\(\int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx\) [328]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=-\frac {3 c^3 x \sqrt {1-a^2 x^2}}{16 a^2}+\frac {3}{8} c^3 x^3 \sqrt {1-a^2 x^2}+\frac {2 c^3 \left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {1}{6} c^3 x^3 \left (1-a^2 x^2\right )^{3/2}-\frac {2 c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^3}+\frac {3 c^3 \arcsin (a x)}{16 a^3} \] Output:

-3/16*c^3*x*(-a^2*x^2+1)^(1/2)/a^2+3/8*c^3*x^3*(-a^2*x^2+1)^(1/2)+2/3*c^3* 
(-a^2*x^2+1)^(3/2)/a^3-1/6*c^3*x^3*(-a^2*x^2+1)^(3/2)-2/5*c^3*(-a^2*x^2+1) 
^(5/2)/a^3+3/16*c^3*arcsin(a*x)/a^3
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.61 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\frac {c^3 \left (\sqrt {1-a^2 x^2} \left (64-45 a x+32 a^2 x^2+50 a^3 x^3-96 a^4 x^4+40 a^5 x^5\right )-90 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{240 a^3} \] Input:

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

Output:

(c^3*(Sqrt[1 - a^2*x^2]*(64 - 45*a*x + 32*a^2*x^2 + 50*a^3*x^3 - 96*a^4*x^ 
4 + 40*a^5*x^5) - 90*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(240*a^3)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {6678, 27, 541, 27, 533, 25, 27, 533, 25, 27, 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.

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

\(\Big \downarrow \) 6678

\(\displaystyle c \int c^2 x^2 (1-a x)^2 \sqrt {1-a^2 x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \int x^2 (1-a x)^2 \sqrt {1-a^2 x^2}dx\)

\(\Big \downarrow \) 541

\(\displaystyle c^3 \left (-\frac {\int -3 a^2 x^2 (3-4 a x) \sqrt {1-a^2 x^2}dx}{6 a^2}-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \left (\frac {1}{2} \int x^2 (3-4 a x) \sqrt {1-a^2 x^2}dx-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {\int -a (15-32 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}+\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int a (15-32 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int (15-32 a x) \sqrt {1-a^2 x^2}dx}{4 a}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {15 \int \sqrt {1-a^2 x^2}dx+\frac {32 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {15 \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {32 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^3 \left (\frac {1}{2} \left (\frac {4 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {15 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {15 \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {32 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}\right )-\frac {1}{6} x^3 \left (1-a^2 x^2\right )^{3/2}\right )\)

Input:

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

Output:

c^3*(-1/6*(x^3*(1 - a^2*x^2)^(3/2)) + ((4*x^2*(1 - a^2*x^2)^(3/2))/(5*a) - 
 ((15*x*(1 - a^2*x^2)^(3/2))/(4*a) - ((32*(1 - a^2*x^2)^(3/2))/(3*a) + 15* 
((x*Sqrt[1 - a^2*x^2])/2 + ArcSin[a*x]/(2*a)))/(4*a))/(5*a))/2)
 

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 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 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 6678

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* 
(x_))^(m_.), x_Symbol] :> Simp[c^n   Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - 
 a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 
0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 
, 0]) && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (40 a^{5} x^{5}-96 a^{4} x^{4}+50 a^{3} x^{3}+32 a^{2} x^{2}-45 a x +64\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{240 a^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{16 a^{2} \sqrt {a^{2}}}\)	\(102\)
meijerg	\(\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (56 a^{4} x^{4}+70 a^{2} x^{2}+105\right ) \sqrt {-a^{2} x^{2}+1}}{168 a^{6}}+\frac {5 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{2 a^{2} \sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {c^{3} \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}\right )}{a^{3} \sqrt {\pi }}-\frac {c^{3} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{a^{3} \sqrt {\pi }}-\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{2 a^{2} \sqrt {\pi }\, \sqrt {-a^{2}}}\)	\(247\)
default	\(-c^{3} \left (a^{4} \left (-\frac {x^{5} \sqrt {-a^{2} x^{2}+1}}{6 a^{2}}+\frac {-\frac {5 x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}+\frac {5 \left (-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}\right )}{6 a^{2}}}{a^{2}}\right )+\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}+2 a \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )-2 a^{3} \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )\right )\)	\(270\)

Input:

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

Output:

-1/240*(40*a^5*x^5-96*a^4*x^4+50*a^3*x^3+32*a^2*x^2-45*a*x+64)*(a^2*x^2-1) 
/a^3/(-a^2*x^2+1)^(1/2)*c^3+3/16/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^ 
2*x^2+1)^(1/2))*c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=-\frac {90 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (40 \, a^{5} c^{3} x^{5} - 96 \, a^{4} c^{3} x^{4} + 50 \, a^{3} c^{3} x^{3} + 32 \, a^{2} c^{3} x^{2} - 45 \, a c^{3} x + 64 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{240 \, a^{3}} \] Input:

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

Output:

-1/240*(90*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (40*a^5*c^3*x^5 - 
96*a^4*c^3*x^4 + 50*a^3*c^3*x^3 + 32*a^2*c^3*x^2 - 45*a*c^3*x + 64*c^3)*sq 
rt(-a^2*x^2 + 1))/a^3
 

