\(\int \frac {\sqrt {1-d x} \sqrt {1+d x} (A+B x+C x^2)}{(e+f x)^3} \, dx\) [37]

Optimal result

Integrand size = 37, antiderivative size = 301 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{(e+f x)^3} \, dx=\frac {\left (2 (d e-f) (d e+f) (3 C e-B f)-f \left (d^2 f (B e-A f)-C \left (3 d^2 e^2-2 f^2\right )\right ) x\right ) \sqrt {1-d^2 x^2}}{2 f^3 \left (d^2 e^2-f^2\right ) (e+f x)}+\frac {\left (C e^2-B e f+A f^2\right ) \left (1-d^2 x^2\right )^{3/2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}+\frac {d (3 C e-B f) \arcsin (d x)}{f^4}+\frac {\left (d^2 f \left (2 B d^2 e^3-3 B e f^2+A f^3\right )-C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right ) \arctan \left (\frac {f+d^2 e x}{\sqrt {d^2 e^2-f^2} \sqrt {1-d^2 x^2}}\right )}{2 f^4 \left (d^2 e^2-f^2\right )^{3/2}} \] Output:

1/2*(2*(d*e-f)*(d*e+f)*(-B*f+3*C*e)-f*(d^2*f*(-A*f+B*e)-C*(3*d^2*e^2-2*f^2 
))*x)*(-d^2*x^2+1)^(1/2)/f^3/(d^2*e^2-f^2)/(f*x+e)+1/2*(A*f^2-B*e*f+C*e^2) 
*(-d^2*x^2+1)^(3/2)/f/(d^2*e^2-f^2)/(f*x+e)^2+d*(-B*f+3*C*e)*arcsin(d*x)/f 
^4+1/2*(d^2*f*(2*B*d^2*e^3+A*f^3-3*B*e*f^2)-C*(6*d^4*e^4-9*d^2*e^2*f^2+2*f 
^4))*arctan((d^2*e*x+f)/(d^2*e^2-f^2)^(1/2)/(-d^2*x^2+1)^(1/2))/f^4/(d^2*e 
^2-f^2)^(3/2)
 

Mathematica [A] (verified)

Time = 2.03 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{(e+f x)^3} \, dx=\frac {\frac {f \sqrt {1-d^2 x^2} \left (A f^3 \left (f+d^2 e x\right )+B f^3 (e+2 f x)-B d^2 e^2 f (2 e+3 f x)-C f^2 \left (5 e^2+8 e f x+2 f^2 x^2\right )+C d^2 e^2 \left (6 e^2+9 e f x+2 f^2 x^2\right )\right )}{(d e-f) (d e+f) (e+f x)^2}+4 d (3 C e-B f) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )+\frac {2 \sqrt {d^2 e^2-f^2} \left (-d^2 f \left (2 B d^2 e^3-3 B e f^2+A f^3\right )+C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right ) \arctan \left (\frac {\sqrt {d^2 e^2-f^2} x}{e+f x-e \sqrt {1-d^2 x^2}}\right )}{(-d e+f)^2 (d e+f)^2}}{2 f^4} \] Input:

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

Output:

((f*Sqrt[1 - d^2*x^2]*(A*f^3*(f + d^2*e*x) + B*f^3*(e + 2*f*x) - B*d^2*e^2 
*f*(2*e + 3*f*x) - C*f^2*(5*e^2 + 8*e*f*x + 2*f^2*x^2) + C*d^2*e^2*(6*e^2 
+ 9*e*f*x + 2*f^2*x^2)))/((d*e - f)*(d*e + f)*(e + f*x)^2) + 4*d*(3*C*e - 
B*f)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])] + (2*Sqrt[d^2*e^2 - f^2]*(-(d^ 
2*f*(2*B*d^2*e^3 - 3*B*e*f^2 + A*f^3)) + C*(6*d^4*e^4 - 9*d^2*e^2*f^2 + 2* 
f^4))*ArcTan[(Sqrt[d^2*e^2 - f^2]*x)/(e + f*x - e*Sqrt[1 - d^2*x^2])])/((- 
(d*e) + f)^2*(d*e + f)^2))/(2*f^4)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {2112, 2182, 681, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2182

