\(\int e^{-\coth ^{-1}(a x)} (c-\frac {c}{a x})^4 \, dx\) [428]

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

Optimal result

Integrand size = 22, antiderivative size = 130 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-\frac {11 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{3 a}+\frac {5 c^4 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a^2 x}+c^4 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {25 c^4 \csc ^{-1}(a x)}{2 a}-\frac {5 c^4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \] Output:

-11*c^4*(1-1/a^2/x^2)^(1/2)/a+1/3*c^4*(1-1/a^2/x^2)^(3/2)/a+5/2*c^4*(1-1/a 
^2/x^2)^(1/2)/a^2/x+c^4*(1-1/a^2/x^2)^(1/2)*x-25/2*c^4*arccsc(a*x)/a-5*c^4 
*arctanh((1-1/a^2/x^2)^(1/2))/a
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^4 \left (2-15 a x+62 a^2 x^2+9 a^3 x^3-64 a^4 x^4+6 a^5 x^5+90 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^4 \arcsin \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-30 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^4 \arcsin \left (\frac {1}{a x}\right )-30 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{6 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4} \] Input:

Integrate[(c - c/(a*x))^4/E^ArcCoth[a*x],x]
 

Output:

(c^4*(2 - 15*a*x + 62*a^2*x^2 + 9*a^3*x^3 - 64*a^4*x^4 + 6*a^5*x^5 + 90*a^ 
4*Sqrt[1 - 1/(a^2*x^2)]*x^4*ArcSin[Sqrt[1 - 1/(a*x)]/Sqrt[2]] - 30*a^4*Sqr 
t[1 - 1/(a^2*x^2)]*x^4*ArcSin[1/(a*x)] - 30*a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4* 
ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))/(6*a^5*Sqrt[1 - 1/(a^2*x^2)]*x^4)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {6731, 27, 540, 2340, 25, 2340, 25, 2340, 27, 538, 223, 243, 73, 221}

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.

\(\displaystyle \int \left (c-\frac {c}{a x}\right )^4 e^{-\coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6731

\(\displaystyle -\frac {\int \frac {c^5 \left (a-\frac {1}{x}\right )^5 x^2}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{c}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

\(\displaystyle -\frac {c^4 \left (a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )-\int \frac {\left (5 a^4-\frac {10 a^3}{x}+\frac {10 a^2}{x^2}-\frac {5 a}{x^3}+\frac {1}{x^4}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )}{a^5}\)

\(\Big \downarrow \) 2340

\(\displaystyle -\frac {c^4 \left (\frac {1}{3} a^2 \int -\frac {\left (15 a^2-\frac {30 a}{x}+\frac {32}{x^2}-\frac {15}{x^3 a}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \int \frac {\left (15 a^2-\frac {30 a}{x}+\frac {32}{x^2}-\frac {15}{x^3 a}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 2340

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}-\frac {1}{2} a^2 \int -\frac {\left (30-\frac {75}{a x}+\frac {64}{a^2 x^2}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \int \frac {\left (30-\frac {75}{a x}+\frac {64}{a^2 x^2}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 2340

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (a^2 \left (-\int -\frac {15 \left (2 a-\frac {5}{x}\right ) x}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \int \frac {\left (2 a-\frac {5}{x}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 538

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \left (2 a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-5 \int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \left (2 a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-5 a \arcsin \left (\frac {1}{a x}\right )\right )}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \left (a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}-5 a \arcsin \left (\frac {1}{a x}\right )\right )}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \left (-2 a^3 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}-5 a \arcsin \left (\frac {1}{a x}\right )\right )}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {c^4 \left (-\frac {1}{3} a^2 \left (\frac {1}{2} a^2 \left (\frac {15 \left (-2 a \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-5 a \arcsin \left (\frac {1}{a x}\right )\right )}{a}-64 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {15 a \sqrt {1-\frac {1}{a^2 x^2}}}{2 x}\right )+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+a^5 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )}{a^5}\)

Input:

Int[(c - c/(a*x))^4/E^ArcCoth[a*x],x]
 

Output:

-((c^4*((a^2*Sqrt[1 - 1/(a^2*x^2)])/(3*x^2) - a^5*Sqrt[1 - 1/(a^2*x^2)]*x 
- (a^2*((15*a*Sqrt[1 - 1/(a^2*x^2)])/(2*x) + (a^2*(-64*Sqrt[1 - 1/(a^2*x^2 
)] + (15*(-5*a*ArcSin[1/(a*x)] - 2*a*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))/a))/ 
2))/3))/a^5)
 

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 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 243
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 538
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 540
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2340
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

rule 6731
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> S 
imp[-c^n   Subst[Int[(c + d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^2), x], x, 1/ 
x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && IntegerQ[(n - 1)/2] 
&& IntegerQ[2*p]
 
Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.21

method result size
risch \(-\frac {\left (a x +1\right ) \left (64 a^{2} x^{2}-15 a x +2\right ) c^{4} \sqrt {\frac {a x -1}{a x +1}}}{6 x^{3} a^{4}}+\frac {\left (-\frac {25 a^{3} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )}{2}-\frac {5 a^{4} \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+a^{3} \sqrt {\left (a x -1\right ) \left (a x +1\right )}\right ) c^{4} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{a^{4} \left (a x -1\right )}\) \(157\)
default \(\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c^{4} \left (-66 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a^{4} x^{4}+66 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a^{2} x^{2}-75 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a^{3} x^{3}-75 a^{3} \sqrt {a^{2}}\, x^{3} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+66 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+96 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{3} x^{3}-96 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}-15 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x +2 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right )}{6 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{4} x^{3} \sqrt {a^{2}}}\) \(290\)

Input:

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

Output:

-1/6*(a*x+1)*(64*a^2*x^2-15*a*x+2)/x^3*c^4/a^4*((a*x-1)/(a*x+1))^(1/2)+(-2 
5/2*a^3*arctan(1/(a^2*x^2-1)^(1/2))-5*a^4*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1) 
^(1/2))/(a^2)^(1/2)+a^3*((a*x-1)*(a*x+1))^(1/2))*c^4/a^4/(a*x-1)*((a*x-1)/ 
(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.20 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {150 \, a^{3} c^{4} x^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 30 \, a^{3} c^{4} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 30 \, a^{3} c^{4} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (6 \, a^{4} c^{4} x^{4} - 58 \, a^{3} c^{4} x^{3} - 49 \, a^{2} c^{4} x^{2} + 13 \, a c^{4} x - 2 \, c^{4}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, a^{4} x^{3}} \] Input:

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

Output:

1/6*(150*a^3*c^4*x^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - 30*a^3*c^4*x^3*lo 
g(sqrt((a*x - 1)/(a*x + 1)) + 1) + 30*a^3*c^4*x^3*log(sqrt((a*x - 1)/(a*x 
+ 1)) - 1) + (6*a^4*c^4*x^4 - 58*a^3*c^4*x^3 - 49*a^2*c^4*x^2 + 13*a*c^4*x 
 - 2*c^4)*sqrt((a*x - 1)/(a*x + 1)))/(a^4*x^3)
 

Sympy [F]

\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^{4} \left (\int a^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}\, dx + \int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{4}}\, dx + \int \left (- \frac {4 a \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{3}}\right )\, dx + \int \frac {6 a^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x^{2}}\, dx + \int \left (- \frac {4 a^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{x}\right )\, dx\right )}{a^{4}} \] Input:

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

Output:

c**4*(Integral(a**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)), x) + Integral(sqrt( 
a*x/(a*x + 1) - 1/(a*x + 1))/x**4, x) + Integral(-4*a*sqrt(a*x/(a*x + 1) - 
 1/(a*x + 1))/x**3, x) + Integral(6*a**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) 
/x**2, x) + Integral(-4*a**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/x, x))/a**4
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.72 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {1}{3} \, {\left (\frac {75 \, c^{4} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} - \frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {87 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 61 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 55 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 45 \, c^{4} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {2 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} + a^{2}}\right )} a \] Input:

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

Output:

1/3*(75*c^4*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 - 15*c^4*log(sqrt((a*x - 
 1)/(a*x + 1)) + 1)/a^2 + 15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + 
(87*c^4*((a*x - 1)/(a*x + 1))^(7/2) + 61*c^4*((a*x - 1)/(a*x + 1))^(5/2) - 
 55*c^4*((a*x - 1)/(a*x + 1))^(3/2) - 45*c^4*sqrt((a*x - 1)/(a*x + 1)))/(2 
*(a*x - 1)*a^2/(a*x + 1) - 2*(a*x - 1)^3*a^2/(a*x + 1)^3 - (a*x - 1)^4*a^2 
/(a*x + 1)^4 + a^2))*a
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (114) = 228\).

