\(\int e^{2 \text {arctanh}(a x)} x (c-a^2 c x^2)^{5/2} \, dx\) [1122]

Optimal result

Integrand size = 25, antiderivative size = 157 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 x \sqrt {c-a^2 c x^2}}{8 a}+\frac {1}{4} a c^2 x^3 \sqrt {c-a^2 c x^2}+\frac {1}{3} a c x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {9 \left (c-a^2 c x^2\right )^{5/2}}{35 a^2}-\frac {1}{7} x^2 \left (c-a^2 c x^2\right )^{5/2}+\frac {c^{5/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^2} \] Output:

-1/8*c^2*x*(-a^2*c*x^2+c)^(1/2)/a+1/4*a*c^2*x^3*(-a^2*c*x^2+c)^(1/2)+1/3*a 
*c*x^3*(-a^2*c*x^2+c)^(3/2)-9/35*(-a^2*c*x^2+c)^(5/2)/a^2-1/7*x^2*(-a^2*c* 
x^2+c)^(5/2)+1/8*c^(5/2)*arctan(a*c^(1/2)*x/(-a^2*c*x^2+c)^(1/2))/a^2
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.73 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 \left (\sqrt {c-a^2 c x^2} \left (216+105 a x-312 a^2 x^2-490 a^3 x^3-24 a^4 x^4+280 a^5 x^5+120 a^6 x^6\right )+105 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )\right )}{840 a^2} \] Input:

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

Output:

-1/840*(c^2*(Sqrt[c - a^2*c*x^2]*(216 + 105*a*x - 312*a^2*x^2 - 490*a^3*x^ 
3 - 24*a^4*x^4 + 280*a^5*x^5 + 120*a^6*x^6) + 105*Sqrt[c]*ArcTan[(a*x*Sqrt 
[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))]))/a^2
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {6701, 541, 25, 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.

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

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 541

\(\displaystyle c \left (-\frac {\int -a^2 c x (14 a x+9) \left (c-a^2 c x^2\right )^{3/2}dx}{7 a^2 c}-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int a^2 c x (14 a x+9) \left (c-a^2 c x^2\right )^{3/2}dx}{7 a^2 c}-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{7} \int x (14 a x+9) \left (c-a^2 c x^2\right )^{3/2}dx-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c \left (\frac {1}{7} \left (\frac {\int 2 a c (27 a x+7) \left (c-a^2 c x^2\right )^{3/2}dx}{6 a^2 c}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{7} \left (\frac {\int (27 a x+7) \left (c-a^2 c x^2\right )^{3/2}dx}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c \left (\frac {1}{7} \left (\frac {7 \int \left (c-a^2 c x^2\right )^{3/2}dx-\frac {27 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c \left (\frac {1}{7} \left (\frac {7 \left (\frac {3}{4} c \int \sqrt {c-a^2 c x^2}dx+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {27 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c \left (\frac {1}{7} \left (\frac {7 \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c-a^2 c x^2}}dx+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {27 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle c \left (\frac {1}{7} \left (\frac {7 \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {27 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle c \left (\frac {1}{7} \left (\frac {7 \left (\frac {3}{4} c \left (\frac {\sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2}\right )+\frac {1}{4} x \left (c-a^2 c x^2\right )^{3/2}\right )-\frac {27 \left (c-a^2 c x^2\right )^{5/2}}{5 a c}}{3 a}-\frac {7 x \left (c-a^2 c x^2\right )^{5/2}}{3 a c}\right )-\frac {x^2 \left (c-a^2 c x^2\right )^{5/2}}{7 c}\right )\)

Input:

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

Output:

c*(-1/7*(x^2*(c - a^2*c*x^2)^(5/2))/c + ((-7*x*(c - a^2*c*x^2)^(5/2))/(3*a 
*c) + ((-27*(c - a^2*c*x^2)^(5/2))/(5*a*c) + 7*((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))/(3*a))/7)
 

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 = 0.98 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {\left (120 x^{6} a^{6}+280 a^{5} x^{5}-24 a^{4} x^{4}-490 a^{3} x^{3}-312 a^{2} x^{2}+105 a x +216\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{840 a^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c^{3}}{8 a \sqrt {a^{2} c}}\)	\(117\)
default	\(\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{7 a^{2} c}-\frac {2 \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )}{6}\right )}{a}-\frac {2 \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}{5}-a c \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right ) \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}{8 a^{2} c}+\frac {3 c \left (-\frac {\left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{4 a^{2} c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )\right )}{a^{2}}\)	\(323\)

Input:

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

Output:

1/840*(120*a^6*x^6+280*a^5*x^5-24*a^4*x^4-490*a^3*x^3-312*a^2*x^2+105*a*x+ 
216)*(a^2*x^2-1)/a^2/(-c*(a^2*x^2-1))^(1/2)*c^3+1/8/a/(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 = 0.10 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.68 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=\left [\frac {105 \, \sqrt {-c} c^{2} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, {\left (120 \, a^{6} c^{2} x^{6} + 280 \, a^{5} c^{2} x^{5} - 24 \, a^{4} c^{2} x^{4} - 490 \, a^{3} c^{2} x^{3} - 312 \, a^{2} c^{2} x^{2} + 105 \, a c^{2} x + 216 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{1680 \, a^{2}}, -\frac {105 \, c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + {\left (120 \, a^{6} c^{2} x^{6} + 280 \, a^{5} c^{2} x^{5} - 24 \, a^{4} c^{2} x^{4} - 490 \, a^{3} c^{2} x^{3} - 312 \, a^{2} c^{2} x^{2} + 105 \, a c^{2} x + 216 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{840 \, a^{2}}\right ] \] Input:

