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

Optimal. Leaf size=28 \[ \frac{x}{2 a}-\frac{1}{2 d (a \tanh (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.0128619, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3479, 8} \[ \frac{x}{2 a}-\frac{1}{2 d (a \tanh (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tanh[c + d*x])^(-1),x]

[Out]

x/(2*a) - 1/(2*d*(a + a*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)} \, dx &=-\frac{1}{2 d (a+a \tanh (c+d x))}+\frac{\int 1 \, dx}{2 a}\\ &=\frac{x}{2 a}-\frac{1}{2 d (a+a \tanh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.100759, size = 39, normalized size = 1.39 \[ \frac{(2 d x+1) \tanh (c+d x)+2 d x-1}{4 a d (\tanh (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tanh[c + d*x])^(-1),x]

[Out]

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

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Maple [B]  time = 0.017, size = 54, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,da \left ( \tanh \left ( dx+c \right ) +1 \right ) }}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{4\,da}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{4\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08422, size = 42, normalized size = 1.5 \begin{align*} \frac{d x + c}{2 \, a d} - \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.19028, size = 136, normalized size = 4.86 \begin{align*} \frac{{\left (2 \, d x - 1\right )} \cosh \left (d x + c\right ) +{\left (2 \, d x + 1\right )} \sinh \left (d x + c\right )}{4 \,{\left (a d \cosh \left (d x + c\right ) + a d \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.8272, size = 73, normalized size = 2.61 \begin{align*} \begin{cases} \frac{d x \tanh{\left (c + d x \right )}}{2 a d \tanh{\left (c + d x \right )} + 2 a d} + \frac{d x}{2 a d \tanh{\left (c + d x \right )} + 2 a d} - \frac{1}{2 a d \tanh{\left (c + d x \right )} + 2 a d} & \text{for}\: d \neq 0 \\\frac{x}{a \tanh{\left (c \right )} + a} & \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)),x)

[Out]

Piecewise((d*x*tanh(c + d*x)/(2*a*d*tanh(c + d*x) + 2*a*d) + d*x/(2*a*d*tanh(c + d*x) + 2*a*d) - 1/(2*a*d*tanh
(c + d*x) + 2*a*d), Ne(d, 0)), (x/(a*tanh(c) + a), True))

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Giac [A]  time = 1.18865, size = 36, normalized size = 1.29 \begin{align*} \frac{2 \, d x + 2 \, c - e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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