3.4.32 \(\int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx\) [332]

3.4.32.1 Optimal result

Integrand size = 12, antiderivative size = 58 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=-\frac {\sqrt {2} \left (1+\frac {1}{x}\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {-\frac {1-x}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (1+x)^{3/2}} \]

output

-(1+1/x)^(3/2)*arctan(2^(1/2)*(1/x)^(1/2)/((-1+x)/x)^(1/2))*2^(1/2)/(1/x)^ 
(3/2)/(1+x)^(3/2)
 

3.4.32.2 Mathematica [A] (warning: unable to verify)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=\sqrt {2} \sqrt {\frac {1}{1+x}} \sqrt {1+x} \arctan \left (\frac {\sqrt {\frac {-1+x}{x^2}} x}{\sqrt {2}}\right ) \]

input

Integrate[E^ArcCoth[x]/(1 + x)^(3/2),x]
 

output

Sqrt[2]*Sqrt[(1 + x)^(-1)]*Sqrt[1 + x]*ArcTan[(Sqrt[(-1 + x)/x^2]*x)/Sqrt[ 
2]]
 

3.4.32.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6727, 104, 216}

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.

\(\displaystyle \int \frac {e^{\coth ^{-1}(x)}}{(x+1)^{3/2}} \, dx\)

\(\Big \downarrow \) 6727

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\sqrt {2} \left (\frac {1}{x}+1\right )^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {1-\frac {1}{x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (x+1)^{3/2}}\)

input

Int[E^ArcCoth[x]/(1 + x)^(3/2),x]
 

output

-((Sqrt[2]*(1 + x^(-1))^(3/2)*ArcTan[(Sqrt[2]*Sqrt[x^(-1)])/Sqrt[1 - x^(-1 
)]])/((x^(-1))^(3/2)*(1 + x)^(3/2)))
 

3.4.32.3.1 Defintions of rubi rules used

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 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 6727

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

3.4.32.4 Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x -1}\, \sqrt {2}}{2}\right ) \sqrt {x -1}}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {1+x}}\)	\(37\)

input

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

output

2^(1/2)*arctan(1/2*(x-1)^(1/2)*2^(1/2))/((x-1)/(1+x))^(1/2)/(1+x)^(1/2)*(x 
-1)^(1/2)
 

3.4.32.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.45 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1} \sqrt {\frac {x - 1}{x + 1}}\right ) \]

input

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

output

sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x + 1)*sqrt((x - 1)/(x + 1)))
 

3.4.32.6 Sympy [A] (verification not implemented)

Time = 40.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=2 \left (\begin {cases} \frac {\sqrt {2} \operatorname {acos}{\left (\frac {\sqrt {2}}{\sqrt {x + 1}} \right )}}{2} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) \]

input

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

output

2*Piecewise((sqrt(2)*acos(sqrt(2)/sqrt(x + 1))/2, (sqrt(x + 1) < sqrt(2)) 
& (sqrt(x + 1) > -sqrt(2))))
 

3.4.32.7 Maxima [F]

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

input

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

output

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

3.4.32.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\coth ^{-1}(x)}}{(1+x)^{3/2}} \, dx=\text {Exception raised: NotImplementedError} \]

input

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

output

Exception raised: NotImplementedError >> unable to parse Giac output: 2*(- 
1/2*sqrt(2)*atan(i)+1/2*sqrt(2)*atan(sqrt(sageVARx-1)/sqrt(2)))*sign(sageV 
ARx)/sign(sageVARx+1)
 

3.4.32.9 Mupad [F(-1)]

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

input

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

output

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