\(\int \sqrt {1-d x} \sqrt {1+d x} (A+B x+C x^2) \, dx\) [34]

Optimal result

Integrand size = 30, antiderivative size = 95 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {\left (C+4 A d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^2}-\frac {B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac {C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2}+\frac {\left (C+4 A d^2\right ) \arcsin (d x)}{8 d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {d \sqrt {1-d^2 x^2} \left (-8 B-3 C x+12 A d^2 x+8 B d^2 x^2+6 C d^2 x^3\right )+6 \left (C+4 A d^2\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{24 d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1188, 2346, 25, 455, 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.

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

\(\Big \downarrow \) 1188

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 1188

Int[((d_) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d*f + e*g*x^2)^m*(a + b*x + c*x^2 
)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m, n] && EqQ[e*f 
 + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63

method	result	size
risch	\(-\frac {\left (6 C \,d^{2} x^{3}+8 B \,d^{2} x^{2}+12 A \,d^{2} x -3 C x -8 B \right ) \sqrt {x d +1}\, \left (x d -1\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{24 d^{2} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}+\frac {\left (4 d^{2} A +C \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{8 d^{2} \sqrt {d^{2}}\, \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(155\)
default	\(\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \left (6 C \,\operatorname {csgn}\left (d \right ) d^{3} x^{3} \sqrt {-d^{2} x^{2}+1}+8 B \,\operatorname {csgn}\left (d \right ) d^{3} x^{2} \sqrt {-d^{2} x^{2}+1}+12 A \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} x -3 C \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d x +12 A \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) d^{2}-8 B \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d +3 C \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \operatorname {csgn}\left (d \right )}{24 \sqrt {-d^{2} x^{2}+1}\, d^{3}}\)	\(185\)

Input:

int((-d*x+1)^(1/2)*(d*x+1)^(1/2)*(C*x^2+B*x+A),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(6*C*d^2*x^3+8*B*d^2*x^2+12*A*d^2*x-3*C*x-8*B)*(d*x+1)^(1/2)*(d*x-1) 
/d^2/(-(d*x+1)*(d*x-1))^(1/2)*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)+1/8* 
(4*A*d^2+C)/d^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 = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {{\left (6 \, C d^{3} x^{3} + 8 \, B d^{3} x^{2} - 8 \, B d + 3 \, {\left (4 \, A d^{3} - C d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 6 \, {\left (4 \, A d^{2} + C\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {1}{2} \, \sqrt {-d^{2} x^{2} + 1} A x - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C x}{4 \, d^{2}} + \frac {A \arcsin \left (d x\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} B}{3 \, d^{2}} + \frac {\sqrt {-d^{2} x^{2} + 1} C x}{8 \, d^{2}} + \frac {C \arcsin \left (d x\right )}{8 \, d^{3}} \] Input:

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

Output:

1/2*sqrt(-d^2*x^2 + 1)*A*x - 1/4*(-d^2*x^2 + 1)^(3/2)*C*x/d^2 + 1/2*A*arcs 
in(d*x)/d - 1/3*(-d^2*x^2 + 1)^(3/2)*B/d^2 + 1/8*sqrt(-d^2*x^2 + 1)*C*x/d^ 
2 + 1/8*C*arcsin(d*x)/d^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (81) = 162\).

Time = 0.16 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.99 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {12 \, {\left (\sqrt {d x + 1} {\left (d x - 2\right )} \sqrt {-d x + 1} - 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} A d^{2} + 24 \, {\left (\sqrt {d x + 1} \sqrt {-d x + 1} + 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} A d^{2} + 4 \, {\left ({\left ({\left (2 \, d x - 5\right )} {\left (d x + 1\right )} + 9\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} B d + 12 \, {\left (\sqrt {d x + 1} {\left (d x - 2\right )} \sqrt {-d x + 1} - 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} B d + {\left ({\left ({\left (2 \, {\left (3 \, d x - 10\right )} {\left (d x + 1\right )} + 43\right )} {\left (d x + 1\right )} - 39\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 18 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} C + 4 \, {\left ({\left ({\left (2 \, d x - 5\right )} {\left (d x + 1\right )} + 9\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )\right )} C}{24 \, d^{3}} \] Input:

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

Output:

1/24*(12*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sq 
rt(d*x + 1)))*A*d^2 + 24*(sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*arcsin(1/2*sqrt 
(2)*sqrt(d*x + 1)))*A*d^2 + 4*(((2*d*x - 5)*(d*x + 1) + 9)*sqrt(d*x + 1)*s 
qrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B*d + 12*(sqrt(d*x + 
1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B*d + ( 
((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 + 4*(((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^3
 

Mupad [B] (verification not implemented)

Time = 7.48 (sec) , antiderivative size = 361, normalized size of antiderivative = 3.80 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {A\,x\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{2}-\frac {\frac {35\,C\,{\left (\sqrt {1-d\,x}-1\right )}^3}{2\,{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {273\,C\,{\left (\sqrt {1-d\,x}-1\right )}^5}{2\,{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {715\,C\,{\left (\sqrt {1-d\,x}-1\right )}^7}{2\,{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {715\,C\,{\left (\sqrt {1-d\,x}-1\right )}^9}{2\,{\left (\sqrt {d\,x+1}-1\right )}^9}+\frac {273\,C\,{\left (\sqrt {1-d\,x}-1\right )}^{11}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{11}}-\frac {35\,C\,{\left (\sqrt {1-d\,x}-1\right )}^{13}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{13}}+\frac {C\,{\left (\sqrt {1-d\,x}-1\right )}^{15}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{15}}-\frac {C\,\left (\sqrt {1-d\,x}-1\right )}{2\,\left (\sqrt {d\,x+1}-1\right )}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^8}-\frac {C\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2\,d^3}-\frac {A\,\sqrt {d}\,\ln \left (\sqrt {-d}\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}-d^{3/2}\,x\right )}{2\,{\left (-d\right )}^{3/2}}+\frac {B\,\left (d^2\,x^2-1\right )\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{3\,d^2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.47 \[ \int \sqrt {1-d x} \sqrt {1+d x} \left (A+B x+C x^2\right ) \, dx=\frac {-24 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) a \,d^{2}-6 \mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right ) c +12 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{3} x +8 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{3} x^{2}-8 \sqrt {d x +1}\, \sqrt {-d x +1}\, b d +6 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,d^{3} x^{3}-3 \sqrt {d x +1}\, \sqrt {-d x +1}\, c d x}{24 d^{3}} \] Input:

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

Output:

( - 24*asin(sqrt( - d*x + 1)/sqrt(2))*a*d**2 - 6*asin(sqrt( - d*x + 1)/sqr 
t(2))*c + 12*sqrt(d*x + 1)*sqrt( - d*x + 1)*a*d**3*x + 8*sqrt(d*x + 1)*sqr 
t( - d*x + 1)*b*d**3*x**2 - 8*sqrt(d*x + 1)*sqrt( - d*x + 1)*b*d + 6*sqrt( 
d*x + 1)*sqrt( - d*x + 1)*c*d**3*x**3 - 3*sqrt(d*x + 1)*sqrt( - d*x + 1)*c 
*d*x)/(24*d**3)