3.291 $$\int \frac{e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx$$

Optimal. Leaf size=45 $\tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right )-\frac{\sqrt{\frac{x-1}{x}}}{\sqrt{\frac{1}{x}+1}}$

[Out]

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

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Rubi [A]  time = 0.0826707, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.454, Rules used = {6175, 6180, 96, 92, 206} $\tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right )-\frac{\sqrt{\frac{x-1}{x}}}{\sqrt{\frac{1}{x}+1}}$

Antiderivative was successfully veriﬁed.

[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 6175

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 6180

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]

Rule 96

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 92

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 206

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

Rubi steps

\begin{align*} \int \frac{e^{\coth ^{-1}(x)} x}{(1+x)^2} \, dx &=\int \frac{e^{\coth ^{-1}(x)}}{\left (1+\frac{1}{x}\right )^2 x} \, dx\\ &=-\operatorname{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}}}-\operatorname{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}}}+\operatorname{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}}}+\tanh ^{-1}\left (\sqrt{1+\frac{1}{x}} \sqrt{\frac{-1+x}{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0429387, size = 36, normalized size = 0.8 $\log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right )-\frac{\sqrt{1-\frac{1}{x^2}} x}{x+1}$

Warning: Unable to verify antiderivative.

[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]

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Maple [B]  time = 0.123, size = 110, normalized size = 2.4 \begin{align*}{\frac{-1+x}{2\, \left ( 1+x \right ) ^{2}} \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 ){\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}{\frac{1}{\sqrt{ \left ( 1+x \right ) \left ( -1+x \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-1+x)*((x^2-1)^(3/2)-x^2*(x^2-1)^(1/2)+2*ln(x+(x^2-1)^(1/2))*x^2-2*x*(x^2-1)^(1/2)+4*ln(x+(x^2-1)^(1/2))*
x-(x^2-1)^(1/2)+2*ln(x+(x^2-1)^(1/2)))/((-1+x)/(1+x))^(1/2)/((1+x)*(-1+x))^(1/2)/(1+x)^2

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Maxima [A]  time = 1.00741, size = 59, normalized size = 1.31 \begin{align*} -\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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.62333, size = 122, normalized size = 2.71 \begin{align*} -\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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{x - 1}{x + 1}} \left (x + 1\right )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.13298, size = 61, normalized size = 1.36 \begin{align*} -\sqrt{\frac{x - 1}{x + 1}} + \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \log \left ({\left | \sqrt{\frac{x - 1}{x + 1}} - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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