\(\int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx\) [123]

Optimal result

Integrand size = 10, antiderivative size = 186 \[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}-\frac {5 \sqrt [4]{1-a x} (1+a x)^{3/4}}{a}+\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{\sqrt {2} a}-\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{\sqrt {2} a}+\frac {5 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x} \left (1+\frac {\sqrt {1+a x}}{\sqrt {1-a x}}\right )}\right )}{\sqrt {2} a} \] Output:

-4*(-a*x+1)^(5/4)/a/(a*x+1)^(1/4)-5*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/a+5/2*arc 
tan(1-2^(1/2)*(a*x+1)^(1/4)/(-a*x+1)^(1/4))*2^(1/2)/a-5/2*arctan(1+2^(1/2) 
*(a*x+1)^(1/4)/(-a*x+1)^(1/4))*2^(1/2)/a+5/2*arctanh(2^(1/2)*(a*x+1)^(1/4) 
/(-a*x+1)^(1/4)/(1+(a*x+1)^(1/2)/(-a*x+1)^(1/2)))*2^(1/2)/a
 

Mathematica [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.18 \[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=-\frac {8 e^{-\frac {1}{2} \text {arctanh}(a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},-e^{2 \text {arctanh}(a x)}\right )}{a} \] Input:

Integrate[E^((-5*ArcTanh[a*x])/2),x]
 

Output:

(-8*Hypergeometric2F1[-1/4, 2, 3/4, -E^(2*ArcTanh[a*x])])/(a*E^(ArcTanh[a* 
x]/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.25, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.300, Rules used = {6675, 57, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx\)

\(\Big \downarrow \) 6675

\(\displaystyle \int \frac {(1-a x)^{5/4}}{(a x+1)^{5/4}}dx\)

\(\Big \downarrow \) 57

\(\displaystyle -5 \int \frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}dx-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 60

\(\displaystyle -5 \left (\frac {1}{2} \int \frac {1}{(1-a x)^{3/4} \sqrt [4]{a x+1}}dx+\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

\(\displaystyle -5 \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 1479

\(\displaystyle -5 \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -5 \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -5 \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 1103

\(\displaystyle -5 \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}\right )-\frac {4 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}}\)

Input:

Int[E^((-5*ArcTanh[a*x])/2),x]
 

Output:

(-4*(1 - a*x)^(5/4))/(a*(1 + a*x)^(1/4)) - 5*(((1 - a*x)^(1/4)*(1 + a*x)^( 
3/4))/a - (2*((-(ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqr 
t[2]) + ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqrt[2])/2 + 
 (-1/2*Log[1 + Sqrt[1 - a*x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/ 
Sqrt[2] + Log[1 + Sqrt[1 - a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4 
)]/(2*Sqrt[2]))/2))/a)
 

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 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 6675

Int[E^(ArcTanh[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 + a*x)^(n/2)/(1 - a*x 
)^(n/2), x] /; FreeQ[{a, n}, x] &&  !IntegerQ[(n - 1)/2]
 

Maple [F]

\[\int \frac {1}{{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}d x\]

Input:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
 

Output:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.48 \[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=-\frac {10 \, \sqrt {2} {\left (a x + 1\right )} \arctan \left (\sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 10 \, \sqrt {2} {\left (a x + 1\right )} \arctan \left (\sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 5 \, \sqrt {2} {\left (a x + 1\right )} \log \left (\frac {a x + \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}\right ) + 5 \, \sqrt {2} {\left (a x + 1\right )} \log \left (\frac {a x - \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} {\left (a x + 9\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{4 \, {\left (a^{2} x + a\right )}} \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="fricas")
 

Output:

-1/4*(10*sqrt(2)*(a*x + 1)*arctan(sqrt(2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 
1)) + 1) + 10*sqrt(2)*(a*x + 1)*arctan(sqrt(2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a 
*x - 1)) - 1) - 5*sqrt(2)*(a*x + 1)*log((a*x + sqrt(2)*(a*x - 1)*sqrt(-sqr 
t(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)) + 5*sqrt(2 
)*(a*x + 1)*log((a*x - sqrt(2)*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1 
)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)) + 4*sqrt(-a^2*x^2 + 1)*(a*x + 9)*s 
qrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/(a^2*x + a)
 

Sympy [F]

\[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=\int \frac {1}{\left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2),x)
 

Output:

Integral(((a*x + 1)/sqrt(-a**2*x**2 + 1))**(-5/2), x)
 

Maxima [F]

\[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=\int { \frac {1}{\left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="maxima")
 

Output:

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(-5/2), x)
 

Giac [F]

\[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=\int { \frac {1}{\left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="giac")
 

Output:

integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=\int \frac {1}{{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \] Input:

int(1/((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2),x)
 

Output:

int(1/((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2), x)
 

Reduce [F]

\[ \int e^{-\frac {5}{2} \text {arctanh}(a x)} \, dx=\int \frac {1}{{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}d x \] Input:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
 

Output:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)