3.46 \(\int \frac{1}{(a+a \tanh (c+d x))^2} \, dx\)

Optimal. Leaf size=51 \[ -\frac{1}{4 d \left (a^2 \tanh (c+d x)+a^2\right )}+\frac{x}{4 a^2}-\frac{1}{4 d (a \tanh (c+d x)+a)^2} \]

[Out]

x/(4*a^2) - 1/(4*d*(a + a*Tanh[c + d*x])^2) - 1/(4*d*(a^2 + a^2*Tanh[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0275757, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3479, 8} \[ -\frac{1}{4 d \left (a^2 \tanh (c+d x)+a^2\right )}+\frac{x}{4 a^2}-\frac{1}{4 d (a \tanh (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tanh[c + d*x])^(-2),x]

[Out]

x/(4*a^2) - 1/(4*d*(a + a*Tanh[c + d*x])^2) - 1/(4*d*(a^2 + a^2*Tanh[c + d*x]))

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \tanh (c+d x))^2} \, dx &=-\frac{1}{4 d (a+a \tanh (c+d x))^2}+\frac{\int \frac{1}{a+a \tanh (c+d x)} \, dx}{2 a}\\ &=-\frac{1}{4 d (a+a \tanh (c+d x))^2}-\frac{1}{4 d \left (a^2+a^2 \tanh (c+d x)\right )}+\frac{\int 1 \, dx}{4 a^2}\\ &=\frac{x}{4 a^2}-\frac{1}{4 d (a+a \tanh (c+d x))^2}-\frac{1}{4 d \left (a^2+a^2 \tanh (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.170535, size = 60, normalized size = 1.18 \[ \frac{\text{sech}^2(c+d x) ((4 d x+1) \sinh (2 (c+d x))+(4 d x-1) \cosh (2 (c+d x))-4)}{16 a^2 d (\tanh (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tanh[c + d*x])^(-2),x]

[Out]

(Sech[c + d*x]^2*(-4 + (-1 + 4*d*x)*Cosh[2*(c + d*x)] + (1 + 4*d*x)*Sinh[2*(c + d*x)]))/(16*a^2*d*(1 + Tanh[c
+ d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.018, size = 72, normalized size = 1.4 \begin{align*} -{\frac{1}{4\,d{a}^{2} \left ( \tanh \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{1}{4\,d{a}^{2} \left ( \tanh \left ( dx+c \right ) +1 \right ) }}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{8\,d{a}^{2}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{8\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*tanh(d*x+c))^2,x)

[Out]

-1/4/d/a^2/(tanh(d*x+c)+1)^2-1/4/d/a^2/(tanh(d*x+c)+1)+1/8/d/a^2*ln(tanh(d*x+c)+1)-1/8/d/a^2*ln(tanh(d*x+c)-1)

________________________________________________________________________________________

Maxima [A]  time = 1.08075, size = 58, normalized size = 1.14 \begin{align*} \frac{d x + c}{4 \, a^{2} d} - \frac{4 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^2,x, algorithm="maxima")

[Out]

1/4*(d*x + c)/(a^2*d) - 1/16*(4*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))/(a^2*d)

________________________________________________________________________________________

Fricas [B]  time = 2.22352, size = 270, normalized size = 5.29 \begin{align*} \frac{{\left (4 \, d x - 1\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (4 \, d x + 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (4 \, d x - 1\right )} \sinh \left (d x + c\right )^{2} - 4}{16 \,{\left (a^{2} d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*((4*d*x - 1)*cosh(d*x + c)^2 + 2*(4*d*x + 1)*cosh(d*x + c)*sinh(d*x + c) + (4*d*x - 1)*sinh(d*x + c)^2 -
4)/(a^2*d*cosh(d*x + c)^2 + 2*a^2*d*cosh(d*x + c)*sinh(d*x + c) + a^2*d*sinh(d*x + c)^2)

________________________________________________________________________________________

Sympy [A]  time = 1.64479, size = 223, normalized size = 4.37 \begin{align*} \begin{cases} \frac{d x \tanh ^{2}{\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh{\left (c + d x \right )} + 4 a^{2} d} + \frac{2 d x \tanh{\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh{\left (c + d x \right )} + 4 a^{2} d} + \frac{d x}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh{\left (c + d x \right )} + 4 a^{2} d} - \frac{\tanh{\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh{\left (c + d x \right )} + 4 a^{2} d} - \frac{2}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh{\left (c + d x \right )} + 4 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x}{\left (a \tanh{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))**2,x)

[Out]

Piecewise((d*x*tanh(c + d*x)**2/(4*a**2*d*tanh(c + d*x)**2 + 8*a**2*d*tanh(c + d*x) + 4*a**2*d) + 2*d*x*tanh(c
+ d*x)/(4*a**2*d*tanh(c + d*x)**2 + 8*a**2*d*tanh(c + d*x) + 4*a**2*d) + d*x/(4*a**2*d*tanh(c + d*x)**2 + 8*a
**2*d*tanh(c + d*x) + 4*a**2*d) - tanh(c + d*x)/(4*a**2*d*tanh(c + d*x)**2 + 8*a**2*d*tanh(c + d*x) + 4*a**2*d
) - 2/(4*a**2*d*tanh(c + d*x)**2 + 8*a**2*d*tanh(c + d*x) + 4*a**2*d), Ne(d, 0)), (x/(a*tanh(c) + a)**2, True)
)

________________________________________________________________________________________

Giac [A]  time = 1.16991, size = 54, normalized size = 1.06 \begin{align*} \frac{4 \, d x -{\left (4 \, e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, c}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(4*d*x - (4*e^(2*d*x + 2*c) + 1)*e^(-4*d*x - 4*c) + 4*c)/(a^2*d)