3.439 \(\int \frac{\tanh (e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=19 \[ -\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]

[Out]

-(1/(f*Sqrt[a*Cosh[e + f*x]^2]))

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Rubi [A]  time = 0.0696189, antiderivative size = 19, normalized size of antiderivative = 1., 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 = {3176, 3205, 16, 32} \[ -\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(f*Sqrt[a*Cosh[e + f*x]^2]))

Rule 3176

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 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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh (e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx &=\int \frac{\tanh (e+f x)}{\sqrt{a \cosh ^2(e+f x)}} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a x)^{3/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac{1}{f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0317421, size = 19, normalized size = 1. \[ -\frac{1}{f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(f*Sqrt[a*Cosh[e + f*x]^2]))

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Maple [A]  time = 0.024, size = 20, normalized size = 1.1 \begin{align*} -{\frac{1}{f}{\frac{1}{\sqrt{a+a \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.77428, size = 45, normalized size = 2.37 \begin{align*} -\frac{2 \, e^{\left (-f x - e\right )}}{{\left (\sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt{a}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.75363, size = 428, normalized size = 22.53 \begin{align*} -\frac{2 \, \sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a}{\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 )} e^{\left (-f x - e\right )}}{a f \cosh \left (f x + e\right )^{2} +{\left (a f e^{\left (2 \, f x + 2 \, e\right )} + a f\right )} \sinh \left (f x + e\right )^{2} + a f +{\left (a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \,{\left (a f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25999, size = 39, normalized size = 2.05 \begin{align*} -\frac{2 \, e^{\left (f x + e\right )}}{\sqrt{a} f{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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