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

Output:

[1/1680*(105*sqrt(-c)*c^2*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt( 
-c)*x - c) - 2*(120*a^6*c^2*x^6 + 280*a^5*c^2*x^5 - 24*a^4*c^2*x^4 - 490*a 
^3*c^2*x^3 - 312*a^2*c^2*x^2 + 105*a*c^2*x + 216*c^2)*sqrt(-a^2*c*x^2 + c) 
)/a^2, -1/840*(105*c^(5/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c* 
x^2 - c)) + (120*a^6*c^2*x^6 + 280*a^5*c^2*x^5 - 24*a^4*c^2*x^4 - 490*a^3* 
c^2*x^3 - 312*a^2*c^2*x^2 + 105*a*c^2*x + 216*c^2)*sqrt(-a^2*c*x^2 + c))/a 
^2]
 

Sympy [A] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.23 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=- a^{4} c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{6}}{7} - \frac {x^{4}}{35 a^{2}} - \frac {4 x^{2}}{105 a^{4}} - \frac {8}{105 a^{6}}\right ) & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{6}}{6} & \text {otherwise} \end {cases}\right ) - 2 a^{3} c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {x^{5}}{6} - \frac {x^{3}}{24 a^{2}} - \frac {x}{16 a^{4}}\right ) + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{16 a^{4}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{5}}{5} & \text {otherwise} \end {cases}\right ) + 2 a c^{2} \left (\begin {cases} \left (\frac {x^{3}}{4} - \frac {x}{8 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{8 a^{2}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \left (\frac {x^{2}}{3} - \frac {1}{3 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a**4*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, True)) - 
 2*a**3*c**2*Piecewise((sqrt(-a**2*c*x**2 + c)*(x**5/6 - x**3/(24*a**2) - 
x/(16*a**4)) + 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*log(x)/sqrt(-a**2*c*x**2), True)) 
/(16*a**4), Ne(a**2*c, 0)), (sqrt(c)*x**5/5, True)) + 2*a*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)) + c**2*Piecewise(((x**2/3 - 1/(3*a**2))*sqrt(-a**2*c*x**2 + c), N 
e(a**2*c, 0)), (sqrt(c)*x**2/2, True))
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.23 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {1}{840} \, {\left (\frac {280 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x}{a^{2}} - \frac {70 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c x}{a^{2}} - \frac {630 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c^{2} x}{a^{2}} + \frac {525 \, \sqrt {-a^{2} c x^{2} + c} c^{2} x}{a^{2}} + \frac {525 \, c^{\frac {5}{2}} \arcsin \left (a x\right )}{a^{3}} + \frac {336 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{a^{3}} - \frac {120 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}}{a^{3} c} + \frac {1260 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c^{2}}{a^{3}} - \frac {630 \, c^{4} \arcsin \left (a x - 2\right )}{a^{6} \left (-\frac {c}{a^{2}}\right )^{\frac {3}{2}}}\right )} a \] Input:

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

Output:

-1/840*(280*(-a^2*c*x^2 + c)^(5/2)*x/a^2 - 70*(-a^2*c*x^2 + c)^(3/2)*c*x/a 
^2 - 630*sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c^2*x/a^2 + 525*sqrt(-a^2*c*x^2 + 
 c)*c^2*x/a^2 + 525*c^(5/2)*arcsin(a*x)/a^3 + 336*(-a^2*c*x^2 + c)^(5/2)/a 
^3 - 120*(-a^2*c*x^2 + c)^(7/2)/(a^3*c) + 1260*sqrt(a^2*c*x^2 - 4*a*c*x + 
3*c)*c^2/a^3 - 630*c^4*arcsin(a*x - 2)/(a^6*(-c/a^2)^(3/2)))*a
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{840} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left (156 \, c^{2} + {\left (245 \, a c^{2} + 4 \, {\left (3 \, a^{2} c^{2} - 5 \, {\left (3 \, a^{4} c^{2} x + 7 \, a^{3} c^{2}\right )} x\right )} x\right )} x\right )} x - \frac {105 \, c^{2}}{a}\right )} x - \frac {216 \, c^{2}}{a^{2}}\right )} - \frac {c^{3} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{8 \, a \sqrt {-c} {\left | a \right |}} \] Input:

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

Output:

1/840*sqrt(-a^2*c*x^2 + c)*((2*(156*c^2 + (245*a*c^2 + 4*(3*a^2*c^2 - 5*(3 
*a^4*c^2*x + 7*a^3*c^2)*x)*x)*x)*x - 105*c^2/a)*x - 216*c^2/a^2) - 1/8*c^3 
*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a*sqrt(-c)*abs(a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.90 \[ \int e^{2 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (105 \mathit {asin} \left (a x \right )-120 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{6}-280 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}+24 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+490 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+312 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-105 \sqrt {-a^{2} x^{2}+1}\, a x -216 \sqrt {-a^{2} x^{2}+1}+216\right )}{840 a^{2}} \] Input:

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

Output:

(sqrt(c)*c**2*(105*asin(a*x) - 120*sqrt( - a**2*x**2 + 1)*a**6*x**6 - 280* 
sqrt( - a**2*x**2 + 1)*a**5*x**5 + 24*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 4 
90*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 312*sqrt( - a**2*x**2 + 1)*a**2*x**2 
 - 105*sqrt( - a**2*x**2 + 1)*a*x - 216*sqrt( - a**2*x**2 + 1) + 216))/(84 
0*a**2)