
*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 30*(12*B*d^3*e^2*f + 3*B*d*f^3 + 4 
*(2*A*d^5 + C*d^3)*e^3 + 3*(4*A*d^3 + 3*C*d)*e*f^2)*arctan((sqrt(d*x + 1)* 
sqrt(-d*x + 1) - 1)/(d*x)))/d^6
 

Sympy [F(-1)]

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

integrate((f*x+e)**3*(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 = 0.12 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.07 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\sqrt {-d^{2} x^{2} + 1} C f^{3} x^{4}}{5 \, d^{2}} + \frac {A e^{3} \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} B e^{3}}{d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} A e^{2} f}{d^{2}} - \frac {4 \, \sqrt {-d^{2} x^{2} + 1} C f^{3} x^{2}}{15 \, d^{4}} - \frac {{\left (3 \, C e f^{2} + B f^{3}\right )} \sqrt {-d^{2} x^{2} + 1} x^{3}}{4 \, d^{2}} - \frac {{\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} \sqrt {-d^{2} x^{2} + 1} x^{2}}{3 \, d^{2}} - \frac {{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \sqrt {-d^{2} x^{2} + 1} x}{2 \, d^{2}} + \frac {{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \arcsin \left (d x\right )}{2 \, d^{3}} - \frac {8 \, \sqrt {-d^{2} x^{2} + 1} C f^{3}}{15 \, d^{6}} - \frac {3 \, {\left (3 \, C e f^{2} + B f^{3}\right )} \sqrt {-d^{2} x^{2} + 1} x}{8 \, d^{4}} - \frac {2 \, {\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} \sqrt {-d^{2} x^{2} + 1}}{3 \, d^{4}} + \frac {3 \, {\left (3 \, C e f^{2} + B f^{3}\right )} \arcsin \left (d x\right )}{8 \, d^{5}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.13 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (120 \, B d^{4} e^{3} + 360 \, A d^{4} e^{2} f - 60 \, C d^{3} e^{3} - 180 \, B d^{3} e^{2} f - 180 \, A d^{3} e f^{2} + 360 \, C d^{2} e^{2} f + 360 \, B d^{2} e f^{2} + 120 \, A d^{2} f^{3} - 225 \, C d e f^{2} - 75 \, B d f^{3} + 120 \, C f^{3} + {\left (60 \, C d^{3} e^{3} + 180 \, B d^{3} e^{2} f + 180 \, A d^{3} e f^{2} - 240 \, C d^{2} e^{2} f - 240 \, B d^{2} e f^{2} - 80 \, A d^{2} f^{3} + 405 \, C d e f^{2} + 135 \, B d f^{3} - 160 \, C f^{3} + 2 \, {\left (60 \, C d^{2} e^{2} f + 60 \, B d^{2} e f^{2} + 20 \, A d^{2} f^{3} - 135 \, C d e f^{2} - 45 \, B d f^{3} + 88 \, C f^{3} + 3 \, {\left (15 \, C d e f^{2} + 4 \, {\left (d x + 1\right )} C f^{3} + 5 \, B d f^{3} - 16 \, C f^{3}\right )} {\left (d x + 1\right )}\right )} {\left (d x + 1\right )}\right )} {\left (d x + 1\right )}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 30 \, {\left (8 \, A d^{5} e^{3} + 4 \, C d^{3} e^{3} + 12 \, B d^{3} e^{2} f + 12 \, A d^{3} e f^{2} + 9 \, C d e f^{2} + 3 \, B d f^{3}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{120 \, d^{6}} \] Input:

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

Output:

-1/120*((120*B*d^4*e^3 + 360*A*d^4*e^2*f - 60*C*d^3*e^3 - 180*B*d^3*e^2*f 
- 180*A*d^3*e*f^2 + 360*C*d^2*e^2*f + 360*B*d^2*e*f^2 + 120*A*d^2*f^3 - 22 
5*C*d*e*f^2 - 75*B*d*f^3 + 120*C*f^3 + (60*C*d^3*e^3 + 180*B*d^3*e^2*f + 1 
80*A*d^3*e*f^2 - 240*C*d^2*e^2*f - 240*B*d^2*e*f^2 - 80*A*d^2*f^3 + 405*C* 
d*e*f^2 + 135*B*d*f^3 - 160*C*f^3 + 2*(60*C*d^2*e^2*f + 60*B*d^2*e*f^2 + 2 
0*A*d^2*f^3 - 135*C*d*e*f^2 - 45*B*d*f^3 + 88*C*f^3 + 3*(15*C*d*e*f^2 + 4* 
(d*x + 1)*C*f^3 + 5*B*d*f^3 - 16*C*f^3)*(d*x + 1))*(d*x + 1))*(d*x + 1))*s 
qrt(d*x + 1)*sqrt(-d*x + 1) - 30*(8*A*d^5*e^3 + 4*C*d^3*e^3 + 12*B*d^3*e^2 
*f + 12*A*d^3*e*f^2 + 9*C*d*e*f^2 + 3*B*d*f^3)*arcsin(1/2*sqrt(2)*sqrt(d*x 
 + 1)))/d^6
 

Mupad [B] (verification not implemented)

