3.3.20 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx\) [220]

3.3.20.1 Optimal result
3.3.20.2 Mathematica [A] (verified)
3.3.20.3 Rubi [A] (verified)
3.3.20.4 Maple [B] (verified)
3.3.20.5 Fricas [A] (verification not implemented)
3.3.20.6 Sympy [F]
3.3.20.7 Maxima [A] (verification not implemented)
3.3.20.8 Giac [F]
3.3.20.9 Mupad [B] (verification not implemented)

3.3.20.1 Optimal result

Integrand size = 18, antiderivative size = 53 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx=\frac {2 \left (a-\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \]

output
-arctanh((1-1/a^2/x^2)^(1/2))/a/c+2*(a-1/x)/a^2/c/(1-1/a^2/x^2)^(1/2)
 
3.3.20.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx=\frac {\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{1+a x}-\frac {\log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a}}{c} \]

input
Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)),x]
 
output
((2*Sqrt[1 - 1/(a^2*x^2)]*x)/(1 + a*x) - Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)]) 
*x]/a)/c
 
3.3.20.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6724, 27, 528, 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.

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

\(\Big \downarrow \) 6724

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 528

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input
Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)),x]
 
output
((2*a*(a - x^(-1)))/Sqrt[1 - 1/(a^2*x^2)] - a^2*ArcTanh[Sqrt[1 - 1/(a^2*x^ 
2)]])/(a^3*c)
 

3.3.20.3.1 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 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 6724
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S 
imp[-d^n   Subst[Int[(d + c*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(p + 2)), x], 
 x, 1/x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0] && IntegerQ[p] && In 
tegerQ[n]
 
3.3.20.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(49)=98\).

Time = 0.41 (sec) , antiderivative size = 248, normalized size of antiderivative = 4.68

method result size
default \(-\frac {\left (\ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+2 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x +\left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-2 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x +a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )-\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \sqrt {a^{2}}\, c \left (a x -1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}\) \(248\)

input
int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x,method=_RETURNVERBOSE)
 
output
-(ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*a^3*x^2-(a^2 
)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*a^2*x^2+2*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*( 
a*x+1))^(1/2))/(a^2)^(1/2))*a^2*x+((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-2*(a 
^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*a*x+a*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a* 
x+1))^(1/2))/(a^2)^(1/2))-(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/a*((a*x-1)/ 
(a*x+1))^(3/2)/(a^2)^(1/2)/c/(a*x-1)/((a*x-1)*(a*x+1))^(1/2)
 
3.3.20.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx=\frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}} - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a c} \]

input
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="fricas")
 
output
(2*sqrt((a*x - 1)/(a*x + 1)) - log(sqrt((a*x - 1)/(a*x + 1)) + 1) + log(sq 
rt((a*x - 1)/(a*x + 1)) - 1))/(a*c)
 
3.3.20.6 Sympy [F]

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

input
integrate(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c),x)
 
output
-(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**2*x**2 - 1), x) + Integr 
al(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**2*x**2 - 1), x))/c
 
3.3.20.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx=-a {\left (\frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c} - \frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c}\right )} \]

input
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="maxima")
 
output
-a*(log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c) - log(sqrt((a*x - 1)/(a*x + 
 1)) - 1)/(a^2*c) - 2*sqrt((a*x - 1)/(a*x + 1))/(a^2*c))
 
3.3.20.8 Giac [F]

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

input
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c),x, algorithm="giac")
 
output
undef
 
3.3.20.9 Mupad [B] (verification not implemented)

Time = 4.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{c-a c x} \, dx=\frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a\,c}-\frac {2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a\,c} \]

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