3.5 \(\int \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=13 \[ x-\frac {\tanh (a+b x)}{b} \]

[Out]

x-tanh(b*x+a)/b

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3473, 8} \[ x-\frac {\tanh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[a + b*x]^2,x]

[Out]

x - Tanh[a + b*x]/b

Rule 8

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \tanh ^2(a+b x) \, dx &=-\frac {\tanh (a+b x)}{b}+\int 1 \, dx\\ &=x-\frac {\tanh (a+b x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 1.77 \[ \frac {\tanh ^{-1}(\tanh (a+b x))}{b}-\frac {\tanh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + b*x]^2,x]

[Out]

ArcTanh[Tanh[a + b*x]]/b - Tanh[a + b*x]/b

________________________________________________________________________________________

fricas [B]  time = 0.52, size = 33, normalized size = 2.54 \[ \frac {{\left (b x + 1\right )} \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}{b \cosh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

((b*x + 1)*cosh(b*x + a) - sinh(b*x + a))/(b*cosh(b*x + a))

________________________________________________________________________________________

giac [A]  time = 0.12, size = 24, normalized size = 1.85 \[ \frac {b x + a + \frac {2}{e^{\left (2 \, b x + 2 \, a\right )} + 1}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(b*x+a)^2,x, algorithm="giac")

[Out]

(b*x + a + 2/(e^(2*b*x + 2*a) + 1))/b

________________________________________________________________________________________

maple [B]  time = 0.01, size = 41, normalized size = 3.15 \[ -\frac {\tanh \left (b x +a \right )}{b}-\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{2 b}+\frac {\ln \left (1+\tanh \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(b*x+a)^2,x)

[Out]

-tanh(b*x+a)/b-1/2/b*ln(-1+tanh(b*x+a))+1/2*ln(1+tanh(b*x+a))/b

________________________________________________________________________________________

maxima [A]  time = 0.30, size = 25, normalized size = 1.92 \[ x + \frac {a}{b} - \frac {2}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 13, normalized size = 1.00 \[ x-\frac {\mathrm {tanh}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + b*x)^2,x)

[Out]

x - tanh(a + b*x)/b

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 15, normalized size = 1.15 \[ \begin {cases} x - \frac {\tanh {\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \tanh ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(b*x+a)**2,x)

[Out]

Piecewise((x - tanh(a + b*x)/b, Ne(b, 0)), (x*tanh(a)**2, True))

________________________________________________________________________________________