\(\int e^{-3 \text {arctanh}(a x)} (c-a^2 c x^2)^4 \, dx\) [1291]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {55}{128} c^4 x \sqrt {1-a^2 x^2}+\frac {55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac {11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac {11}{56} c^4 x \left (1-a^2 x^2\right )^{7/2}+\frac {11 c^4 \left (1-a^2 x^2\right )^{9/2}}{63 a}+\frac {2 c^4 \left (1-a^2 x^2\right )^{11/2}}{7 a (1+a x)^2}+\frac {55 c^4 \arcsin (a x)}{128 a} \] Output:

55/128*c^4*x*(-a^2*x^2+1)^(1/2)+55/192*c^4*x*(-a^2*x^2+1)^(3/2)+11/48*c^4* 
x*(-a^2*x^2+1)^(5/2)+11/56*c^4*x*(-a^2*x^2+1)^(7/2)+11/63*c^4*(-a^2*x^2+1) 
^(9/2)/a+2/7*c^4*(-a^2*x^2+1)^(11/2)/a/(a*x+1)^2+55/128*c^4*arcsin(a*x)/a
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.68 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {c^4 \left (\sqrt {1-a^2 x^2} \left (-3712-4599 a x+10240 a^2 x^2-3066 a^3 x^3-8448 a^4 x^4+7224 a^5 x^5+1024 a^6 x^6-3024 a^7 x^7+896 a^8 x^8\right )+6930 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{8064 a} \] Input:

Integrate[(c - a^2*c*x^2)^4/E^(3*ArcTanh[a*x]),x]
 

Output:

-1/8064*(c^4*(Sqrt[1 - a^2*x^2]*(-3712 - 4599*a*x + 10240*a^2*x^2 - 3066*a 
^3*x^3 - 8448*a^4*x^4 + 7224*a^5*x^5 + 1024*a^6*x^6 - 3024*a^7*x^7 + 896*a 
^8*x^8) + 6930*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/a
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6689, 469, 469, 455, 211, 211, 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.

\(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx\)

\(\Big \downarrow \) 6689

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

\(\Big \downarrow \) 469

\(\displaystyle c^4 \left (\frac {11}{9} \int (1-a x)^2 \left (1-a^2 x^2\right )^{5/2}dx+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 469

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \int (1-a x) \left (1-a^2 x^2\right )^{5/2}dx+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\int \left (1-a^2 x^2\right )^{5/2}dx+\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \int \left (1-a^2 x^2\right )^{3/2}dx+\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \int \sqrt {1-a^2 x^2}dx+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )+\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )+\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^4 \left (\frac {11}{9} \left (\frac {9}{8} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2}\right )+\frac {\left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{7/2}}{8 a}\right )+\frac {(1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}\right )\)

Input:

Int[(c - a^2*c*x^2)^4/E^(3*ArcTanh[a*x]),x]
 

Output:

c^4*(((1 - a*x)^2*(1 - a^2*x^2)^(7/2))/(9*a) + (11*(((1 - a*x)*(1 - a^2*x^ 
2)^(7/2))/(8*a) + (9*((x*(1 - a^2*x^2)^(5/2))/6 + (1 - a^2*x^2)^(7/2)/(7*a 
) + (5*((x*(1 - a^2*x^2)^(3/2))/4 + (3*((x*Sqrt[1 - a^2*x^2])/2 + ArcSin[a 
*x]/(2*a)))/4))/6))/8))/9)
 

Defintions of rubi rules used

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 469

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

rule 6689

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> 
Simp[c^p   Int[(1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n, x], x] /; FreeQ[{a, c, 
d, p}, x] && EqQ[a^2*c + d, 0] && IntegerQ[p] && ILtQ[(n - 1)/2, 0] &&  !In 
tegerQ[p - n/2]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (896 a^{8} x^{8}-3024 a^{7} x^{7}+1024 x^{6} a^{6}+7224 a^{5} x^{5}-8448 a^{4} x^{4}-3066 a^{3} x^{3}+10240 a^{2} x^{2}-4599 a x -3712\right ) \left (a^{2} x^{2}-1\right ) c^{4}}{8064 a \sqrt {-a^{2} x^{2}+1}}+\frac {55 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{4}}{128 \sqrt {a^{2}}}\)	\(123\)
default	\(c^{4} \left (\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}}+a^{5} \left (-\frac {x^{4} \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{9 a^{2}}+\frac {-\frac {4 x^{2} \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{63 a^{2}}-\frac {8 \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{315 a^{4}}}{a^{2}}\right )+\frac {3 \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{5 a}+2 a^{2} \left (-\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{6 a^{2}}+\frac {\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}}}{6 a^{2}}\right )+2 a^{3} \left (-\frac {x^{2} \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{7 a^{2}}-\frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{35 a^{4}}\right )-3 a^{4} \left (-\frac {x^{3} \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{8 a^{2}}+\frac {-\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{16 a^{2}}+\frac {\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}}}{16 a^{2}}}{a^{2}}\right )\right )\)	\(391\)