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

\(\Big \downarrow \) 681

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

\(\displaystyle \frac {\frac {\sqrt {1-d^2 x^2} \left (2 (d e-f) (d e+f) (3 C e-B f)-f x \left (d^2 f (B e-A f)-C \left (3 d^2 e^2-2 f^2\right )\right )\right )}{f^3 (e+f x)}-\frac {-\frac {\left (d^2 f \left (A f^3+2 B d^2 e^3-3 B e f^2\right )-C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right ) \int \frac {1}{(e+f x) \sqrt {1-d^2 x^2}}dx}{f}-\frac {2 d^2 (d e-f) (d e+f) (3 C e-B f) \int \frac {1}{\sqrt {1-d^2 x^2}}dx}{f}}{f^3}}{2 \left (d^2 e^2-f^2\right )}+\frac {\left (1-d^2 x^2\right )^{3/2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {\sqrt {1-d^2 x^2} \left (2 (d e-f) (d e+f) (3 C e-B f)-f x \left (d^2 f (B e-A f)-C \left (3 d^2 e^2-2 f^2\right )\right )\right )}{f^3 (e+f x)}-\frac {-\frac {\left (d^2 f \left (A f^3+2 B d^2 e^3-3 B e f^2\right )-C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right ) \int \frac {1}{(e+f x) \sqrt {1-d^2 x^2}}dx}{f}-\frac {2 d \arcsin (d x) (d e-f) (d e+f) (3 C e-B f)}{f}}{f^3}}{2 \left (d^2 e^2-f^2\right )}+\frac {\left (1-d^2 x^2\right )^{3/2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\frac {\sqrt {1-d^2 x^2} \left (2 (d e-f) (d e+f) (3 C e-B f)-f x \left (d^2 f (B e-A f)-C \left (3 d^2 e^2-2 f^2\right )\right )\right )}{f^3 (e+f x)}-\frac {\frac {\left (d^2 f \left (A f^3+2 B d^2 e^3-3 B e f^2\right )-C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right ) \int \frac {1}{-d^2 e^2+f^2-\frac {\left (e x d^2+f\right )^2}{1-d^2 x^2}}d\frac {e x d^2+f}{\sqrt {1-d^2 x^2}}}{f}-\frac {2 d \arcsin (d x) (d e-f) (d e+f) (3 C e-B f)}{f}}{f^3}}{2 \left (d^2 e^2-f^2\right )}+\frac {\left (1-d^2 x^2\right )^{3/2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\sqrt {1-d^2 x^2} \left (2 (d e-f) (d e+f) (3 C e-B f)-f x \left (d^2 f (B e-A f)-C \left (3 d^2 e^2-2 f^2\right )\right )\right )}{f^3 (e+f x)}-\frac {-\frac {\arctan \left (\frac {d^2 e x+f}{\sqrt {1-d^2 x^2} \sqrt {d^2 e^2-f^2}}\right ) \left (d^2 f \left (A f^3+2 B d^2 e^3-3 B e f^2\right )-C \left (6 d^4 e^4-9 d^2 e^2 f^2+2 f^4\right )\right )}{f \sqrt {d^2 e^2-f^2}}-\frac {2 d \arcsin (d x) (d e-f) (d e+f) (3 C e-B f)}{f}}{f^3}}{2 \left (d^2 e^2-f^2\right )}+\frac {\left (1-d^2 x^2\right )^{3/2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\)

Input:

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

Output:

