3.5.40 \(\int \frac {\coth (e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\) [440]

Optimal. Leaf size=31 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]

[Out]

-arctanh((a*cosh(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3255, 3284, 65, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

-(ArcTanh[Sqrt[a*Cosh[e + f*x]^2]/Sqrt[a]]/(Sqrt[a]*f))

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 3255

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3284

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
]

Rubi steps

\begin {align*} \int \frac {\coth (e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx &=\int \frac {\coth (e+f x)}{\sqrt {a \cosh ^2(e+f x)}} \, dx\\ &=-\frac {\text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac {\text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a \cosh ^2(e+f x)}\right )}{a f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 49, normalized size = 1.58 \begin {gather*} \frac {\cosh (e+f x) \left (-\log \left (\cosh \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sinh \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.95, size = 33, normalized size = 1.06

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {1}{\sinh \left (f x +e \right ) \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(33\)
risch	\(-\frac {\ln \left ({\mathrm e}^{f x}+{\mathrm e}^{-e}\right ) \left ({\mathrm e}^{2 f x +2 e}+1\right ) {\mathrm e}^{-f x -e}}{f \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}+\frac {\ln \left ({\mathrm e}^{f x}-{\mathrm e}^{-e}\right ) \left ({\mathrm e}^{2 f x +2 e}+1\right ) {\mathrm e}^{-f x -e}}{f \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)/(a+a*sinh(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

`int/indef0`(1/sinh(f*x+e)/(a*cosh(f*x+e)^2)^(1/2),sinh(f*x+e))/f

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Maxima [A]
time = 0.51, size = 42, normalized size = 1.35 \begin {gather*} -\frac {\log \left (e^{\left (-f x - e\right )} + 1\right )}{\sqrt {a} f} + \frac {\log \left (e^{\left (-f x - e\right )} - 1\right )}{\sqrt {a} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (25) = 50\).
time = 0.45, size = 174, normalized size = 5.61 \begin {gather*} \left [\frac {\sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} \log \left (\frac {\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - 1}{\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1}\right )}{a f e^{\left (2 \, f x + 2 \, e\right )} + a f}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} \sqrt {-a}}{a \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a \cosh \left (f x + e\right ) + {\left (a e^{\left (2 \, f x + 2 \, e\right )} + a\right )} \sinh \left (f x + e\right )}\right )}{a f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth {\left (e + f x \right )}}{\sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)/(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(coth(e + f*x)/sqrt(a*(sinh(e + f*x)**2 + 1)), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\mathrm {coth}\left (e+f\,x\right )}{\sqrt {a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(e + f*x)/(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

int(coth(e + f*x)/(a + a*sinh(e + f*x)^2)^(1/2), x)

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