\(\int \frac {\coth ^{-1}(\sqrt {x})}{x^3} \, dx\) [88]

Optimal result

Integrand size = 10, antiderivative size = 42 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {1}{6 x^{3/2}}-\frac {1}{2 \sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\text {arctanh}\left (\sqrt {x}\right )}{2} \]

[Out]

-1/6/x^(3/2)-1/2*arccoth(x^(1/2))/x^2+1/2*arctanh(x^(1/2))-1/2/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6038, 53, 65, 212} \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {\text {arctanh}\left (\sqrt {x}\right )}{2}-\frac {1}{6 x^{3/2}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {1}{2 \sqrt {x}} \]

[In]

Int[ArcCoth[Sqrt[x]]/x^3,x]

[Out]

-1/6*1/x^(3/2) - 1/(2*Sqrt[x]) - ArcCoth[Sqrt[x]]/(2*x^2) + ArcTanh[Sqrt[x]]/2

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 6038

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCoth[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = -\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{4} \int \frac {1}{(1-x) x^{5/2}} \, dx \\ & = -\frac {1}{6 x^{3/2}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{4} \int \frac {1}{(1-x) x^{3/2}} \, dx \\ & = -\frac {1}{6 x^{3/2}}-\frac {1}{2 \sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{4} \int \frac {1}{(1-x) \sqrt {x}} \, dx \\ & = -\frac {1}{6 x^{3/2}}-\frac {1}{2 \sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {1}{6 x^{3/2}}-\frac {1}{2 \sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\text {arctanh}\left (\sqrt {x}\right )}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {1}{6 x^{3/2}}-\frac {1}{2 \sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {1}{4} \log \left (1-\sqrt {x}\right )+\frac {1}{4} \log \left (1+\sqrt {x}\right ) \]

[In]

Integrate[ArcCoth[Sqrt[x]]/x^3,x]

[Out]

-1/6*1/x^(3/2) - 1/(2*Sqrt[x]) - ArcCoth[Sqrt[x]]/(2*x^2) - Log[1 - Sqrt[x]]/4 + Log[1 + Sqrt[x]]/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {\operatorname {arccoth}\left (\sqrt {x}\right )}{2 x^{2}}-\frac {1}{6 x^{\frac {3}{2}}}-\frac {1}{2 \sqrt {x}}-\frac {\ln \left (\sqrt {x}-1\right )}{4}+\frac {\ln \left (\sqrt {x}+1\right )}{4}\)	\(37\)
default	\(-\frac {\operatorname {arccoth}\left (\sqrt {x}\right )}{2 x^{2}}-\frac {1}{6 x^{\frac {3}{2}}}-\frac {1}{2 \sqrt {x}}-\frac {\ln \left (\sqrt {x}-1\right )}{4}+\frac {\ln \left (\sqrt {x}+1\right )}{4}\)	\(37\)
parts	\(-\frac {\operatorname {arccoth}\left (\sqrt {x}\right )}{2 x^{2}}-\frac {1}{6 x^{\frac {3}{2}}}-\frac {1}{2 \sqrt {x}}-\frac {\ln \left (\sqrt {x}-1\right )}{4}+\frac {\ln \left (\sqrt {x}+1\right )}{4}\)	\(37\)

[In]

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

[Out]

-1/2*arccoth(x^(1/2))/x^2-1/6/x^(3/2)-1/2/x^(1/2)-1/4*ln(x^(1/2)-1)+1/4*ln(x^(1/2)+1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {3 \, {\left (x^{2} - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) - 2 \, {\left (3 \, x + 1\right )} \sqrt {x}}{12 \, x^{2}} \]

[In]

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

[Out]

1/12*(3*(x^2 - 1)*log((x + 2*sqrt(x) + 1)/(x - 1)) - 2*(3*x + 1)*sqrt(x))/x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (36) = 72\).

Time = 1.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.81 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {3 x^{\frac {7}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 x^{\frac {5}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 x^{\frac {3}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 \sqrt {x} \operatorname {acoth}{\left (\sqrt {x} \right )}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 x^{3}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {2 x^{2}}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {x}{6 x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} \]

[In]

integrate(acoth(x**(1/2))/x**3,x)

[Out]

3*x**(7/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*x**(5/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*
x**(3/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) + 3*sqrt(x)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*x**
3/(6*x**(7/2) - 6*x**(5/2)) + 2*x**2/(6*x**(7/2) - 6*x**(5/2)) + x/(6*x**(7/2) - 6*x**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {3 \, x + 1}{6 \, x^{\frac {3}{2}}} - \frac {\operatorname {arcoth}\left (\sqrt {x}\right )}{2 \, x^{2}} + \frac {1}{4} \, \log \left (\sqrt {x} + 1\right ) - \frac {1}{4} \, \log \left (\sqrt {x} - 1\right ) \]

[In]

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

[Out]

-1/6*(3*x + 1)/x^(3/2) - 1/2*arccoth(sqrt(x))/x^2 + 1/4*log(sqrt(x) + 1) - 1/4*log(sqrt(x) - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (26) = 52\).

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.71 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\sqrt {x} + 1\right )}^{2}}{{\left (\sqrt {x} - 1\right )}^{2}} + \frac {3 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1} + 2\right )}}{3 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} + 1\right )}^{3}} + \frac {2 \, {\left (\frac {{\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )} \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{{\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} + 1\right )}^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.70 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {\ln \left (1-\frac {1}{\sqrt {x}}\right )}{4\,x^2}-\frac {\frac {x}{2}+\frac {1}{6}}{x^{3/2}}-\frac {\ln \left (\frac {1}{\sqrt {x}}+1\right )}{4\,x^2}-\frac {\mathrm {atan}\left (\sqrt {x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

[In]

int(acoth(x^(1/2))/x^3,x)

[Out]

log(1 - 1/x^(1/2))/(4*x^2) - (atan(x^(1/2)*1i)*1i)/2 - (x/2 + 1/6)/x^(3/2) - log(1/x^(1/2) + 1)/(4*x^2)