\(\int \frac {(a+c x^2)^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx\) [3]

Optimal result

Integrand size = 29, antiderivative size = 149 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=-\frac {c \left (5 c^2+18 a c e^2+24 a^2 e^4\right ) x \sqrt {1-e^2 x^2}}{16 e^6}-\frac {c^2 \left (5 c+18 a e^2\right ) x^3 \sqrt {1-e^2 x^2}}{24 e^4}-\frac {c^3 x^5 \sqrt {1-e^2 x^2}}{6 e^2}+\frac {\left (c+2 a e^2\right ) \left (5 c^2+8 a c e^2+8 a^2 e^4\right ) \arcsin (e x)}{16 e^7} \] Output:

-1/16*c*(24*a^2*e^4+18*a*c*e^2+5*c^2)*x*(-e^2*x^2+1)^(1/2)/e^6-1/24*c^2*(1 
8*a*e^2+5*c)*x^3*(-e^2*x^2+1)^(1/2)/e^4-1/6*c^3*x^5*(-e^2*x^2+1)^(1/2)/e^2 
+1/16*(2*a*e^2+c)*(8*a^2*e^4+8*a*c*e^2+5*c^2)*arcsin(e*x)/e^7
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=\frac {-c e x \sqrt {1-e^2 x^2} \left (72 a^2 e^4+18 a c e^2 \left (3+2 e^2 x^2\right )+c^2 \left (15+10 e^2 x^2+8 e^4 x^4\right )\right )+6 \left (5 c^3+18 a c^2 e^2+24 a^2 c e^4+16 a^3 e^6\right ) \arctan \left (\frac {e x}{-1+\sqrt {1-e^2 x^2}}\right )}{48 e^7} \] Input:

Integrate[(a + c*x^2)^3/(Sqrt[1 - e*x]*Sqrt[1 + e*x]),x]
 

Output:

(-(c*e*x*Sqrt[1 - e^2*x^2]*(72*a^2*e^4 + 18*a*c*e^2*(3 + 2*e^2*x^2) + c^2* 
(15 + 10*e^2*x^2 + 8*e^4*x^4))) + 6*(5*c^3 + 18*a*c^2*e^2 + 24*a^2*c*e^4 + 
 16*a^3*e^6)*ArcTan[(e*x)/(-1 + Sqrt[1 - e^2*x^2])])/(48*e^7)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {643, 318, 25, 403, 25, 299, 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.

\(\displaystyle \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {e x+1}} \, dx\)

\(\Big \downarrow \) 643

\(\displaystyle \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e^2 x^2}}dx\)

\(\Big \downarrow \) 318

\(\displaystyle -\frac {\int -\frac {\left (c x^2+a\right ) \left (5 c \left (2 a e^2+c\right ) x^2+a \left (6 a e^2+c\right )\right )}{\sqrt {1-e^2 x^2}}dx}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (c x^2+a\right ) \left (5 c \left (2 a e^2+c\right ) x^2+a \left (6 a e^2+c\right )\right )}{\sqrt {1-e^2 x^2}}dx}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {-\frac {\int -\frac {c \left (44 a^2 e^4+44 a c e^2+15 c^2\right ) x^2+a \left (24 a^2 e^4+14 a c e^2+5 c^2\right )}{\sqrt {1-e^2 x^2}}dx}{4 e^2}-\frac {5 c x \sqrt {1-e^2 x^2} \left (2 a e^2+c\right ) \left (a+c x^2\right )}{4 e^2}}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {c \left (44 a^2 e^4+44 a c e^2+15 c^2\right ) x^2+a \left (24 a^2 e^4+14 a c e^2+5 c^2\right )}{\sqrt {1-e^2 x^2}}dx}{4 e^2}-\frac {5 c x \sqrt {1-e^2 x^2} \left (2 a e^2+c\right ) \left (a+c x^2\right )}{4 e^2}}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {3 \left (2 a e^2+c\right ) \left (8 a^2 e^4+8 a c e^2+5 c^2\right ) \int \frac {1}{\sqrt {1-e^2 x^2}}dx}{2 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (44 a^2 e^4+44 a c e^2+15 c^2\right )}{2 e^2}}{4 e^2}-\frac {5 c x \sqrt {1-e^2 x^2} \left (2 a e^2+c\right ) \left (a+c x^2\right )}{4 e^2}}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {\frac {3 \left (2 a e^2+c\right ) \left (8 a^2 e^4+8 a c e^2+5 c^2\right ) \arcsin (e x)}{2 e^3}-\frac {c x \sqrt {1-e^2 x^2} \left (44 a^2 e^4+44 a c e^2+15 c^2\right )}{2 e^2}}{4 e^2}-\frac {5 c x \sqrt {1-e^2 x^2} \left (2 a e^2+c\right ) \left (a+c x^2\right )}{4 e^2}}{6 e^2}-\frac {c x \sqrt {1-e^2 x^2} \left (a+c x^2\right )^2}{6 e^2}\)

