\(\int e^{-3 \coth ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [446]

Optimal result

Integrand size = 22, antiderivative size = 105 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {5 c^2 \csc ^{-1}(a x)}{a}-\frac {5 c^2 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \] Output:

c^2*(1-1/a^2/x^2)^(1/2)/a+16*c^2*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+c^2*(1-1/ 
a^2/x^2)^(1/2)*x+5*c^2*arccsc(a*x)/a-5*c^2*arctanh((1-1/a^2/x^2)^(1/2))/a
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.45 (sec) , antiderivative size = 424, normalized size of antiderivative = 4.04 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {c^2 \left (-35 a^2 \sqrt {1+\frac {1}{a x}} x^2+315 a^3 \sqrt {1+\frac {1}{a x}} x^3+280 a^4 \sqrt {1+\frac {1}{a x}} x^4-595 a^5 \sqrt {1+\frac {1}{a x}} x^5+35 a^6 \sqrt {1+\frac {1}{a x}} x^6+910 a^4 \sqrt {1-\frac {1}{a x}} x^4 \arcsin \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )+910 a^5 \sqrt {1-\frac {1}{a x}} x^5 \arcsin \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-105 a^4 \sqrt {1-\frac {1}{a x}} x^4 \arcsin \left (\frac {1}{a x}\right )-105 a^5 \sqrt {1-\frac {1}{a x}} x^5 \arcsin \left (\frac {1}{a x}\right )-175 a^5 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1+\frac {1}{a x}} x^5 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )+7 \sqrt {2} a x (-1+a x)^3 (1+a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )+5 \sqrt {2} (-1+a x)^4 (1+a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )\right )}{35 a^5 \sqrt {1-\frac {1}{a x}} x^4 (1+a x)} \] Input:

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

Output:

(c^2*(-35*a^2*Sqrt[1 + 1/(a*x)]*x^2 + 315*a^3*Sqrt[1 + 1/(a*x)]*x^3 + 280* 
a^4*Sqrt[1 + 1/(a*x)]*x^4 - 595*a^5*Sqrt[1 + 1/(a*x)]*x^5 + 35*a^6*Sqrt[1 
+ 1/(a*x)]*x^6 + 910*a^4*Sqrt[1 - 1/(a*x)]*x^4*ArcSin[Sqrt[1 - 1/(a*x)]/Sq 
rt[2]] + 910*a^5*Sqrt[1 - 1/(a*x)]*x^5*ArcSin[Sqrt[1 - 1/(a*x)]/Sqrt[2]] - 
 105*a^4*Sqrt[1 - 1/(a*x)]*x^4*ArcSin[1/(a*x)] - 105*a^5*Sqrt[1 - 1/(a*x)] 
*x^5*ArcSin[1/(a*x)] - 175*a^5*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 + 1/(a*x)]*x^5 
*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]] + 7*Sqrt[2]*a*x*(-1 + a*x)^3*(1 + a*x)*Hyp 
ergeometric2F1[3/2, 5/2, 7/2, (1 - 1/(a*x))/2] + 5*Sqrt[2]*(-1 + a*x)^4*(1 
 + a*x)*Hypergeometric2F1[3/2, 7/2, 9/2, (1 - 1/(a*x))/2]))/(35*a^5*Sqrt[1 
 - 1/(a*x)]*x^4*(1 + a*x))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6731, 27, 528, 2338, 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.

\(\displaystyle \int \left (c-\frac {c}{a x}\right )^2 e^{-3 \coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6731

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 528

\(\displaystyle -\frac {c^2 \left (a^2 \int \frac {\left (a^3-\frac {5 a^2}{x}-\frac {5 a}{x^2}+\frac {1}{x^3}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 2338

\(\displaystyle -\frac {c^2 \left (a^2 \left (a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )-\int \frac {\left (5 a^2+\frac {5 a}{x}-\frac {1}{x^2}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 2340

\(\displaystyle -\frac {c^2 \left (a^2 \left (a^2 \int -\frac {5 \left (a+\frac {1}{x}\right ) x}{a \sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \int \frac {\left (a+\frac {1}{x}\right ) x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 538

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \left (\int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \left (a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+a \arcsin \left (\frac {1}{a x}\right )\right )-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \left (\frac {1}{2} a \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}+a \arcsin \left (\frac {1}{a x}\right )\right )-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \left (a \arcsin \left (\frac {1}{a x}\right )-a^3 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}\right )-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {c^2 \left (a^2 \left (-5 a \left (a \arcsin \left (\frac {1}{a x}\right )-a \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-a^2 \sqrt {1-\frac {1}{a^2 x^2}}+a^3 x \left (-\sqrt {1-\frac {1}{a^2 x^2}}\right )\right )-\frac {16 a^3 \left (a-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )}{a^5}\)

Input:

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

Output:

-((c^2*((-16*a^3*(a - x^(-1)))/Sqrt[1 - 1/(a^2*x^2)] + a^2*(-(a^2*Sqrt[1 - 
 1/(a^2*x^2)]) - a^3*Sqrt[1 - 1/(a^2*x^2)]*x - 5*a*(a*ArcSin[1/(a*x)] - a* 
ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))))/a^5)
 

Defintions of rubi rules used

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 528

Int[((x_)^(m_)*((c_) + (d_.)*(x_))^(n_.))/((a_) + (b_.)*(x_)^2)^(3/2), x_Sy 
mbol] :> Simp[(-2^(n - 1))*c^(m + n - 2)*((c + d*x)/(b*d^(m - 1)*Sqrt[a + b 
*x^2])), x] + Simp[c^2/a   Int[(x^m/Sqrt[a + b*x^2])*ExpandToSum[((c + d*x) 
^(n - 1) - (2^(n - 1)*c^(m + n - 1))/(d^m*x^m))/(c - d*x), x], x], x] /; Fr 
eeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && EqQ[b*c^2 + a*d^2, 0]
 

