3.201 \(\int \frac {\sqrt {\coth (a+b \log (c x^n))}}{x} \, dx\)

Optimal. Leaf size=48 \[ \frac {\tanh ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {\tan ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \]

[Out]

-arctan(coth(a+b*ln(c*x^n))^(1/2))/b/n+arctanh(coth(a+b*ln(c*x^n))^(1/2))/b/n

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3476, 329, 298, 203, 206} \[ \frac {\tanh ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {\tan ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Coth[a + b*Log[c*x^n]]]/x,x]

[Out]

-(ArcTan[Sqrt[Coth[a + b*Log[c*x^n]]]]/(b*n)) + ArcTanh[Sqrt[Coth[a + b*Log[c*x^n]]]]/(b*n)

Rule 203

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {\coth (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{-1+x^2} \, dx,x,\coth \left (a+b \log \left (c x^n\right )\right )\right )}{b n}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}\\ &=-\frac {\tan ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}+\frac {\tanh ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 48, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {\tan ^{-1}\left (\sqrt {\coth \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Coth[a + b*Log[c*x^n]]]/x,x]

[Out]

-(ArcTan[Sqrt[Coth[a + b*Log[c*x^n]]]]/(b*n)) + ArcTanh[Sqrt[Coth[a + b*Log[c*x^n]]]]/(b*n)

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fricas [B]  time = 0.45, size = 305, normalized size = 6.35 \[ \frac {2 \, \arctan \left (-\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 2 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + {\left (\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 1\right )} \sqrt {\frac {\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{\sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}}\right ) - \log \left (-\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 2 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) - \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + {\left (\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} + 2 \, \cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + \sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2} - 1\right )} \sqrt {\frac {\cosh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}{\sinh \left (b n \log \relax (x) + b \log \relax (c) + a\right )}}\right )}{2 \, b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+b*log(c*x^n))^(1/2)/x,x, algorithm="fricas")

[Out]

1/2*(2*arctan(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c
) + a) - sinh(b*n*log(x) + b*log(c) + a)^2 + (cosh(b*n*log(x) + b*log(c) + a)^2 + 2*cosh(b*n*log(x) + b*log(c)
+ a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 - 1)*sqrt(cosh(b*n*log(x) + b*log(c)
+ a)/sinh(b*n*log(x) + b*log(c) + a))) - log(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*log(c
) + a)*sinh(b*n*log(x) + b*log(c) + a) - sinh(b*n*log(x) + b*log(c) + a)^2 + (cosh(b*n*log(x) + b*log(c) + a)^
2 + 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 - 1)
*sqrt(cosh(b*n*log(x) + b*log(c) + a)/sinh(b*n*log(x) + b*log(c) + a))))/(b*n)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+b*log(c*x^n))^(1/2)/x,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.13, size = 72, normalized size = 1.50 \[ -\frac {\ln \left (\sqrt {\coth }\left (a +b \ln \left (c \,x^{n}\right )\right )-1\right )}{2 b n}+\frac {\ln \left (\sqrt {\coth }\left (a +b \ln \left (c \,x^{n}\right )\right )+1\right )}{2 b n}-\frac {\arctan \left (\sqrt {\coth }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a+b*ln(c*x^n))^(1/2)/x,x)

[Out]

-1/2/b/n*ln(coth(a+b*ln(c*x^n))^(1/2)-1)+1/2/b/n*ln(coth(a+b*ln(c*x^n))^(1/2)+1)-arctan(coth(a+b*ln(c*x^n))^(1
/2))/b/n

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\coth \left (b \log \left (c x^{n}\right ) + a\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+b*log(c*x^n))^(1/2)/x,x, algorithm="maxima")

[Out]

integrate(sqrt(coth(b*log(c*x^n) + a))/x, x)

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mupad [B]  time = 1.50, size = 39, normalized size = 0.81 \[ -\frac {\mathrm {atan}\left (\sqrt {\mathrm {coth}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )-\mathrm {atanh}\left (\sqrt {\mathrm {coth}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*log(c*x^n))^(1/2)/x,x)

[Out]

-(atan(coth(a + b*log(c*x^n))^(1/2)) - atanh(coth(a + b*log(c*x^n))^(1/2)))/(b*n)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\coth {\left (a + b \log {\left (c x^{n} \right )} \right )}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(a+b*ln(c*x**n))**(1/2)/x,x)

[Out]

Integral(sqrt(coth(a + b*log(c*x**n)))/x, x)

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