Input:

Int[(a + c*x^2)^3/(Sqrt[1 - e*x]*Sqrt[1 + e*x]),x]
 

Output:

-1/6*(c*x*(a + c*x^2)^2*Sqrt[1 - e^2*x^2])/e^2 + ((-5*c*(c + 2*a*e^2)*x*(a 
 + c*x^2)*Sqrt[1 - e^2*x^2])/(4*e^2) + (-1/2*(c*(15*c^2 + 44*a*c*e^2 + 44* 
a^2*e^4)*x*Sqrt[1 - e^2*x^2])/e^2 + (3*(c + 2*a*e^2)*(5*c^2 + 8*a*c*e^2 + 
8*a^2*e^4)*ArcSin[e*x])/(2*e^3))/(4*e^2))/(6*e^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

rule 643

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

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {c x \left (8 c^{2} x^{4} e^{4}+36 x^{2} a c \,e^{4}+72 a^{2} e^{4}+10 c^{2} e^{2} x^{2}+54 a c \,e^{2}+15 c^{2}\right ) \left (e x -1\right ) \sqrt {e x +1}\, \sqrt {\left (-e x +1\right ) \left (e x +1\right )}}{48 e^{6} \sqrt {-\left (e x -1\right ) \left (e x +1\right )}\, \sqrt {-e x +1}}+\frac {\left (16 e^{6} a^{3}+24 e^{4} a^{2} c +18 c^{2} a \,e^{2}+5 c^{3}\right ) \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+1}}\right ) \sqrt {\left (-e x +1\right ) \left (e x +1\right )}}{16 e^{6} \sqrt {e^{2}}\, \sqrt {-e x +1}\, \sqrt {e x +1}}\)	\(201\)
default	\(-\frac {\sqrt {-e x +1}\, \sqrt {e x +1}\, \left (8 \,\operatorname {csgn}\left (e \right ) c^{3} e^{5} x^{5} \sqrt {-e^{2} x^{2}+1}+36 \,\operatorname {csgn}\left (e \right ) a \,c^{2} e^{5} x^{3} \sqrt {-e^{2} x^{2}+1}+72 \sqrt {-e^{2} x^{2}+1}\, \operatorname {csgn}\left (e \right ) e^{5} a^{2} c x +10 \sqrt {-e^{2} x^{2}+1}\, \operatorname {csgn}\left (e \right ) e^{3} c^{3} x^{3}-48 \arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+1}}\right ) a^{3} e^{6}+54 \sqrt {-e^{2} x^{2}+1}\, \operatorname {csgn}\left (e \right ) e^{3} a \,c^{2} x -72 \arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+1}}\right ) a^{2} c \,e^{4}+15 \sqrt {-e^{2} x^{2}+1}\, \operatorname {csgn}\left (e \right ) e \,c^{3} x -54 \arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+1}}\right ) a \,c^{2} e^{2}-15 \arctan \left (\frac {\operatorname {csgn}\left (e \right ) e x}{\sqrt {-e^{2} x^{2}+1}}\right ) c^{3}\right ) \operatorname {csgn}\left (e \right )}{48 e^{7} \sqrt {-e^{2} x^{2}+1}}\)	\(284\)

