3.5.39 \(\int \frac {\tanh (e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\) [439]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 19, 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, 16, 32} \begin {gather*} -\frac {1}{f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

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 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]

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 {\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 {\text {Subst}\left (\int \frac {1}{x \sqrt {a x}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac {a \text {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*}

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Mathematica [A]
time = 0.04, size = 19, normalized size = 1.00 \begin {gather*} -\frac {1}{f \sqrt {a \cosh ^2(e+f x)}} \end {gather*}

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 = 0.60, size = 20, normalized size = 1.05

method	result	size
derivativedivides	\(-\frac {1}{\sqrt {a +a \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(20\)
default	\(-\frac {1}{\sqrt {a +a \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(20\)
risch	\(-\frac {2}{\sqrt {\left ({\mathrm e}^{2 f x +2 e}+1\right )^{2} a \,{\mathrm e}^{-2 f x -2 e}}\, f}\)	\(32\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 35, normalized size = 1.84 \begin {gather*} -\frac {2 \, e^{\left (-f x - e\right )}}{{\left (\sqrt {a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt {a}\right )} f} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (17) = 34\).
time = 0.47, size = 168, normalized size = 8.84 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh {\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(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 [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(tanh(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 [B]
time = 0.85, size = 30, normalized size = 1.58 \begin {gather*} -\frac {\sqrt {a\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}}{a\,f\,{\mathrm {cosh}\left (e+f\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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