\(\int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx\) [75]

Optimal result

Integrand size = 14, antiderivative size = 110 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=-\frac {3 a \sqrt [4]{1-a x} (1+a x)^{3/4}}{4 x}-\frac {\sqrt [4]{1-a x} (1+a x)^{7/4}}{2 x^2}+\frac {9}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {9}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \] Output:

-3/4*a*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/x-1/2*(-a*x+1)^(1/4)*(a*x+1)^(7/4)/x^2 
+9/4*a^2*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))-9/4*a^2*arctanh((a*x+1)^(1/4 
)/(-a*x+1)^(1/4))
 

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=-\frac {\sqrt [4]{1-a x} \left (2+7 a x+5 a^2 x^2+18 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1-a x}{1+a x}\right )\right )}{4 x^2 \sqrt [4]{1+a x}} \] Input:

Integrate[E^((3*ArcTanh[a*x])/2)/x^3,x]
 

Output:

-1/4*((1 - a*x)^(1/4)*(2 + 7*a*x + 5*a^2*x^2 + 18*a^2*x^2*Hypergeometric2F 
1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(x^2*(1 + a*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6676, 107, 105, 104, 25, 827, 216, 219}

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.

\(\displaystyle \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx\)

\(\Big \downarrow \) 6676

\(\displaystyle \int \frac {(a x+1)^{3/4}}{x^3 (1-a x)^{3/4}}dx\)

\(\Big \downarrow \) 107

\(\displaystyle \frac {3}{4} a \int \frac {(a x+1)^{3/4}}{x^2 (1-a x)^{3/4}}dx-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

\(\displaystyle \frac {3}{4} a \left (6 a \int -\frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3}{4} a \left (-6 a \int \frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {a x+1}}{\sqrt {1-a x}}+1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{4} a \left (6 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{7/4}}{2 x^2}\)

Input:

Int[E^((3*ArcTanh[a*x])/2)/x^3,x]
 

Output:

-1/2*((1 - a*x)^(1/4)*(1 + a*x)^(7/4))/x^2 + (3*a*(-(((1 - a*x)^(1/4)*(1 + 
 a*x)^(3/4))/x) + 6*a*(ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2 - ArcTanh 
[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2)))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 216

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

rule 219

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

rule 827

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

rule 6676

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

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\frac {18 \, a^{2} x^{2} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 9 \, a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 9 \, a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} {\left (5 \, a x + 2\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{8 \, x^{2}} \] Input:

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

Output:

1/8*(18*a^2*x^2*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 9*a^2*x^2*lo 
g(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + 9*a^2*x^2*log(sqrt(-sqrt(-a^2 
*x^2 + 1)/(a*x - 1)) - 1) - 2*sqrt(-a^2*x^2 + 1)*(5*a*x + 2)*sqrt(-sqrt(-a 
^2*x^2 + 1)/(a*x - 1)))/x^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {3}{2} \text {arctanh}(a x)}}{x^3} \, dx=\int \frac {{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{3/2}}{x^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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