Time = 0.17 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.04 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {25 \, c^{4} \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right ) \mathrm {sgn}\left (a x + 1\right )}{a} + \frac {5 \, c^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{{\left | a \right |}} + \frac {\sqrt {a^{2} x^{2} - 1} c^{4} \mathrm {sgn}\left (a x + 1\right )}{a} - \frac {15 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{5} c^{4} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) + 60 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{4} a c^{4} \mathrm {sgn}\left (a x + 1\right ) + 132 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} a c^{4} \mathrm {sgn}\left (a x + 1\right ) - 15 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )} c^{4} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) + 64 \, a c^{4} \mathrm {sgn}\left (a x + 1\right )}{3 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{3} a {\left | a \right |}} \] Input:

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

Output:

25*c^4*arctan(-x*abs(a) + sqrt(a^2*x^2 - 1))*sgn(a*x + 1)/a + 5*c^4*log(ab 
s(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/abs(a) + sqrt(a^2*x^2 - 1)* 
c^4*sgn(a*x + 1)/a - 1/3*(15*(x*abs(a) - sqrt(a^2*x^2 - 1))^5*c^4*abs(a)*s 
gn(a*x + 1) + 60*(x*abs(a) - sqrt(a^2*x^2 - 1))^4*a*c^4*sgn(a*x + 1) + 132 
*(x*abs(a) - sqrt(a^2*x^2 - 1))^2*a*c^4*sgn(a*x + 1) - 15*(x*abs(a) - sqrt 
(a^2*x^2 - 1))*c^4*abs(a)*sgn(a*x + 1) + 64*a*c^4*sgn(a*x + 1))/(((x*abs(a 
) - sqrt(a^2*x^2 - 1))^2 + 1)^3*a*abs(a))
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 13.66 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.42 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {25\,c^4\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}-\frac {15\,c^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}+\frac {55\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}-\frac {61\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{3}-29\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{a+\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}-\frac {2\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}}-\frac {10\,c^4\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \] Input:

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

Output:

(25*c^4*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a - (15*c^4*((a*x - 1)/(a*x + 1 
))^(1/2) + (55*c^4*((a*x - 1)/(a*x + 1))^(3/2))/3 - (61*c^4*((a*x - 1)/(a* 
x + 1))^(5/2))/3 - 29*c^4*((a*x - 1)/(a*x + 1))^(7/2))/(a + (2*a*(a*x - 1) 
)/(a*x + 1) - (2*a*(a*x - 1)^3)/(a*x + 1)^3 - (a*(a*x - 1)^4)/(a*x + 1)^4) 
 - (10*c^4*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.25 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^{4} \left (150 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}-1\right ) a^{3} x^{3}-150 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}+1\right ) a^{3} x^{3}+6 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{3} x^{3}-64 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{2} x^{2}+15 \sqrt {a x +1}\, \sqrt {a x -1}\, a x -2 \sqrt {a x +1}\, \sqrt {a x -1}-60 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right ) a^{3} x^{3}+24 a^{3} x^{3}\right )}{6 a^{4} x^{3}} \] Input:

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

Output:

(c**4*(150*atan(sqrt(a*x - 1) + sqrt(a*x + 1) - 1)*a**3*x**3 - 150*atan(sq 
rt(a*x - 1) + sqrt(a*x + 1) + 1)*a**3*x**3 + 6*sqrt(a*x + 1)*sqrt(a*x - 1) 
*a**3*x**3 - 64*sqrt(a*x + 1)*sqrt(a*x - 1)*a**2*x**2 + 15*sqrt(a*x + 1)*s 
qrt(a*x - 1)*a*x - 2*sqrt(a*x + 1)*sqrt(a*x - 1) - 60*log((sqrt(a*x - 1) + 
 sqrt(a*x + 1))/sqrt(2))*a**3*x**3 + 24*a**3*x**3))/(6*a**4*x**3)