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.26 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (\frac {a^{2} c^{3} x^{5}}{6} - \frac {2 a c^{3} x^{4}}{5} + \frac {5 c^{3} x^{3}}{24} + \frac {2 c^{3} x^{2}}{15 a} - \frac {3 c^{3} x}{16 a^{2}} + \frac {4 c^{3}}{15 a^{3}}\right ) + \frac {3 c^{3} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{16 a^{2} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\- \frac {a^{4} c^{3} x^{7}}{7} + \frac {a^{3} c^{3} x^{6}}{3} - \frac {a c^{3} x^{4}}{2} + \frac {c^{3} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(-a**2*x**2 + 1)*(a**2*c**3*x**5/6 - 2*a*c**3*x**4/5 + 5*c* 
*3*x**3/24 + 2*c**3*x**2/(15*a) - 3*c**3*x/(16*a**2) + 4*c**3/(15*a**3)) + 
 3*c**3*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(16*a**2*sqrt( 
-a**2)), Ne(a**2, 0)), (-a**4*c**3*x**7/7 + a**3*c**3*x**6/3 - a*c**3*x**4 
/2 + c**3*x**3/3, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{5} - \frac {2}{5} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{4} + \frac {5}{24} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{3} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{2}}{15 \, a} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} c^{3} x}{16 \, a^{2}} + \frac {3 \, c^{3} \arcsin \left (a x\right )}{16 \, a^{3}} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{15 \, a^{3}} \] Input:

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

Output:

1/6*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^5 - 2/5*sqrt(-a^2*x^2 + 1)*a*c^3*x^4 + 5/ 
24*sqrt(-a^2*x^2 + 1)*c^3*x^3 + 2/15*sqrt(-a^2*x^2 + 1)*c^3*x^2/a - 3/16*s 
qrt(-a^2*x^2 + 1)*c^3*x/a^2 + 3/16*c^3*arcsin(a*x)/a^3 + 4/15*sqrt(-a^2*x^ 
2 + 1)*c^3/a^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.68 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\frac {3 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{16 \, a^{2} {\left | a \right |}} + \frac {1}{240} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (\frac {16 \, c^{3}}{a} + {\left (25 \, c^{3} + 4 \, {\left (5 \, a^{2} c^{3} x - 12 \, a c^{3}\right )} x\right )} x\right )} x - \frac {45 \, c^{3}}{a^{2}}\right )} x + \frac {64 \, c^{3}}{a^{3}}\right )} \] Input:

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

Output:

3/16*c^3*arcsin(a*x)*sgn(a)/(a^2*abs(a)) + 1/240*sqrt(-a^2*x^2 + 1)*((2*(1 
6*c^3/a + (25*c^3 + 4*(5*a^2*c^3*x - 12*a*c^3)*x)*x)*x - 45*c^3/a^2)*x + 6 
4*c^3/a^3)
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\frac {4\,c^3\,\sqrt {1-a^2\,x^2}}{15\,a^3}+\frac {5\,c^3\,x^3\,\sqrt {1-a^2\,x^2}}{24}-\frac {3\,c^3\,x\,\sqrt {1-a^2\,x^2}}{16\,a^2}-\frac {2\,a\,c^3\,x^4\,\sqrt {1-a^2\,x^2}}{5}+\frac {3\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{16\,a^2\,\sqrt {-a^2}}+\frac {2\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a}+\frac {a^2\,c^3\,x^5\,\sqrt {1-a^2\,x^2}}{6} \] Input:

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

Output:

(4*c^3*(1 - a^2*x^2)^(1/2))/(15*a^3) + (5*c^3*x^3*(1 - a^2*x^2)^(1/2))/24 
- (3*c^3*x*(1 - a^2*x^2)^(1/2))/(16*a^2) - (2*a*c^3*x^4*(1 - a^2*x^2)^(1/2 
))/5 + (3*c^3*asinh(x*(-a^2)^(1/2)))/(16*a^2*(-a^2)^(1/2)) + (2*c^3*x^2*(1 
 - a^2*x^2)^(1/2))/(15*a) + (a^2*c^3*x^5*(1 - a^2*x^2)^(1/2))/6
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.88 \[ \int e^{\text {arctanh}(a x)} x^2 (c-a c x)^3 \, dx=\frac {c^{3} \left (45 \mathit {asin} \left (a x \right )+40 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}-96 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+50 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+32 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-45 \sqrt {-a^{2} x^{2}+1}\, a x +64 \sqrt {-a^{2} x^{2}+1}-64\right )}{240 a^{3}} \] Input:

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

Output:

(c**3*(45*asin(a*x) + 40*sqrt( - a**2*x**2 + 1)*a**5*x**5 - 96*sqrt( - a** 
2*x**2 + 1)*a**4*x**4 + 50*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 32*sqrt( - a 
**2*x**2 + 1)*a**2*x**2 - 45*sqrt( - a**2*x**2 + 1)*a*x + 64*sqrt( - a**2* 
x**2 + 1) - 64))/(240*a**3)