3.5.29 \(\int \coth (e+f x) \sqrt {a+a \sinh ^2(e+f x)} \, dx\) [429]

Optimal. Leaf size=50 \[ -\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3255, 3284, 52, 65, 212} \begin {gather*} \frac {\sqrt {a \cosh ^2(e+f x)}}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 \coth (e+f x) \sqrt {a+a \sinh ^2(e+f x)} \, dx &=\int \sqrt {a \cosh ^2(e+f x)} \coth (e+f x) \, dx\\ &=-\frac {\text {Subst}\left (\int \frac {\sqrt {a x}}{1-x} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {\sqrt {a \cosh ^2(e+f x)}}{f}-\frac {a \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {\sqrt {a \cosh ^2(e+f x)}}{f}-\frac {\text {Subst}\left (\int \frac {1}{1-\frac {x^2}{a}} \, dx,x,\sqrt {a \cosh ^2(e+f x)}\right )}{f}\\ &=-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \cosh ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a \cosh ^2(e+f x)}}{f}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 42, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a \cosh ^2(e+f x)} \left (\cosh (e+f x)+\log \left (\tanh \left (\frac {1}{2} (e+f x)\right )\right )\right ) \text {sech}(e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.99, size = 42, normalized size = 0.84

method	result	size
default	\(\frac {\mathit {`\,int/indef0`\,}\left (\frac {a \left (\cosh ^{2}\left (f x +e \right )\right )}{\sinh \left (f x +e \right ) \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f}\)	\(42\)
risch	\(\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{2 f x +2 e}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}}{2 f \left ({\mathrm e}^{2 f x +2 e}+1\right )}-\frac {\ln \left ({\mathrm e}^{f x}+{\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {\ln \left ({\mathrm e}^{f x}-{\mathrm e}^{-e}\right ) \sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, {\mathrm e}^{f x +e}}{f \left ({\mathrm e}^{2 f x +2 e}+1\right )}\)	\(220\)

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

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Maxima [A]
time = 0.52, size = 72, normalized size = 1.44 \begin {gather*} \frac {{\left (\sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt {a}\right )} e^{\left (f x + e\right )}}{2 \, f} - \frac {\sqrt {a} \log \left (e^{\left (-f x - e\right )} + 1\right )}{f} + \frac {\sqrt {a} \log \left (e^{\left (-f x - e\right )} - 1\right )}{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]

1/2*(sqrt(a)*e^(-2*f*x - 2*e) + sqrt(a))*e^(f*x + e)/f - sqrt(a)*log(e^(-f*x - e) + 1)/f + sqrt(a)*log(e^(-f*x
- e) - 1)/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (42) = 84\).
time = 0.49, size = 200, normalized size = 4.00 \begin {gather*} \frac {{\left (2 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + {\left (\cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} + 2 \, {\left (\cosh \left (f x + e\right ) e^{\left (f x + e\right )} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )\right )} \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 )\right )} \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{2 \, {\left (f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right ) + {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )\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]

1/2*(2*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e) + e^(f*x + e)*sinh(f*x + e)^2 + (cosh(f*x + e)^2 + 1)*e^(f*x +
e) + 2*(cosh(f*x + e)*e^(f*x + e) + e^(f*x + e)*sinh(f*x + e))*log((cosh(f*x + e) + sinh(f*x + e) - 1)/(cosh(f
*x + e) + sinh(f*x + e) + 1)))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)/(f*cosh(f*x + e)
*e^(2*f*x + 2*e) + f*cosh(f*x + e) + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e))

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

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Giac [A]
time = 0.42, size = 47, normalized size = 0.94 \begin {gather*} \frac {\sqrt {a} {\left (e^{\left (f x + e\right )} + e^{\left (-f x - e\right )} - 2 \, \log \left (e^{\left (f x + e\right )} + 1\right ) + 2 \, \log \left ({\left | e^{\left (f x + e\right )} - 1 \right |}\right )\right )}}{2 \, 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="giac")

[Out]

1/2*sqrt(a)*(e^(f*x + e) + e^(-f*x - e) - 2*log(e^(f*x + e) + 1) + 2*log(abs(e^(f*x + e) - 1)))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \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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