Input:

int((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8064*(896*a^8*x^8-3024*a^7*x^7+1024*a^6*x^6+7224*a^5*x^5-8448*a^4*x^4-30 
66*a^3*x^3+10240*a^2*x^2-4599*a*x-3712)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c 
^4+55/128/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {6930 \, c^{4} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (896 \, a^{8} c^{4} x^{8} - 3024 \, a^{7} c^{4} x^{7} + 1024 \, a^{6} c^{4} x^{6} + 7224 \, a^{5} c^{4} x^{5} - 8448 \, a^{4} c^{4} x^{4} - 3066 \, a^{3} c^{4} x^{3} + 10240 \, a^{2} c^{4} x^{2} - 4599 \, a c^{4} x - 3712 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{8064 \, a} \] Input:

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fric 
as")
 

Output:

-1/8064*(6930*c^4*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (896*a^8*c^4*x^ 
8 - 3024*a^7*c^4*x^7 + 1024*a^6*c^4*x^6 + 7224*a^5*c^4*x^5 - 8448*a^4*c^4* 
x^4 - 3066*a^3*c^4*x^3 + 10240*a^2*c^4*x^2 - 4599*a*c^4*x - 3712*c^4)*sqrt 
(-a^2*x^2 + 1))/a
 

Sympy [A] (verification not implemented)

Time = 2.96 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.46 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx =\text {Too large to display} \] Input:

integrate((-a**2*c*x**2+c)**4/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
 

Output:

-a**7*c**4*Piecewise((sqrt(-a**2*x**2 + 1)*(x**8/9 - x**6/(63*a**2) - 2*x* 
*4/(105*a**4) - 8*x**2/(315*a**6) - 16/(315*a**8)), Ne(a**2, 0)), (x**8/8, 
 True)) + 3*a**6*c**4*Piecewise((sqrt(-a**2*x**2 + 1)*(x**7/8 - x**5/(48*a 
**2) - 5*x**3/(192*a**4) - 5*x/(128*a**6)) + 5*log(-2*a**2*x + 2*sqrt(-a** 
2)*sqrt(-a**2*x**2 + 1))/(128*a**6*sqrt(-a**2)), Ne(a**2, 0)), (x**7/7, Tr 
ue)) - a**5*c**4*Piecewise((sqrt(-a**2*x**2 + 1)*(x**6/7 - x**4/(35*a**2) 
- 4*x**2/(105*a**4) - 8/(105*a**6)), Ne(a**2, 0)), (x**6/6, True)) - 5*a** 
4*c**4*Piecewise((sqrt(-a**2*x**2 + 1)*(x**5/6 - x**3/(24*a**2) - x/(16*a* 
*4)) + log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(16*a**4*sqrt(- 
a**2)), Ne(a**2, 0)), (x**5/5, True)) + 5*a**3*c**4*Piecewise((sqrt(-a**2* 
x**2 + 1)*(x**4/5 - x**2/(15*a**2) - 2/(15*a**4)), Ne(a**2, 0)), (x**4/4, 
True)) + a**2*c**4*Piecewise(((x**3/4 - x/(8*a**2))*sqrt(-a**2*x**2 + 1) + 
 log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(8*a**2*sqrt(-a**2)), 
 Ne(a**2, 0)), (x**3/3, True)) - 3*a*c**4*Piecewise(((x**2/3 - 1/(3*a**2)) 
*sqrt(-a**2*x**2 + 1), Ne(a**2, 0)), (x**2/2, True)) + c**4*Piecewise((x*s 
qrt(-a**2*x**2 + 1)/2 + log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1) 
)/(2*sqrt(-a**2)), Ne(a**2, 0)), (x, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {1}{9} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{3} c^{4} x^{4} + \frac {3}{8} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{2} c^{4} x^{3} - \frac {22}{63} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a c^{4} x^{2} - \frac {7}{48} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} c^{4} x + \frac {55}{192} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{4} x + \frac {29 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} c^{4}}{63 \, a} + \frac {55}{128} \, \sqrt {-a^{2} x^{2} + 1} c^{4} x + \frac {55 \, c^{4} \arcsin \left (a x\right )}{128 \, a} \] Input:

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxi 
ma")
 

Output:

-1/9*(-a^2*x^2 + 1)^(5/2)*a^3*c^4*x^4 + 3/8*(-a^2*x^2 + 1)^(5/2)*a^2*c^4*x 
^3 - 22/63*(-a^2*x^2 + 1)^(5/2)*a*c^4*x^2 - 7/48*(-a^2*x^2 + 1)^(5/2)*c^4* 
x + 55/192*(-a^2*x^2 + 1)^(3/2)*c^4*x + 29/63*(-a^2*x^2 + 1)^(5/2)*c^4/a + 
 55/128*sqrt(-a^2*x^2 + 1)*c^4*x + 55/128*c^4*arcsin(a*x)/a
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {55 \, c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{128 \, {\left | a \right |}} + \frac {1}{8064} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {3712 \, c^{4}}{a} + {\left (4599 \, c^{4} - 2 \, {\left (5120 \, a c^{4} - {\left (1533 \, a^{2} c^{4} + 4 \, {\left (1056 \, a^{3} c^{4} - {\left (903 \, a^{4} c^{4} + 2 \, {\left (64 \, a^{5} c^{4} + 7 \, {\left (8 \, a^{7} c^{4} x - 27 \, a^{6} c^{4}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \] Input:

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac 
")
 

Output:

55/128*c^4*arcsin(a*x)*sgn(a)/abs(a) + 1/8064*sqrt(-a^2*x^2 + 1)*(3712*c^4 
/a + (4599*c^4 - 2*(5120*a*c^4 - (1533*a^2*c^4 + 4*(1056*a^3*c^4 - (903*a^ 
4*c^4 + 2*(64*a^5*c^4 + 7*(8*a^7*c^4*x - 27*a^6*c^4)*x)*x)*x)*x)*x)*x)*x)
 

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.39 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {73\,c^4\,x\,\sqrt {1-a^2\,x^2}}{128}+\frac {55\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{128\,\sqrt {-a^2}}+\frac {29\,c^4\,\sqrt {1-a^2\,x^2}}{63\,a}-\frac {80\,a\,c^4\,x^2\,\sqrt {1-a^2\,x^2}}{63}+\frac {73\,a^2\,c^4\,x^3\,\sqrt {1-a^2\,x^2}}{192}+\frac {22\,a^3\,c^4\,x^4\,\sqrt {1-a^2\,x^2}}{21}-\frac {43\,a^4\,c^4\,x^5\,\sqrt {1-a^2\,x^2}}{48}-\frac {8\,a^5\,c^4\,x^6\,\sqrt {1-a^2\,x^2}}{63}+\frac {3\,a^6\,c^4\,x^7\,\sqrt {1-a^2\,x^2}}{8}-\frac {a^7\,c^4\,x^8\,\sqrt {1-a^2\,x^2}}{9} \] Input:

int(((c - a^2*c*x^2)^4*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
 

Output:

(73*c^4*x*(1 - a^2*x^2)^(1/2))/128 + (55*c^4*asinh(x*(-a^2)^(1/2)))/(128*( 
-a^2)^(1/2)) + (29*c^4*(1 - a^2*x^2)^(1/2))/(63*a) - (80*a*c^4*x^2*(1 - a^ 
2*x^2)^(1/2))/63 + (73*a^2*c^4*x^3*(1 - a^2*x^2)^(1/2))/192 + (22*a^3*c^4* 
x^4*(1 - a^2*x^2)^(1/2))/21 - (43*a^4*c^4*x^5*(1 - a^2*x^2)^(1/2))/48 - (8 
*a^5*c^4*x^6*(1 - a^2*x^2)^(1/2))/63 + (3*a^6*c^4*x^7*(1 - a^2*x^2)^(1/2)) 
/8 - (a^7*c^4*x^8*(1 - a^2*x^2)^(1/2))/9
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.12 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^{4} \left (3465 \mathit {asin} \left (a x \right )-896 \sqrt {-a^{2} x^{2}+1}\, a^{8} x^{8}+3024 \sqrt {-a^{2} x^{2}+1}\, a^{7} x^{7}-1024 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{6}-7224 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}+8448 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+3066 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-10240 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+4599 \sqrt {-a^{2} x^{2}+1}\, a x +3712 \sqrt {-a^{2} x^{2}+1}-3712\right )}{8064 a} \] Input:

int((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
 

Output:

(c**4*(3465*asin(a*x) - 896*sqrt( - a**2*x**2 + 1)*a**8*x**8 + 3024*sqrt( 
- a**2*x**2 + 1)*a**7*x**7 - 1024*sqrt( - a**2*x**2 + 1)*a**6*x**6 - 7224* 
sqrt( - a**2*x**2 + 1)*a**5*x**5 + 8448*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 
 3066*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 10240*sqrt( - a**2*x**2 + 1)*a**2 
*x**2 + 4599*sqrt( - a**2*x**2 + 1)*a*x + 3712*sqrt( - a**2*x**2 + 1) - 37 
12))/(8064*a)