((C*e^2 - B*e*f + A*f^2)*(1 - d^2*x^2)^(3/2))/(2*f*(d^2*e^2 - f^2)*(e + f* 
x)^2) + (((2*(d*e - f)*(d*e + f)*(3*C*e - B*f) - f*(d^2*f*(B*e - A*f) - C* 
(3*d^2*e^2 - 2*f^2))*x)*Sqrt[1 - d^2*x^2])/(f^3*(e + f*x)) - ((-2*d*(d*e - 
 f)*(d*e + f)*(3*C*e - B*f)*ArcSin[d*x])/f - ((d^2*f*(2*B*d^2*e^3 - 3*B*e* 
f^2 + A*f^3) - C*(6*d^4*e^4 - 9*d^2*e^2*f^2 + 2*f^4))*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]))/f^3)/( 
2*(d^2*e^2 - f^2))
 

Defintions of rubi rules used

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 681

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) 
 + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/ 
(e^2*(m + 1)*(m + 2*p + 2))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Sim 
p[g*(2*a*e + 2*a*e*m) + (g*(2*c*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x] 
, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, -1] || 
EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&  !ILtQ[m + 2 
*p + 1, 0] && (IntegerQ[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 2182

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, 
 d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* 
d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2))   Int[(d + e*x)^(m + 
1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b 
*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, 
 x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1155\) vs. \(2(285)=570\).

Time = 0.81 (sec) , antiderivative size = 1156, normalized size of antiderivative = 3.84

method	result	size
risch	\(\text {Expression too large to display}\)	\(1156\)
default	\(\text {Expression too large to display}\)	\(2385\)

Input:

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

Output:

-C/f^3*(d*x+1)^(1/2)*(d*x-1)/(-(d*x+1)*(d*x-1))^(1/2)*((-d*x+1)*(d*x+1))^( 
1/2)/(-d*x+1)^(1/2)-1/f^3*(d^2*(B*f-3*C*e)/f/(d^2)^(1/2)*arctan((d^2)^(1/2 
)*x/(-d^2*x^2+1)^(1/2))-1/f^2*(A*d^2*f^2-3*B*d^2*e*f+6*C*d^2*e^2-C*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))-1/f^3*(2*A*d^2*e*f^2-3*B*d^2*e^2*f+4*C*d^2*e^3+B*f^3-2*C 
*e*f^2)*(1/(d^2*e^2-f^2)*f^2/(x+e/f)*(-d^2*(x+e/f)^2+2/f*d^2*e*(x+e/f)-(d^ 
2*e^2-f^2)/f^2)^(1/2)-f*d^2*e/(d^2*e^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)))+1/f^4*(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)*(1/2/(d^2*e^2-f 
^2)*f^2/(x+e/f)^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)+3/2*f*d^2*e/(d^2*e^2-f^2)*(1/(d^2*e^2-f^2)*f^2/(x+e/f)*(-d^2*(x+e/f)^2+ 
2/f*d^2*e*(x+e/f)-(d^2*e^2-f^2)/f^2)^(1/2)-f*d^2*e/(d^2*e^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)))+1/2*d^2/(d^2*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1089 vs. \(2 (285) = 570\).

