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

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 = {6310, 6315, 98, 94, 212} \begin {gather*} \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}} \end {gather*}

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 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*} \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\\ &=-\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}}}+\tanh ^{-1}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 36, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {1-\frac {1}{x^2}} x}{1+x}+\log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(109\) vs. \(2(37)=74\).
time = 0.12, size = 110, normalized size = 2.44

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	\(-\frac {-1+x}{\sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right )}+\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (1+x \right ) \left (-1+x \right )}}{\sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right )}\)	\(59\)
default	\(\frac {\left (-1+x \right ) \left (\left (x^{2}-1\right )^{\frac {3}{2}}-\sqrt {x^{2}-1}\, x^{2}+2 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}-2 x \sqrt {x^{2}-1}+4 \ln \left (x +\sqrt {x^{2}-1}\right ) x -\sqrt {x^{2}-1}+2 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}\, \left (1+x \right )^{2}}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-1+x)*((x^2-1)^(3/2)-(x^2-1)^(1/2)*x^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 = 0.26, size = 44, normalized size = 0.98 \begin {gather*} -\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 {gather*}

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 = 0.33, size = 44, normalized size = 0.98 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {\frac {x - 1}{x + 1}} \left (x + 1\right )^{2}}\, dx \end {gather*}

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 = 0.41, size = 44, normalized size = 0.98 \begin {gather*} -\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 )} \end {gather*}

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]

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

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Mupad [B]
time = 0.03, size = 28, normalized size = 0.62 \begin {gather*} 2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\sqrt {\frac {x-1}{x+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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