### 3.1038 $$\int \frac{\coth (c+d x)}{\sqrt{a \cosh ^2(c+d x)}} \, dx$$

Optimal. Leaf size=31 $-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}$

[Out]

-(ArcTanh[Sqrt[a*Cosh[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d))

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Rubi [A]  time = 0.0394633, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3205, 63, 206} $-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[c + d*x]/Sqrt[a*Cosh[c + d*x]^2],x]

[Out]

-(ArcTanh[Sqrt[a*Cosh[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d))

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 63

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 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{\coth (c+d x)}{\sqrt{a \cosh ^2(c+d x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a x}} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a \cosh ^2(c+d x)}\right )}{a d}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cosh ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}\\ \end{align*}

Mathematica [A]  time = 0.0661061, size = 49, normalized size = 1.58 $\frac{\cosh (c+d x) \left (\log \left (\sinh \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cosh \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d \sqrt{a \cosh ^2(c+d x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[c + d*x]/Sqrt[a*Cosh[c + d*x]^2],x]

[Out]

(Cosh[c + d*x]*(-Log[Cosh[(c + d*x)/2]] + Log[Sinh[(c + d*x)/2]]))/(d*Sqrt[a*Cosh[c + d*x]^2])

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Maple [A]  time = 0.077, size = 31, normalized size = 1. \begin{align*} -{\frac{\cosh \left ( dx+c \right ){\it Artanh} \left ( \cosh \left ( dx+c \right ) \right ) }{d}{\frac{1}{\sqrt{a \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)/(a*cosh(d*x+c)^2)^(1/2),x)

[Out]

-1/(a*cosh(d*x+c)^2)^(1/2)*cosh(d*x+c)*arctanh(cosh(d*x+c))/d

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Maxima [A]  time = 1.71249, size = 54, normalized size = 1.74 \begin{align*} -\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{\sqrt{a} d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{\sqrt{a} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a*cosh(d*x+c)^2)^(1/2),x, algorithm="maxima")

[Out]

-log(e^(-d*x - c) + 1)/(sqrt(a)*d) + log(e^(-d*x - c) - 1)/(sqrt(a)*d)

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Fricas [B]  time = 1.85583, size = 455, normalized size = 14.68 \begin{align*} \left [\frac{\sqrt{a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + a} \log \left (\frac{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1}{\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1}\right )}{a d e^{\left (2 \, d x + 2 \, c\right )} + a d}, \frac{2 \, \sqrt{-a} \arctan \left (\frac{\sqrt{a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + a} \sqrt{-a}}{a \cosh \left (d x + c\right ) e^{\left (2 \, d x + 2 \, c\right )} + a \cosh \left (d x + c\right ) +{\left (a e^{\left (2 \, d x + 2 \, c\right )} + a\right )} \sinh \left (d x + c\right )}\right )}{a d}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a*cosh(d*x+c)^2)^(1/2),x, algorithm="fricas")

[Out]

[sqrt(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + a)*log((cosh(d*x + c) + sinh(d*x + c) - 1)/(cosh(d*x + c) + si
nh(d*x + c) + 1))/(a*d*e^(2*d*x + 2*c) + a*d), 2*sqrt(-a)*arctan(sqrt(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c)
+ a)*sqrt(-a)/(a*cosh(d*x + c)*e^(2*d*x + 2*c) + a*cosh(d*x + c) + (a*e^(2*d*x + 2*c) + a)*sinh(d*x + c)))/(a*
d)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a*cosh(d*x+c)**2)**(1/2),x)

[Out]

Integral(coth(c + d*x)/sqrt(a*cosh(c + d*x)**2), x)

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Giac [A]  time = 1.20626, size = 46, normalized size = 1.48 \begin{align*} -\frac{\frac{\log \left (e^{\left (d x + c\right )} + 1\right )}{\sqrt{a}} - \frac{\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{\sqrt{a}}}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)/(a*cosh(d*x+c)^2)^(1/2),x, algorithm="giac")

[Out]

-(log(e^(d*x + c) + 1)/sqrt(a) - log(abs(e^(d*x + c) - 1))/sqrt(a))/d