Time = 31.51 (sec) , antiderivative size = 2606, normalized size of antiderivative = 7.85 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\text {Too large to display} \] Input:

int(((e + f*x)^3*(A + B*x + C*x^2))/((1 - d*x)^(1/2)*(d*x + 1)^(1/2)),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)^ 
14)/((d*x + 1)^(1/2) - 1)^14 - (((4096*C*f^3)/3 - 832*C*d^2*e^2*f)*((1 - d 
*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 - (((4096*C*f^3)/3 - 832*C*d^2*e 
^2*f)*((1 - d*x)^(1/2) - 1)^12)/((d*x + 1)^(1/2) - 1)^12 + (((12288*C*f^3) 
/5 + 768*C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + 
 (((1 - d*x)^(1/2) - 1)^3*(2*C*d^3*e^3 - (87*C*d*e*f^2)/2))/((d*x + 1)^(1/ 
2) - 1)^3 - (((1 - d*x)^(1/2) - 1)^17*(2*C*d^3*e^3 - (87*C*d*e*f^2)/2))/(( 
d*x + 1)^(1/2) - 1)^17 + (((1 - d*x)^(1/2) - 1)^7*(88*C*d^3*e^3 - 42*C*d*e 
*f^2))/((d*x + 1)^(1/2) - 1)^7 - (((1 - d*x)^(1/2) - 1)^13*(88*C*d^3*e^3 - 
 42*C*d*e*f^2))/((d*x + 1)^(1/2) - 1)^13 + (((1 - d*x)^(1/2) - 1)^5*(40*C* 
d^3*e^3 + 426*C*d*e*f^2))/((d*x + 1)^(1/2) - 1)^5 - (((1 - d*x)^(1/2) - 1) 
^15*(40*C*d^3*e^3 + 426*C*d*e*f^2))/((d*x + 1)^(1/2) - 1)^15 + (((1 - d*x) 
^(1/2) - 1)^9*(52*C*d^3*e^3 - 507*C*d*e*f^2))/((d*x + 1)^(1/2) - 1)^9 - (( 
(1 - d*x)^(1/2) - 1)^11*(52*C*d^3*e^3 - 507*C*d*e*f^2))/((d*x + 1)^(1/2) - 
 1)^11 - (d*(4*C*d^2*e^3 + 9*C*e*f^2)*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1) 
^(1/2) - 1)) + (d*(4*C*d^2*e^3 + 9*C*e*f^2)*((1 - d*x)^(1/2) - 1)^19)/(2*( 
(d*x + 1)^(1/2) - 1)^19) + (192*C*d^2*e^2*f*((1 - d*x)^(1/2) - 1)^4)/((d*x 
 + 1)^(1/2) - 1)^4 + (192*C*d^2*e^2*f*((1 - d*x)^(1/2) - 1)^16)/((d*x + 1) 
^(1/2) - 1)^16)/(d^6 + (10*d^6*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 565, normalized size of antiderivative = 1.70 \[ \int \frac {(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-240 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{5} e^{3}-360 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{3} e \,f^{2}-360 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) b \,d^{3} e^{2} f -90 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) b d \,f^{3}-120 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c \,d^{3} e^{3}-270 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c d e \,f^{2}-360 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{4} e^{2} f -180 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{4} e \,f^{2} x -40 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{4} f^{3} x^{2}-80 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{2} f^{3}-120 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} e^{3}-180 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} e^{2} f x -120 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} e \,f^{2} x^{2}-30 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{4} f^{3} x^{3}-240 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{2} e \,f^{2}-45 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{2} f^{3} x -60 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} e^{3} x -120 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} e^{2} f \,x^{2}-90 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} e \,f^{2} x^{3}-24 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{4} f^{3} x^{4}-240 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{2} e^{2} f -135 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{2} e \,f^{2} x -32 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{2} f^{3} x^{2}-64 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,f^{3}}{120 d^{6}} \] Input:

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

Output:

( - 240*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**5*e**3 - 360*asin(sqrt( - d*x 
+ 1)/sqrt(2))*a*d**3*e*f**2 - 360*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**3*e* 
*2*f - 90*asin(sqrt( - d*x + 1)/sqrt(2))*b*d*f**3 - 120*asin(sqrt( - d*x + 
 1)/sqrt(2))*c*d**3*e**3 - 270*asin(sqrt( - d*x + 1)/sqrt(2))*c*d*e*f**2 - 
 360*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**4*e**2*f - 180*sqrt(d*x + 1)*sqrt 
( - d*x + 1)*a*d**4*e*f**2*x - 40*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**4*f* 
*3*x**2 - 80*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**2*f**3 - 120*sqrt(d*x + 1 
)*sqrt( - d*x + 1)*b*d**4*e**3 - 180*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**4 
*e**2*f*x - 120*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**4*e*f**2*x**2 - 30*sqr 
t(d*x + 1)*sqrt( - d*x + 1)*b*d**4*f**3*x**3 - 240*sqrt(d*x + 1)*sqrt( - d 
*x + 1)*b*d**2*e*f**2 - 45*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d**2*f**3*x - 
60*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**4*e**3*x - 120*sqrt(d*x + 1)*sqrt( 
- d*x + 1)*c*d**4*e**2*f*x**2 - 90*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**4*e 
*f**2*x**3 - 24*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**4*f**3*x**4 - 240*sqrt 
(d*x + 1)*sqrt( - d*x + 1)*c*d**2*e**2*f - 135*sqrt(d*x + 1)*sqrt( - d*x + 
 1)*c*d**2*e*f**2*x - 32*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*d**2*f**3*x**2 - 
 64*sqrt(d*x + 1)*sqrt( - d*x + 1)*c*f**3)/(120*d**6)