\(\int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx\) [327]

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 [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 164 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {c^3 x \sqrt {1-a^2 x^2}}{8 a^3}+\frac {c^3 x^3 \sqrt {1-a^2 x^2}}{12 a}-\frac {1}{3} a c^3 x^5 \sqrt {1-a^2 x^2}-\frac {11 c^3 \left (1-a^2 x^2\right )^{3/2}}{21 a^4}-\frac {1}{7} c^3 x^4 \left (1-a^2 x^2\right )^{3/2}+\frac {11 c^3 \left (1-a^2 x^2\right )^{5/2}}{35 a^4}-\frac {c^3 \arcsin (a x)}{8 a^4} \] Output:

1/8*c^3*x*(-a^2*x^2+1)^(1/2)/a^3+1/12*c^3*x^3*(-a^2*x^2+1)^(1/2)/a-1/3*a*c 
^3*x^5*(-a^2*x^2+1)^(1/2)-11/21*c^3*(-a^2*x^2+1)^(3/2)/a^4-1/7*c^3*x^4*(-a 
^2*x^2+1)^(3/2)+11/35*c^3*(-a^2*x^2+1)^(5/2)/a^4-1/8*c^3*arcsin(a*x)/a^4
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.55 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {c^3 \left (\sqrt {1-a^2 x^2} \left (-176+105 a x-88 a^2 x^2+70 a^3 x^3+144 a^4 x^4-280 a^5 x^5+120 a^6 x^6\right )+210 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{840 a^4} \] Input:

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

Output:

(c^3*(Sqrt[1 - a^2*x^2]*(-176 + 105*a*x - 88*a^2*x^2 + 70*a^3*x^3 + 144*a^ 
4*x^4 - 280*a^5*x^5 + 120*a^6*x^6) + 210*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/( 
840*a^4)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.842, Rules used = {6678, 27, 541, 25, 27, 533, 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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 541

\(\displaystyle c^3 \left (-\frac {\int -a^2 x^3 (11-14 a x) \sqrt {1-a^2 x^2}dx}{7 a^2}-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c^3 \left (\frac {\int a^2 x^3 (11-14 a x) \sqrt {1-a^2 x^2}dx}{7 a^2}-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \left (\frac {1}{7} \int x^3 (11-14 a x) \sqrt {1-a^2 x^2}dx-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {\int -a x (22-35 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}+\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\int a x (22-35 a x) \sqrt {1-a^2 x^2}dx}{5 a^2}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\int x (22-35 a x) \sqrt {1-a^2 x^2}dx}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {\int -a (35-88 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}+\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int a (35-88 a x) \sqrt {1-a^2 x^2}dx}{4 a^2}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {\int (35-88 a x) \sqrt {1-a^2 x^2}dx}{4 a}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {35 \int \sqrt {1-a^2 x^2}dx+\frac {88 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 211

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {35 \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {88 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^3 \left (\frac {1}{7} \left (\frac {7 x^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {\frac {11 x^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}-\frac {\frac {35 x \left (1-a^2 x^2\right )^{3/2}}{4 a}-\frac {35 \left (\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {88 \left (1-a^2 x^2\right )^{3/2}}{3 a}}{4 a}}{5 a}}{a}\right )-\frac {1}{7} x^4 \left (1-a^2 x^2\right )^{3/2}\right )\)

Input:

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

Output:

c^3*(-1/7*(x^4*(1 - a^2*x^2)^(3/2)) + ((7*x^3*(1 - a^2*x^2)^(3/2))/(3*a) - 
 ((11*x^2*(1 - a^2*x^2)^(3/2))/(5*a) - ((35*x*(1 - a^2*x^2)^(3/2))/(4*a) - 
 ((88*(1 - a^2*x^2)^(3/2))/(3*a) + 35*((x*Sqrt[1 - a^2*x^2])/2 + ArcSin[a* 
x]/(2*a)))/(4*a))/(5*a))/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 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 = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67

