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]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.54, size = 50, normalized size = 1.79 \[ \frac {{\left (2 \, d x - 1\right )} \cosh \left (d x + c\right ) + {\left (2 \, d x + 1\right )} \sinh \left (d x + c\right )}{4 \, {\left (a d \cosh \left (d x + c\right ) + a d \sinh \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.13, size = 30, normalized size = 1.07 \[ \frac {\frac {2 \, {\left (d x + c\right )}}{a} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{a}}{4 \, d} \]

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 + c)/a - e^(-2*d*x - 2*c)/a)/d

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maple [B]  time = 0.10, size = 54, normalized size = 1.93 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{4 d a}-\frac {1}{2 d a \left (1+\tanh \left (d x +c \right )\right )}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 31, normalized size = 1.11 \[ \frac {d x + c}{2 \, a d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, a d} \]

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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mupad [B]  time = 1.07, size = 25, normalized size = 0.89 \[ \frac {x}{2\,a}-\frac {1}{2\,a\,d\,\left (\mathrm {tanh}\left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tanh(c + d*x)),x)

[Out]

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

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sympy [A]  time = 0.50, size = 73, normalized size = 2.61 \[ \begin {cases} \frac {d x \tanh {\left (c + d x \right )}}{2 a d \tanh {\left (c + d x \right )} + 2 a d} + \frac {d x}{2 a d \tanh {\left (c + d x \right )} + 2 a d} - \frac {1}{2 a d \tanh {\left (c + d x \right )} + 2 a d} & \text {for}\: d \neq 0 \\\frac {x}{a \tanh {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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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