\(\int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx\) [291]

Optimal result

Integrand size = 11, antiderivative size = 45 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=-\frac {\sqrt {\frac {-1+x}{x}}}{\sqrt {1+\frac {1}{x}}}+\text {arctanh}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {6310, 6315, 98, 94, 212} \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=\text {arctanh}\left (\sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}}\right )-\frac {\sqrt {\frac {x-1}{x}}}{\sqrt {\frac {1}{x}+1}} \]

[In]

Int[(E^ArcCoth[x]*x)/(1 + x)^2,x]

[Out]

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

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[(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] || Sum
SimplerQ[m, 1])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6310

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

Rule 6315

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\coth ^{-1}(x)}}{\left (1+\frac {1}{x}\right )^2 x} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x (1+x)^{3/2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\sqrt {\frac {-1+x}{x}}}{\sqrt {1+\frac {1}{x}}}-\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x \sqrt {1+x}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\sqrt {\frac {-1+x}{x}}}{\sqrt {1+\frac {1}{x}}}+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \\ & = -\frac {\sqrt {\frac {-1+x}{x}}}{\sqrt {1+\frac {1}{x}}}+\text {arctanh}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=-\frac {\sqrt {1-\frac {1}{x^2}} x}{1+x}+\log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24

method	result	size
trager	\(-\sqrt {-\frac {1-x}{1+x}}-\ln \left (-\sqrt {-\frac {1-x}{1+x}}\, x -\sqrt {-\frac {1-x}{1+x}}+x \right )\)	\(56\)
risch	\(-\frac {x -1}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}+\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\)	\(59\)
default	\(\frac {\left (x -1\right ) \left (\left (x^{2}-1\right )^{\frac {3}{2}}-x^{2} \sqrt {x^{2}-1}+2 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}-2 x \sqrt {x^{2}-1}+4 \ln \left (x +\sqrt {x^{2}-1}\right ) x -\sqrt {x^{2}-1}+2 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}\, \left (1+x \right )^{2}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=-\sqrt {\frac {x - 1}{x + 1}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

-sqrt((x - 1)/(x + 1)) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x + 1)) - 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=-\sqrt {\frac {x - 1}{x + 1}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

-sqrt((x - 1)/(x + 1)) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x + 1)) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=-\frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} - \frac {2}{{\left (x - \sqrt {x^{2} - 1} + 1\right )} \mathrm {sgn}\left (x + 1\right )} \]

[In]

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

[Out]

-log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1) - 2/((x - sqrt(x^2 - 1) + 1)*sgn(x + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx=2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\sqrt {\frac {x-1}{x+1}} \]

[In]

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

[Out]

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