\(\int e^{2 \text {arctanh}(a x)} x^2 (c-a^2 c x^2)^{5/2} \, dx\) [1136]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 180 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {11 c^2 x \sqrt {c-a^2 c x^2}}{128 a^2}+\frac {11}{64} c^2 x^3 \sqrt {c-a^2 c x^2}+\frac {11}{48} c x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {2 \left (c-a^2 c x^2\right )^{5/2}}{5 a^3}-\frac {1}{8} x^3 \left (c-a^2 c x^2\right )^{5/2}+\frac {2 \left (c-a^2 c x^2\right )^{7/2}}{7 a^3 c}+\frac {11 c^{5/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{128 a^3} \] Output:

-11/128*c^2*x*(-a^2*c*x^2+c)^(1/2)/a^2+11/64*c^2*x^3*(-a^2*c*x^2+c)^(1/2)+ 
11/48*c*x^3*(-a^2*c*x^2+c)^(3/2)-2/5*(-a^2*c*x^2+c)^(5/2)/a^3-1/8*x^3*(-a^ 
2*c*x^2+c)^(5/2)+2/7*(-a^2*c*x^2+c)^(7/2)/a^3/c+11/128*c^(5/2)*arctan(a*c^ 
(1/2)*x/(-a^2*c*x^2+c)^(1/2))/a^3
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 \left (\sqrt {c-a^2 c x^2} \left (1536+1155 a x+768 a^2 x^2-3710 a^3 x^3-6144 a^4 x^4-280 a^5 x^5+3840 a^6 x^6+1680 a^7 x^7\right )+1155 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )\right )}{13440 a^3} \] Input:

Integrate[E^(2*ArcTanh[a*x])*x^2*(c - a^2*c*x^2)^(5/2),x]
 

Output:

-1/13440*(c^2*(Sqrt[c - a^2*c*x^2]*(1536 + 1155*a*x + 768*a^2*x^2 - 3710*a 
^3*x^3 - 6144*a^4*x^4 - 280*a^5*x^5 + 3840*a^6*x^6 + 1680*a^7*x^7) + 1155* 
Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))]))/a^3
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.18, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {6701, 541, 25, 27, 533, 27, 533, 27, 455, 211, 211, 224, 216}

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.

\(\displaystyle \int x^2 e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 541

\(\displaystyle c \left (-\frac {\int -a^2 c x^2 (16 a x+11) \left (c-a^2 c x^2\right )^{3/2}dx}{8 a^2 c}-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int a^2 c x^2 (16 a x+11) \left (c-a^2 c x^2\right )^{3/2}dx}{8 a^2 c}-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{8} \int x^2 (16 a x+11) \left (c-a^2 c x^2\right )^{3/2}dx-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\int a c x (77 a x+32) \left (c-a^2 c x^2\right )^{3/2}dx}{7 a^2 c}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\int x (77 a x+32) \left (c-a^2 c x^2\right )^{3/2}dx}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {\int a c (192 a x+77) \left (c-a^2 c x^2\right )^{3/2}dx}{6 a^2 c}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {\int (192 a x+77) \left (c-a^2 c x^2\right )^{3/2}dx}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {77 \int \left (c-a^2 c x^2\right )^{3/2}dx-\frac {192 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {77 \left (\frac {3}{4} c \int \sqrt {c-a^2 c x^2}dx+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {192 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {77 \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c-a^2 c x^2}}dx+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {192 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {77 \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {192 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle c \left (\frac {1}{8} \left (\frac {\frac {77 \left (\frac {3}{4} c \left (\frac {\sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {192 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{6 a}-\frac {77 x \left (c-a^2 c x^2\right )^{5/2}}{6 a c}}{7 a}-\frac {16 x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 a c}\right )-\frac {x^3 \left (c-a^2 c x^2\right )^{5/2}}{8 c}\right )\)

Input:

Int[E^(2*ArcTanh[a*x])*x^2*(c - a^2*c*x^2)^(5/2),x]
 

Output:

c*(-1/8*(x^3*(c - a^2*c*x^2)^(5/2))/c + ((-16*x^2*(c - a^2*c*x^2)^(5/2))/( 
7*a*c) + ((-77*x*(c - a^2*c*x^2)^(5/2))/(6*a*c) + ((-192*(c - a^2*c*x^2)^( 
5/2))/(5*a*c) + 77*((x*(c - a^2*c*x^2)^(3/2))/4 + (3*c*((x*Sqrt[c - a^2*c* 
x^2])/2 + (Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(2*a)))/4))/ 
(6*a))/(7*a))/8)
 

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 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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 533
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 541
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 6701
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[c^(n/2)   Int[x^m*(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] 
 /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ 
[c, 0]) && IGtQ[n/2, 0]
 
Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.69

method result size
risch \(\frac {\left (1680 a^{7} x^{7}+3840 x^{6} a^{6}-280 a^{5} x^{5}-6144 a^{4} x^{4}-3710 a^{3} x^{3}+768 a^{2} x^{2}+1155 a x +1536\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{13440 a^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {11 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c^{3}}{128 a^{2} \sqrt {a^{2} c}}\) \(125\)
default \(\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{8 a^{2} c}-\frac {17 \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )}{6}\right )}{8 a^{2}}+\frac {2 \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{7 a^{3} c}-\frac {2 \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}{5}-a c \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right ) \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}{8 a^{2} c}+\frac {3 c \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{4 a^{2} c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )\right )}{a^{3}}\) \(345\)

Input:

int((a*x+1)^2/(-a^2*x^2+1)*x^2*(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/13440*(1680*a^7*x^7+3840*a^6*x^6-280*a^5*x^5-6144*a^4*x^4-3710*a^3*x^3+7 
68*a^2*x^2+1155*a*x+1536)*(a^2*x^2-1)/a^3/(-c*(a^2*x^2-1))^(1/2)*c^3+11/12 
8/a^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))*c^3
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.58 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=\left [\frac {1155 \, \sqrt {-c} c^{2} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, {\left (1680 \, a^{7} c^{2} x^{7} + 3840 \, a^{6} c^{2} x^{6} - 280 \, a^{5} c^{2} x^{5} - 6144 \, a^{4} c^{2} x^{4} - 3710 \, a^{3} c^{2} x^{3} + 768 \, a^{2} c^{2} x^{2} + 1155 \, a c^{2} x + 1536 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{26880 \, a^{3}}, -\frac {1155 \, c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + {\left (1680 \, a^{7} c^{2} x^{7} + 3840 \, a^{6} c^{2} x^{6} - 280 \, a^{5} c^{2} x^{5} - 6144 \, a^{4} c^{2} x^{4} - 3710 \, a^{3} c^{2} x^{3} + 768 \, a^{2} c^{2} x^{2} + 1155 \, a c^{2} x + 1536 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{13440 \, a^{3}}\right ] \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^2*(-a^2*c*x^2+c)^(5/2),x, algorithm="fr 
icas")
 

Output:

[1/26880*(1155*sqrt(-c)*c^2*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqr 
t(-c)*x - c) - 2*(1680*a^7*c^2*x^7 + 3840*a^6*c^2*x^6 - 280*a^5*c^2*x^5 - 
6144*a^4*c^2*x^4 - 3710*a^3*c^2*x^3 + 768*a^2*c^2*x^2 + 1155*a*c^2*x + 153 
6*c^2)*sqrt(-a^2*c*x^2 + c))/a^3, -1/13440*(1155*c^(5/2)*arctan(sqrt(-a^2* 
c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + (1680*a^7*c^2*x^7 + 3840*a^6*c^2 
*x^6 - 280*a^5*c^2*x^5 - 6144*a^4*c^2*x^4 - 3710*a^3*c^2*x^3 + 768*a^2*c^2 
*x^2 + 1155*a*c^2*x + 1536*c^2)*sqrt(-a^2*c*x^2 + c))/a^3]
 

Sympy [A] (verification not implemented)

Time = 4.34 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.07 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=- a^{4} c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{7}}{8} - \frac {x^{5}}{48 a^{2}} - \frac {5 x^{3}}{192 a^{4}} - \frac {5 x}{128 a^{6}}\right ) + \frac {5 c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{128 a^{6}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{7}}{7} & \text {otherwise} \end {cases}\right ) - 2 a^{3} c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{6}}{7} - \frac {x^{4}}{35 a^{2}} - \frac {4 x^{2}}{105 a^{4}} - \frac {8}{105 a^{6}}\right ) & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{6}}{6} & \text {otherwise} \end {cases}\right ) + 2 a c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{4}}{5} - \frac {x^{2}}{15 a^{2}} - \frac {2}{15 a^{4}}\right ) & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \left (\frac {x^{3}}{4} - \frac {x}{8 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{8 a^{2}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((a*x+1)**2/(-a**2*x**2+1)*x**2*(-a**2*c*x**2+c)**(5/2),x)
 

Output:

-a**4*c**2*Piecewise((sqrt(-a**2*c*x**2 + c)*(x**7/8 - x**5/(48*a**2) - 5* 
x**3/(192*a**4) - 5*x/(128*a**6)) + 5*c*Piecewise((log(-2*a**2*c*x + 2*sqr 
t(-a**2*c)*sqrt(-a**2*c*x**2 + c))/sqrt(-a**2*c), Ne(c, 0)), (x*log(x)/sqr 
t(-a**2*c*x**2), True))/(128*a**6), Ne(a**2*c, 0)), (sqrt(c)*x**7/7, True) 
) - 2*a**3*c**2*Piecewise((sqrt(-a**2*c*x**2 + c)*(x**6/7 - x**4/(35*a**2) 
 - 4*x**2/(105*a**4) - 8/(105*a**6)), Ne(a**2*c, 0)), (sqrt(c)*x**6/6, Tru 
e)) + 2*a*c**2*Piecewise((sqrt(-a**2*c*x**2 + c)*(x**4/5 - x**2/(15*a**2) 
- 2/(15*a**4)), Ne(a**2*c, 0)), (sqrt(c)*x**4/4, True)) + c**2*Piecewise(( 
(x**3/4 - x/(8*a**2))*sqrt(-a**2*c*x**2 + c) + c*Piecewise((log(-2*a**2*c* 
x + 2*sqrt(-a**2*c)*sqrt(-a**2*c*x**2 + c))/sqrt(-a**2*c), Ne(c, 0)), (x*l 
og(x)/sqrt(-a**2*c*x**2), True))/(8*a**2), Ne(a**2*c, 0)), (sqrt(c)*x**3/3 
, True))
 

Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.19 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {1}{13440} \, a {\left (\frac {4760 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x}{a^{3}} - \frac {1680 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} x}{a^{3} c} - \frac {770 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c x}{a^{3}} - \frac {10080 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c^{2} x}{a^{3}} + \frac {8925 \, \sqrt {-a^{2} c x^{2} + c} c^{2} x}{a^{3}} + \frac {8925 \, c^{\frac {5}{2}} \arcsin \left (a x\right )}{a^{4}} + \frac {5376 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{a^{4}} - \frac {3840 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}}{a^{4} c} + \frac {20160 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c^{2}}{a^{4}} - \frac {10080 \, c^{4} \arcsin \left (a x - 2\right )}{a^{7} \left (-\frac {c}{a^{2}}\right )^{\frac {3}{2}}}\right )} \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^2*(-a^2*c*x^2+c)^(5/2),x, algorithm="ma 
xima")
 

Output:

-1/13440*a*(4760*(-a^2*c*x^2 + c)^(5/2)*x/a^3 - 1680*(-a^2*c*x^2 + c)^(7/2 
)*x/(a^3*c) - 770*(-a^2*c*x^2 + c)^(3/2)*c*x/a^3 - 10080*sqrt(a^2*c*x^2 - 
4*a*c*x + 3*c)*c^2*x/a^3 + 8925*sqrt(-a^2*c*x^2 + c)*c^2*x/a^3 + 8925*c^(5 
/2)*arcsin(a*x)/a^4 + 5376*(-a^2*c*x^2 + c)^(5/2)/a^4 - 3840*(-a^2*c*x^2 + 
 c)^(7/2)/(a^4*c) + 20160*sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c^2/a^4 - 10080* 
c^4*arcsin(a*x - 2)/(a^7*(-c/a^2)^(3/2)))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.79 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{13440} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left ({\left (1855 \, c^{2} + 4 \, {\left (768 \, a c^{2} + 5 \, {\left (7 \, a^{2} c^{2} - 6 \, {\left (7 \, a^{4} c^{2} x + 16 \, a^{3} c^{2}\right )} x\right )} x\right )} x\right )} x - \frac {384 \, c^{2}}{a}\right )} x - \frac {1155 \, c^{2}}{a^{2}}\right )} x - \frac {1536 \, c^{2}}{a^{3}}\right )} - \frac {11 \, c^{3} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{128 \, a^{2} \sqrt {-c} {\left | a \right |}} \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^2*(-a^2*c*x^2+c)^(5/2),x, algorithm="gi 
ac")
 

Output:

1/13440*sqrt(-a^2*c*x^2 + c)*((2*((1855*c^2 + 4*(768*a*c^2 + 5*(7*a^2*c^2 
- 6*(7*a^4*c^2*x + 16*a^3*c^2)*x)*x)*x)*x - 384*c^2/a)*x - 1155*c^2/a^2)*x 
 - 1536*c^2/a^3) - 11/128*c^3*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + 
c)))/(a^2*sqrt(-c)*abs(a))
 

Mupad [F(-1)]

Timed out. \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=\int -\frac {x^2\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \] Input:

int(-(x^2*(c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
 

Output:

int(-(x^2*(c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(a^2*x^2 - 1), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89 \[ \int e^{2 \text {arctanh}(a x)} x^2 \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (1155 \mathit {asin} \left (a x \right )-1680 \sqrt {-a^{2} x^{2}+1}\, a^{7} x^{7}-3840 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{6}+280 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}+6144 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+3710 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-768 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-1155 \sqrt {-a^{2} x^{2}+1}\, a x -1536 \sqrt {-a^{2} x^{2}+1}+1536\right )}{13440 a^{3}} \] Input:

int((a*x+1)^2/(-a^2*x^2+1)*x^2*(-a^2*c*x^2+c)^(5/2),x)
 

Output:

(sqrt(c)*c**2*(1155*asin(a*x) - 1680*sqrt( - a**2*x**2 + 1)*a**7*x**7 - 38 
40*sqrt( - a**2*x**2 + 1)*a**6*x**6 + 280*sqrt( - a**2*x**2 + 1)*a**5*x**5 
 + 6144*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 3710*sqrt( - a**2*x**2 + 1)*a** 
3*x**3 - 768*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 1155*sqrt( - a**2*x**2 + 1 
)*a*x - 1536*sqrt( - a**2*x**2 + 1) + 1536))/(13440*a**3)