Time = 63.35 (sec) , antiderivative size = 2201, normalized size of antiderivative = 7.31 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{(e+f x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(6*C*d^4*e^8*f - 2*B*d^4*e^7*f^2 - 11*C*d^2*e^6*f^3 + 3*B*d^2*e^5*f^4 
 + (A*d^2 + 5*C)*e^4*f^5 - B*e^3*f^6 - A*e^2*f^7 + (6*C*d^4*e^6*f^3 - 2*B* 
d^4*e^5*f^4 - 11*C*d^2*e^4*f^5 + 3*B*d^2*e^3*f^6 + (A*d^2 + 5*C)*e^2*f^7 - 
 B*e*f^8 - A*f^9)*x^2 - (6*C*d^4*e^8 - 2*B*d^4*e^7*f - 9*C*d^2*e^6*f^2 + 3 
*B*d^2*e^5*f^3 - (A*d^2 - 2*C)*e^4*f^4 + (6*C*d^4*e^6*f^2 - 2*B*d^4*e^5*f^ 
3 - 9*C*d^2*e^4*f^4 + 3*B*d^2*e^3*f^5 - (A*d^2 - 2*C)*e^2*f^6)*x^2 + 2*(6* 
C*d^4*e^7*f - 2*B*d^4*e^6*f^2 - 9*C*d^2*e^5*f^3 + 3*B*d^2*e^4*f^4 - (A*d^2 
 - 2*C)*e^3*f^5)*x)*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)*sqrt(-d*x + 1)*f - (d^2*e 
^2 - f^2)*sqrt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e)) + (6*C*d^4*e^8*f - 2*B 
*d^4*e^7*f^2 - 11*C*d^2*e^6*f^3 + 3*B*d^2*e^5*f^4 + (A*d^2 + 5*C)*e^4*f^5 
- B*e^3*f^6 - A*e^2*f^7 + 2*(C*d^4*e^6*f^3 - 2*C*d^2*e^4*f^5 + C*e^2*f^7)* 
x^2 + (9*C*d^4*e^7*f^2 - 3*B*d^4*e^6*f^3 + 5*B*d^2*e^4*f^5 + (A*d^4 - 17*C 
*d^2)*e^5*f^4 - (A*d^2 - 8*C)*e^3*f^6 - 2*B*e^2*f^7)*x)*sqrt(d*x + 1)*sqrt 
(-d*x + 1) + 2*(6*C*d^4*e^7*f^2 - 2*B*d^4*e^6*f^3 - 11*C*d^2*e^5*f^4 + 3*B 
*d^2*e^4*f^5 + (A*d^2 + 5*C)*e^3*f^6 - B*e^2*f^7 - A*e*f^8)*x - 4*(3*C*d^5 
*e^9 - B*d^5*e^8*f - 6*C*d^3*e^7*f^2 + 2*B*d^3*e^6*f^3 + 3*C*d*e^5*f^4 - B 
*d*e^4*f^5 + (3*C*d^5*e^7*f^2 - B*d^5*e^6*f^3 - 6*C*d^3*e^5*f^4 + 2*B*d^3* 
e^4*f^5 + 3*C*d*e^3*f^6 - B*d*e^2*f^7)*x^2 + 2*(3*C*d^5*e^8*f - B*d^5*e^7* 
f^2 - 6*C*d^3*e^6*f^3 + 2*B*d^3*e^5*f^4 + 3*C*d*e^4*f^5 - B*d*e^3*f^6)*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^3,x, algorith 
m="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((f-d*e)*(f+d*e)>0)', see `assume 
?` for mor
 

Giac [F(-2)]