method result size
risch \(-\frac {\left (120 x^{6} a^{6}-280 a^{5} x^{5}+144 a^{4} x^{4}+70 a^{3} x^{3}-88 a^{2} x^{2}+105 a x -176\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{840 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{8 a^{3} \sqrt {a^{2}}}\) \(110\)
meijerg \(-\frac {c^{3} \left (\frac {32 \sqrt {\pi }}{35}-\frac {\sqrt {\pi }\, \left (40 x^{6} a^{6}+48 a^{4} x^{4}+64 a^{2} x^{2}+128\right ) \sqrt {-a^{2} x^{2}+1}}{140}\right )}{2 a^{4} \sqrt {\pi }}-\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (56 a^{4} x^{4}+70 a^{2} x^{2}+105\right ) \sqrt {-a^{2} x^{2}+1}}{168 a^{6}}+\frac {5 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{a^{3} \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{3} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{2 a^{4} \sqrt {\pi }}\) \(266\)
default \(-c^{3} \left (a^{4} \left (-\frac {x^{6} \sqrt {-a^{2} x^{2}+1}}{7 a^{2}}+\frac {-\frac {6 x^{4} \sqrt {-a^{2} x^{2}+1}}{35 a^{2}}+\frac {6 \left (-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}\right )}{7 a^{2}}}{a^{2}}\right )+\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}+\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}+2 a \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-2 a^{3} \left (-\frac {x^{5} \sqrt {-a^{2} x^{2}+1}}{6 a^{2}}+\frac {-\frac {5 x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}+\frac {5 \left (-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}\right )}{6 a^{2}}}{a^{2}}\right )\right )\) \(322\)

Input:

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

Output:

-1/840*(120*a^6*x^6-280*a^5*x^5+144*a^4*x^4+70*a^3*x^3-88*a^2*x^2+105*a*x- 
176)*(a^2*x^2-1)/a^4/(-a^2*x^2+1)^(1/2)*c^3-1/8/a^3/(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 = 114, normalized size of antiderivative = 0.70 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {210 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (120 \, a^{6} c^{3} x^{6} - 280 \, a^{5} c^{3} x^{5} + 144 \, a^{4} c^{3} x^{4} + 70 \, a^{3} c^{3} x^{3} - 88 \, a^{2} c^{3} x^{2} + 105 \, a c^{3} x - 176 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{840 \, a^{4}} \] Input:

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

Output:

1/840*(210*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (120*a^6*c^3*x^6 - 
 280*a^5*c^3*x^5 + 144*a^4*c^3*x^4 + 70*a^3*c^3*x^3 - 88*a^2*c^3*x^2 + 105 
*a*c^3*x - 176*c^3)*sqrt(-a^2*x^2 + 1))/a^4
 

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.11 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (\frac {a^{2} c^{3} x^{6}}{7} - \frac {a c^{3} x^{5}}{3} + \frac {6 c^{3} x^{4}}{35} + \frac {c^{3} x^{3}}{12 a} - \frac {11 c^{3} x^{2}}{105 a^{2}} + \frac {c^{3} x}{8 a^{3}} - \frac {22 c^{3}}{105 a^{4}}\right ) - \frac {c^{3} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 a^{3} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\- \frac {a^{4} c^{3} x^{8}}{8} + \frac {2 a^{3} c^{3} x^{7}}{7} - \frac {2 a c^{3} x^{5}}{5} + \frac {c^{3} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(-a**2*x**2 + 1)*(a**2*c**3*x**6/7 - a*c**3*x**5/3 + 6*c**3 
*x**4/35 + c**3*x**3/(12*a) - 11*c**3*x**2/(105*a**2) + c**3*x/(8*a**3) - 
22*c**3/(105*a**4)) - c**3*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 
 1))/(8*a**3*sqrt(-a**2)), Ne(a**2, 0)), (-a**4*c**3*x**8/8 + 2*a**3*c**3* 
x**7/7 - 2*a*c**3*x**5/5 + c**3*x**4/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {1}{7} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{6} - \frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{5} + \frac {6}{35} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{4} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{3} x^{3}}{12 \, a} - \frac {11 \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{2}}{105 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{3} x}{8 \, a^{3}} - \frac {c^{3} \arcsin \left (a x\right )}{8 \, a^{4}} - \frac {22 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{105 \, a^{4}} \] Input:

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

Output:

1/7*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^6 - 1/3*sqrt(-a^2*x^2 + 1)*a*c^3*x^5 + 6/ 
35*sqrt(-a^2*x^2 + 1)*c^3*x^4 + 1/12*sqrt(-a^2*x^2 + 1)*c^3*x^3/a - 11/105 
*sqrt(-a^2*x^2 + 1)*c^3*x^2/a^2 + 1/8*sqrt(-a^2*x^2 + 1)*c^3*x/a^3 - 1/8*c 
^3*arcsin(a*x)/a^4 - 22/105*sqrt(-a^2*x^2 + 1)*c^3/a^4
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.63 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {1}{840} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left ({\left (\frac {35 \, c^{3}}{a} + 4 \, {\left (18 \, c^{3} + 5 \, {\left (3 \, a^{2} c^{3} x - 7 \, a c^{3}\right )} x\right )} x\right )} x - \frac {44 \, c^{3}}{a^{2}}\right )} x + \frac {105 \, c^{3}}{a^{3}}\right )} x - \frac {176 \, c^{3}}{a^{4}}\right )} - \frac {c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, a^{3} {\left | a \right |}} \] Input:

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

Output:

1/840*sqrt(-a^2*x^2 + 1)*((2*((35*c^3/a + 4*(18*c^3 + 5*(3*a^2*c^3*x - 7*a 
*c^3)*x)*x)*x - 44*c^3/a^2)*x + 105*c^3/a^3)*x - 176*c^3/a^4) - 1/8*c^3*ar 
csin(a*x)*sgn(a)/(a^3*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.08 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {6\,c^3\,x^4\,\sqrt {1-a^2\,x^2}}{35}-\frac {22\,c^3\,\sqrt {1-a^2\,x^2}}{105\,a^4}+\frac {c^3\,x\,\sqrt {1-a^2\,x^2}}{8\,a^3}-\frac {a\,c^3\,x^5\,\sqrt {1-a^2\,x^2}}{3}-\frac {c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^3\,\sqrt {-a^2}}+\frac {c^3\,x^3\,\sqrt {1-a^2\,x^2}}{12\,a}-\frac {11\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{105\,a^2}+\frac {a^2\,c^3\,x^6\,\sqrt {1-a^2\,x^2}}{7} \] Input:

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

Output:

(6*c^3*x^4*(1 - a^2*x^2)^(1/2))/35 - (22*c^3*(1 - a^2*x^2)^(1/2))/(105*a^4 
) + (c^3*x*(1 - a^2*x^2)^(1/2))/(8*a^3) - (a*c^3*x^5*(1 - a^2*x^2)^(1/2))/ 
3 - (c^3*asinh(x*(-a^2)^(1/2)))/(8*a^3*(-a^2)^(1/2)) + (c^3*x^3*(1 - a^2*x 
^2)^(1/2))/(12*a) - (11*c^3*x^2*(1 - a^2*x^2)^(1/2))/(105*a^2) + (a^2*c^3* 
x^6*(1 - a^2*x^2)^(1/2))/7
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.85 \[ \int e^{\text {arctanh}(a x)} x^3 (c-a c x)^3 \, dx=\frac {c^{3} \left (-105 \mathit {asin} \left (a x \right )+120 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{6}-280 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}+144 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+70 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-88 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+105 \sqrt {-a^{2} x^{2}+1}\, a x -176 \sqrt {-a^{2} x^{2}+1}+176\right )}{840 a^{4}} \] Input:

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

Output:

(c**3*( - 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 + 144*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 70*sq 
rt( - a**2*x**2 + 1)*a**3*x**3 - 88*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 105 
*sqrt( - a**2*x**2 + 1)*a*x - 176*sqrt( - a**2*x**2 + 1) + 176))/(840*a**4 
)