3.449 \(\int \frac{\tanh (e+f x)}{(a+a \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=21 \[ -\frac{1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]

[Out]

-1/(3*f*(a*Cosh[e + f*x]^2)^(3/2))

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Rubi [A]  time = 0.0774936, antiderivative size = 21, 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} \[ -\frac{1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-1/(3*f*(a*Cosh[e + f*x]^2)^(3/2))

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)}{\left (a+a \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac{\tanh (e+f x)}{\left (a \cosh ^2(e+f x)\right )^{3/2}} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a x)^{3/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{1}{(a x)^{5/2}} \, dx,x,\cosh ^2(e+f x)\right )}{2 f}\\ &=-\frac{1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0388074, size = 21, normalized size = 1. \[ -\frac{1}{3 f \left (a \cosh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]/(a + a*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-1/(3*f*(a*Cosh[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.018, size = 20, normalized size = 1. \begin{align*} -{\frac{1}{3\,f} \left ( a+a \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

-1/3/f/(a+a*sinh(f*x+e)^2)^(3/2)

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Maxima [B]  time = 1.8976, size = 82, normalized size = 3.9 \begin{align*} -\frac{8 \, e^{\left (-3 \, f x - 3 \, e\right )}}{3 \,{\left (3 \, a^{\frac{3}{2}} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \, a^{\frac{3}{2}} e^{\left (-4 \, f x - 4 \, e\right )} + a^{\frac{3}{2}} e^{\left (-6 \, f x - 6 \, e\right )} + a^{\frac{3}{2}}\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)^(3/2),x, algorithm="maxima")

[Out]

-8/3*e^(-3*f*x - 3*e)/((3*a^(3/2)*e^(-2*f*x - 2*e) + 3*a^(3/2)*e^(-4*f*x - 4*e) + a^(3/2)*e^(-6*f*x - 6*e) + a
^(3/2))*f)

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Fricas [B]  time = 1.84687, size = 1511, normalized size = 71.95 \begin{align*} \text{result too large to display} \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)^(3/2),x, algorithm="fricas")

[Out]

-8/3*(cosh(f*x + e)^3*e^(f*x + e) + 3*cosh(f*x + e)^2*e^(f*x + e)*sinh(f*x + e) + 3*cosh(f*x + e)*e^(f*x + e)*
sinh(f*x + e)^2 + e^(f*x + e)*sinh(f*x + e)^3)*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)/
(a^2*f*cosh(f*x + e)^6 + 3*a^2*f*cosh(f*x + e)^4 + (a^2*f*e^(2*f*x + 2*e) + a^2*f)*sinh(f*x + e)^6 + 6*(a^2*f*
cosh(f*x + e)*e^(2*f*x + 2*e) + a^2*f*cosh(f*x + e))*sinh(f*x + e)^5 + 3*a^2*f*cosh(f*x + e)^2 + 3*(5*a^2*f*co
sh(f*x + e)^2 + a^2*f + (5*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^4 + 4*(5*a^2*f*cosh(f
*x + e)^3 + 3*a^2*f*cosh(f*x + e) + (5*a^2*f*cosh(f*x + e)^3 + 3*a^2*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*
x + e)^3 + a^2*f + 3*(5*a^2*f*cosh(f*x + e)^4 + 6*a^2*f*cosh(f*x + e)^2 + a^2*f + (5*a^2*f*cosh(f*x + e)^4 + 6
*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e))*sinh(f*x + e)^2 + (a^2*f*cosh(f*x + e)^6 + 3*a^2*f*cosh(f*x +
e)^4 + 3*a^2*f*cosh(f*x + e)^2 + a^2*f)*e^(2*f*x + 2*e) + 6*(a^2*f*cosh(f*x + e)^5 + 2*a^2*f*cosh(f*x + e)^3
+ a^2*f*cosh(f*x + e) + (a^2*f*cosh(f*x + e)^5 + 2*a^2*f*cosh(f*x + e)^3 + a^2*f*cosh(f*x + e))*e^(2*f*x + 2*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 )}}{\left (a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(tanh(e + f*x)/(a*(sinh(e + f*x)**2 + 1))**(3/2), x)

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Giac [A]  time = 1.31568, size = 43, normalized size = 2.05 \begin{align*} -\frac{8 \, e^{\left (3 \, f x + 3 \, e\right )}}{3 \, a^{\frac{3}{2}} f{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{3}} \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)^(3/2),x, algorithm="giac")

[Out]

-8/3*e^(3*f*x + 3*e)/(a^(3/2)*f*(e^(2*f*x + 2*e) + 1)^3)