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

integrate((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(C*x^2+B*x+A)/(f*x+e)^3,x, algorith 
m="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 = 70.72 (sec) , antiderivative size = 21781, normalized size of antiderivative = 72.36 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{(e+f x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(25*B*d^3*e^2 - 16*B*d*f^2)*((1 - d*x)^(1/2) - 1)^3)/(f*(f^2 - d^2*e^2 
)*((d*x + 1)^(1/2) - 1)^3) - (2*(25*B*d^3*e^2 - 16*B*d*f^2)*((1 - d*x)^(1/ 
2) - 1)^5)/(f*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)^5) + (4*((1 - d*x)^(1/ 
2) - 1)^2*(2*B*d^4*e^4 - 2*B*f^4 + 3*B*d^2*e^2*f^2))/(e*f^2*(f^2 - d^2*e^2 
)*((d*x + 1)^(1/2) - 1)^2) + (8*((1 - d*x)^(1/2) - 1)^4*(2*B*f^4 + 2*B*d^4 
*e^4 - 5*B*d^2*e^2*f^2))/(e*f^2*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)^4) + 
 (4*((1 - d*x)^(1/2) - 1)^6*(2*B*d^4*e^4 - 2*B*f^4 + 3*B*d^2*e^2*f^2))/(e* 
f^2*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)^6) + (2*B*d^3*e^2*((1 - d*x)^(1/ 
2) - 1))/(f*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)) - (2*B*d^3*e^2*((1 - d* 
x)^(1/2) - 1)^7)/(f*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)^7))/(d^2*e^2 + ( 
((1 - d*x)^(1/2) - 1)^2*(16*f^2 + 4*d^2*e^2))/((d*x + 1)^(1/2) - 1)^2 + (( 
(1 - d*x)^(1/2) - 1)^6*(16*f^2 + 4*d^2*e^2))/((d*x + 1)^(1/2) - 1)^6 - ((( 
1 - d*x)^(1/2) - 1)^4*(32*f^2 - 6*d^2*e^2))/((d*x + 1)^(1/2) - 1)^4 + (d^2 
*e^2*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 + (8*d*e*f*((1 - d*x 
)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 - (8*d*e*f*((1 - d*x)^(1/2) - 1)^5 
)/((d*x + 1)^(1/2) - 1)^5 - (8*d*e*f*((1 - d*x)^(1/2) - 1)^7)/((d*x + 1)^( 
1/2) - 1)^7 + (8*d*e*f*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1)) - ((2 
*(7*A*d^3*e^2 + 2*A*d*f^2)*((1 - d*x)^(1/2) - 1)^3)/(e*(f^2 - d^2*e^2)*((d 
*x + 1)^(1/2) - 1)^3) - (2*(7*A*d^3*e^2 + 2*A*d*f^2)*((1 - d*x)^(1/2) - 1) 
^5)/(e*(f^2 - d^2*e^2)*((d*x + 1)^(1/2) - 1)^5) + (4*(2*A*f^3 + A*d^2*e...
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 3422, normalized size of antiderivative = 11.37 \[ \int \frac {\sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right )}{(e+f x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**5*e**6*f + 8*asin(sqrt( - d*x + 1)/ 
sqrt(2))*b*d**5*e**5*f**2*x + 4*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**5*e**4 
*f**3*x**2 - 8*asin(sqrt( - d*x + 1)/sqrt(2))*b*d**3*e**4*f**3 - 16*asin(s 
qrt( - d*x + 1)/sqrt(2))*b*d**3*e**3*f**4*x - 8*asin(sqrt( - d*x + 1)/sqrt 
(2))*b*d**3*e**2*f**5*x**2 + 4*asin(sqrt( - d*x + 1)/sqrt(2))*b*d*e**2*f** 
5 + 8*asin(sqrt( - d*x + 1)/sqrt(2))*b*d*e*f**6*x + 4*asin(sqrt( - d*x + 1 
)/sqrt(2))*b*d*f**7*x**2 - 12*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**5*e**7 - 
 24*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**5*e**6*f*x - 12*asin(sqrt( - d*x + 
 1)/sqrt(2))*c*d**5*e**5*f**2*x**2 + 24*asin(sqrt( - d*x + 1)/sqrt(2))*c*d 
**3*e**5*f**2 + 48*asin(sqrt( - d*x + 1)/sqrt(2))*c*d**3*e**4*f**3*x + 24* 
asin(sqrt( - d*x + 1)/sqrt(2))*c*d**3*e**3*f**4*x**2 - 12*asin(sqrt( - d*x 
 + 1)/sqrt(2))*c*d*e**3*f**4 - 24*asin(sqrt( - d*x + 1)/sqrt(2))*c*d*e**2* 
f**5*x - 12*asin(sqrt( - d*x + 1)/sqrt(2))*c*d*e*f**6*x**2 - 2*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*e**2*f**4 - 4*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*e*f**5*x - 2*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**6*x**2 - 4*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))...