Input:

int((c*x^2+a)^3/(-e*x+1)^(1/2)/(e*x+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*c*x*(8*c^2*e^4*x^4+36*a*c*e^4*x^2+72*a^2*e^4+10*c^2*e^2*x^2+54*a*c*e^ 
2+15*c^2)*(e*x-1)*(e*x+1)^(1/2)/e^6/(-(e*x-1)*(e*x+1))^(1/2)*((-e*x+1)*(e* 
x+1))^(1/2)/(-e*x+1)^(1/2)+1/16*(16*a^3*e^6+24*a^2*c*e^4+18*a*c^2*e^2+5*c^ 
3)/e^6/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+1)^(1/2))*((-e*x+1)*(e*x 
+1))^(1/2)/(-e*x+1)^(1/2)/(e*x+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=-\frac {{\left (8 \, c^{3} e^{5} x^{5} + 2 \, {\left (18 \, a c^{2} e^{5} + 5 \, c^{3} e^{3}\right )} x^{3} + 3 \, {\left (24 \, a^{2} c e^{5} + 18 \, a c^{2} e^{3} + 5 \, c^{3} e\right )} x\right )} \sqrt {e x + 1} \sqrt {-e x + 1} + 6 \, {\left (16 \, a^{3} e^{6} + 24 \, a^{2} c e^{4} + 18 \, a c^{2} e^{2} + 5 \, c^{3}\right )} \arctan \left (\frac {\sqrt {e x + 1} \sqrt {-e x + 1} - 1}{e x}\right )}{48 \, e^{7}} \] Input:

integrate((c*x^2+a)^3/(-e*x+1)^(1/2)/(e*x+1)^(1/2),x, algorithm="fricas")
 

Output:

-1/48*((8*c^3*e^5*x^5 + 2*(18*a*c^2*e^5 + 5*c^3*e^3)*x^3 + 3*(24*a^2*c*e^5 
 + 18*a*c^2*e^3 + 5*c^3*e)*x)*sqrt(e*x + 1)*sqrt(-e*x + 1) + 6*(16*a^3*e^6 
 + 24*a^2*c*e^4 + 18*a*c^2*e^2 + 5*c^3)*arctan((sqrt(e*x + 1)*sqrt(-e*x + 
1) - 1)/(e*x)))/e^7
 

Sympy [F(-1)]

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

integrate((c*x**2+a)**3/(-e*x+1)**(1/2)/(e*x+1)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + 1} c^{3} x^{5}}{6 \, e^{2}} - \frac {3 \, \sqrt {-e^{2} x^{2} + 1} a c^{2} x^{3}}{4 \, e^{2}} + \frac {a^{3} \arcsin \left (\frac {e^{2} x}{\sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} - \frac {3 \, \sqrt {-e^{2} x^{2} + 1} a^{2} c x}{2 \, e^{2}} - \frac {5 \, \sqrt {-e^{2} x^{2} + 1} c^{3} x^{3}}{24 \, e^{4}} + \frac {3 \, a^{2} c \arcsin \left (\frac {e^{2} x}{\sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}} e^{2}} - \frac {9 \, \sqrt {-e^{2} x^{2} + 1} a c^{2} x}{8 \, e^{4}} + \frac {9 \, a c^{2} \arcsin \left (\frac {e^{2} x}{\sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}} e^{4}} - \frac {5 \, \sqrt {-e^{2} x^{2} + 1} c^{3} x}{16 \, e^{6}} + \frac {5 \, c^{3} \arcsin \left (\frac {e^{2} x}{\sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e^{6}} \] Input:

integrate((c*x^2+a)^3/(-e*x+1)^(1/2)/(e*x+1)^(1/2),x, algorithm="maxima")
 

Output:

-1/6*sqrt(-e^2*x^2 + 1)*c^3*x^5/e^2 - 3/4*sqrt(-e^2*x^2 + 1)*a*c^2*x^3/e^2 
 + a^3*arcsin(e^2*x/sqrt(e^2))/sqrt(e^2) - 3/2*sqrt(-e^2*x^2 + 1)*a^2*c*x/ 
e^2 - 5/24*sqrt(-e^2*x^2 + 1)*c^3*x^3/e^4 + 3/2*a^2*c*arcsin(e^2*x/sqrt(e^ 
2))/(sqrt(e^2)*e^2) - 9/8*sqrt(-e^2*x^2 + 1)*a*c^2*x/e^4 + 9/8*a*c^2*arcsi 
n(e^2*x/sqrt(e^2))/(sqrt(e^2)*e^4) - 5/16*sqrt(-e^2*x^2 + 1)*c^3*x/e^6 + 5 
/16*c^3*arcsin(e^2*x/sqrt(e^2))/(sqrt(e^2)*e^6)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=\frac {{\left (72 \, a^{2} c e^{4} + 90 \, a c^{2} e^{2} + 33 \, c^{3} - {\left (72 \, a^{2} c e^{4} + 162 \, a c^{2} e^{2} + 85 \, c^{3} - 2 \, {\left (54 \, a c^{2} e^{2} + 55 \, c^{3} - {\left (18 \, a c^{2} e^{2} + 45 \, c^{3} + 4 \, {\left ({\left (e x + 1\right )} c^{3} - 5 \, c^{3}\right )} {\left (e x + 1\right )}\right )} {\left (e x + 1\right )}\right )} {\left (e x + 1\right )}\right )} {\left (e x + 1\right )}\right )} \sqrt {e x + 1} \sqrt {-e x + 1} + 6 \, {\left (16 \, a^{3} e^{6} + 24 \, a^{2} c e^{4} + 18 \, a c^{2} e^{2} + 5 \, c^{3}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {e x + 1}\right )}{48 \, e^{7}} \] Input:

integrate((c*x^2+a)^3/(-e*x+1)^(1/2)/(e*x+1)^(1/2),x, algorithm="giac")
 

Output:

1/48*((72*a^2*c*e^4 + 90*a*c^2*e^2 + 33*c^3 - (72*a^2*c*e^4 + 162*a*c^2*e^ 
2 + 85*c^3 - 2*(54*a*c^2*e^2 + 55*c^3 - (18*a*c^2*e^2 + 45*c^3 + 4*((e*x + 
 1)*c^3 - 5*c^3)*(e*x + 1))*(e*x + 1))*(e*x + 1))*(e*x + 1))*sqrt(e*x + 1) 
*sqrt(-e*x + 1) + 6*(16*a^3*e^6 + 24*a^2*c*e^4 + 18*a*c^2*e^2 + 5*c^3)*arc 
sin(1/2*sqrt(2)*sqrt(e*x + 1)))/e^7
 

Mupad [B] (verification not implemented)

Time = 14.12 (sec) , antiderivative size = 1044, normalized size of antiderivative = 7.01 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=\text {Too large to display} \] Input:

int((a + c*x^2)^3/((1 - e*x)^(1/2)*(e*x + 1)^(1/2)),x)
 

Output:

((((1 - e*x)^(1/2) - 1)*((5*c^3)/4 + (9*a*c^2*e^2)/2 + 6*a^2*c*e^4))/((e*x 
 + 1)^(1/2) - 1) - (((1 - e*x)^(1/2) - 1)^23*((5*c^3)/4 + (9*a*c^2*e^2)/2 
+ 6*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^23 + (((1 - e*x)^(1/2) - 1)^3*((175* 
c^3)/12 + (105*a*c^2*e^2)/2 + 6*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^3 - (((1 
 - e*x)^(1/2) - 1)^21*((175*c^3)/12 + (105*a*c^2*e^2)/2 + 6*a^2*c*e^4))/(( 
e*x + 1)^(1/2) - 1)^21 - (((1 - e*x)^(1/2) - 1)^5*((669*a*c^2*e^2)/2 - (31 
1*c^3)/4 + 126*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^5 + (((1 - e*x)^(1/2) - 1 
)^19*((669*a*c^2*e^2)/2 - (311*c^3)/4 + 126*a^2*c*e^4))/((e*x + 1)^(1/2) - 
 1)^19 - (((1 - e*x)^(1/2) - 1)^7*((8361*c^3)/4 + (1533*a*c^2*e^2)/2 + 510 
*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^7 + (((1 - e*x)^(1/2) - 1)^17*((8361*c^ 
3)/4 + (1533*a*c^2*e^2)/2 + 510*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^17 - ((( 
1 - e*x)^(1/2) - 1)^11*((25295*c^3)/2 - 549*a*c^2*e^2 + 420*a^2*c*e^4))/(( 
e*x + 1)^(1/2) - 1)^11 + (((1 - e*x)^(1/2) - 1)^13*((25295*c^3)/2 - 549*a* 
c^2*e^2 + 420*a^2*c*e^4))/((e*x + 1)^(1/2) - 1)^13 + (((1 - e*x)^(1/2) - 1 
)^9*((42259*c^3)/6 + 165*a*c^2*e^2 - 804*a^2*c*e^4))/((e*x + 1)^(1/2) - 1) 
^9 - (((1 - e*x)^(1/2) - 1)^15*((42259*c^3)/6 + 165*a*c^2*e^2 - 804*a^2*c* 
e^4))/((e*x + 1)^(1/2) - 1)^15)/(e^7 + (12*e^7*((1 - e*x)^(1/2) - 1)^2)/(( 
e*x + 1)^(1/2) - 1)^2 + (66*e^7*((1 - e*x)^(1/2) - 1)^4)/((e*x + 1)^(1/2) 
- 1)^4 + (220*e^7*((1 - e*x)^(1/2) - 1)^6)/((e*x + 1)^(1/2) - 1)^6 + (495* 
e^7*((1 - e*x)^(1/2) - 1)^8)/((e*x + 1)^(1/2) - 1)^8 + (792*e^7*((1 - e...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {1-e x} \sqrt {1+e x}} \, dx=\frac {-96 \mathit {asin} \left (\frac {\sqrt {-e x +1}}{\sqrt {2}}\right ) a^{3} e^{6}-144 \mathit {asin} \left (\frac {\sqrt {-e x +1}}{\sqrt {2}}\right ) a^{2} c \,e^{4}-108 \mathit {asin} \left (\frac {\sqrt {-e x +1}}{\sqrt {2}}\right ) a \,c^{2} e^{2}-30 \mathit {asin} \left (\frac {\sqrt {-e x +1}}{\sqrt {2}}\right ) c^{3}-72 \sqrt {e x +1}\, \sqrt {-e x +1}\, a^{2} c \,e^{5} x -36 \sqrt {e x +1}\, \sqrt {-e x +1}\, a \,c^{2} e^{5} x^{3}-54 \sqrt {e x +1}\, \sqrt {-e x +1}\, a \,c^{2} e^{3} x -8 \sqrt {e x +1}\, \sqrt {-e x +1}\, c^{3} e^{5} x^{5}-10 \sqrt {e x +1}\, \sqrt {-e x +1}\, c^{3} e^{3} x^{3}-15 \sqrt {e x +1}\, \sqrt {-e x +1}\, c^{3} e x}{48 e^{7}} \] Input:

int((c*x^2+a)^3/(-e*x+1)^(1/2)/(e*x+1)^(1/2),x)
 

Output:

( - 96*asin(sqrt( - e*x + 1)/sqrt(2))*a**3*e**6 - 144*asin(sqrt( - e*x + 1 
)/sqrt(2))*a**2*c*e**4 - 108*asin(sqrt( - e*x + 1)/sqrt(2))*a*c**2*e**2 - 
30*asin(sqrt( - e*x + 1)/sqrt(2))*c**3 - 72*sqrt(e*x + 1)*sqrt( - e*x + 1) 
*a**2*c*e**5*x - 36*sqrt(e*x + 1)*sqrt( - e*x + 1)*a*c**2*e**5*x**3 - 54*s 
qrt(e*x + 1)*sqrt( - e*x + 1)*a*c**2*e**3*x - 8*sqrt(e*x + 1)*sqrt( - e*x 
+ 1)*c**3*e**5*x**5 - 10*sqrt(e*x + 1)*sqrt( - e*x + 1)*c**3*e**3*x**3 - 1 
5*sqrt(e*x + 1)*sqrt( - e*x + 1)*c**3*e*x)/(48*e**7)