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 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

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 = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.61

method	result	size
risch	\(\frac {\left (a x +1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}}{x \,a^{2}}+\frac {\left (a \sqrt {\left (a x -1\right ) \left (a x +1\right )}+5 a \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-\frac {5 a^{2} \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+\frac {16 \sqrt {a^{2} \left (x +\frac {1}{a}\right )^{2}-2 a \left (x +\frac {1}{a}\right )}}{x +\frac {1}{a}}\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{a^{2} \left (a x -1\right )}\)	\(169\)
default	\(-\frac {\left (-\sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a^{4} x^{4}-4 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{3} x^{3}+\sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a^{2} x^{2}-7 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a^{3} x^{3}+\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}-5 a^{3} \sqrt {a^{2}}\, x^{3} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+8 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} a x -8 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}+2 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x -11 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a^{2} x^{2}+2 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-10 a^{2} \sqrt {a^{2}}\, x^{2} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+8 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, a x +\left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}-5 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x +\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x -5 a \sqrt {a^{2}}\, x \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x \right ) c^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a^{2} \sqrt {a^{2}}\, x \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )}\)	\(600\)

Input:

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

Output:

(a*x+1)/x*c^2/a^2*((a*x-1)/(a*x+1))^(1/2)+(a*((a*x-1)*(a*x+1))^(1/2)+5*a*a 
rctan(1/(a^2*x^2-1)^(1/2))-5*a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1)^(1/2))/( 
a^2)^(1/2)+16/(x+1/a)*(a^2*(x+1/a)^2-2*a*(x+1/a))^(1/2))*c^2/a^2/(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 = 0.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.14 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=-\frac {10 \, a c^{2} x \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + 5 \, a c^{2} x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, a c^{2} x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a^{2} c^{2} x^{2} + 18 \, a c^{2} x + c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} x} \] Input:

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

Output:

-(10*a*c^2*x*arctan(sqrt((a*x - 1)/(a*x + 1))) + 5*a*c^2*x*log(sqrt((a*x - 
 1)/(a*x + 1)) + 1) - 5*a*c^2*x*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (a^2* 
c^2*x^2 + 18*a*c^2*x + c^2)*sqrt((a*x - 1)/(a*x + 1)))/(a^2*x)
 

Sympy [F]

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

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

Output:

c**2*(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x**3 + x**2), x) + In 
tegral(3*a*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x**2 + x), x) + Integral(- 
3*a**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1), x) + Integral(a**3*x*s 
qrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1), x))/a**2
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.42 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=-{\left (\frac {4 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} + \frac {10 \, c^{2} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} + \frac {5 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {5 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {16 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} a \] Input:

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

Output:

-(4*c^2*sqrt((a*x - 1)/(a*x + 1))/((a*x - 1)^2*a^2/(a*x + 1)^2 - a^2) + 10 
*c^2*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 + 5*c^2*log(sqrt((a*x - 1)/(a*x 
 + 1)) + 1)/a^2 - 5*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - 16*c^2*sq 
rt((a*x - 1)/(a*x + 1))/a^2)*a
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {16\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}+\frac {4\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}}-\frac {10\,c^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {c^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\,1{}\mathrm {i}\right )\,10{}\mathrm {i}}{a} \] Input:

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

Output:

(16*c^2*((a*x - 1)/(a*x + 1))^(1/2))/a + (4*c^2*((a*x - 1)/(a*x + 1))^(1/2 
))/(a - (a*(a*x - 1)^2)/(a*x + 1)^2) - (10*c^2*atan(((a*x - 1)/(a*x + 1))^ 
(1/2)))/a + (c^2*atan(((a*x - 1)/(a*x + 1))^(1/2)*1i)*10i)/a
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.02 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx=\frac {c^{2} \left (-10 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}-1\right ) a^{2} x^{2}-10 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}-1\right ) a x +10 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}+1\right ) a^{2} x^{2}+10 \mathit {atan} \left (\sqrt {a x -1}+\sqrt {a x +1}+1\right ) a x +\sqrt {a x +1}\, \sqrt {a x -1}\, a^{2} x^{2}+18 \sqrt {a x +1}\, \sqrt {a x -1}\, a x +\sqrt {a x +1}\, \sqrt {a x -1}-10 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right ) a^{2} x^{2}-10 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right ) a x +15 a^{2} x^{2}+15 a x \right )}{a^{2} x \left (a x +1\right )} \] Input:

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

Output:

(c**2*( - 10*atan(sqrt(a*x - 1) + sqrt(a*x + 1) - 1)*a**2*x**2 - 10*atan(s 
qrt(a*x - 1) + sqrt(a*x + 1) - 1)*a*x + 10*atan(sqrt(a*x - 1) + sqrt(a*x + 
 1) + 1)*a**2*x**2 + 10*atan(sqrt(a*x - 1) + sqrt(a*x + 1) + 1)*a*x + sqrt 
(a*x + 1)*sqrt(a*x - 1)*a**2*x**2 + 18*sqrt(a*x + 1)*sqrt(a*x - 1)*a*x + s 
qrt(a*x + 1)*sqrt(a*x - 1) - 10*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2 
))*a**2*x**2 - 10*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2))*a*x + 15*a* 
*2*x**2 + 15*a*x))/(a**2*x*(a*x + 1))