3.142 \(\int \frac{\tanh (c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=23 \[ \frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a d} \]

[Out]

Log[b + a*Cosh[c + d*x]^2]/(2*a*d)

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Rubi [A]  time = 0.0347621, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4138, 260} \[ \frac{\log \left (a \cosh ^2(c+d x)+b\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[c + d*x]/(a + b*Sech[c + d*x]^2),x]

[Out]

Log[b + a*Cosh[c + d*x]^2]/(2*a*d)

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\tanh (c+d x)}{a+b \text{sech}^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{b+a x^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\log \left (b+a \cosh ^2(c+d x)\right )}{2 a d}\\ \end{align*}

Mathematica [A]  time = 0.175158, size = 26, normalized size = 1.13 \[ \frac{\log (a \cosh (2 (c+d x))+a+2 b)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[c + d*x]/(a + b*Sech[c + d*x]^2),x]

[Out]

Log[a + 2*b + a*Cosh[2*(c + d*x)]]/(2*a*d)

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Maple [A]  time = 0.017, size = 38, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( a+b \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) }{2\,da}}-{\frac{\ln \left ({\rm sech} \left (dx+c\right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/a*ln(a+b*sech(d*x+c)^2)-1/d/a*ln(sech(d*x+c))

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Maxima [B]  time = 1.13346, size = 69, normalized size = 3. \begin{align*} \frac{d x + c}{a d} + \frac{\log \left (2 \,{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.18883, size = 198, normalized size = 8.61 \begin{align*} -\frac{2 \, d x - \log \left (\frac{2 \,{\left (a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right )}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*d*x - log(2*(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + a + 2*b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(
d*x + c) + sinh(d*x + c)^2)))/(a*d)

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Sympy [A]  time = 9.33816, size = 124, normalized size = 5.39 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \tanh{\left (c \right )}}{\operatorname{sech}^{2}{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{1}{2 b d \operatorname{sech}^{2}{\left (c + d x \right )}} & \text{for}\: a = 0 \\\frac{x \tanh{\left (c \right )}}{a + b \operatorname{sech}^{2}{\left (c \right )}} & \text{for}\: d = 0 \\\frac{x - \frac{\log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d}}{a} & \text{for}\: b = 0 \\\frac{x}{a} + \frac{\log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \operatorname{sech}{\left (c + d x \right )} \right )}}{2 a d} + \frac{\log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \operatorname{sech}{\left (c + d x \right )} \right )}}{2 a d} - \frac{\log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{a d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tanh(c)/sech(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (1/(2*b*d*sech(c + d*x)**2), Eq(a, 0)),
(x*tanh(c)/(a + b*sech(c)**2), Eq(d, 0)), ((x - log(tanh(c + d*x) + 1)/d)/a, Eq(b, 0)), (x/a + log(-I*sqrt(a)*
sqrt(1/b) + sech(c + d*x))/(2*a*d) + log(I*sqrt(a)*sqrt(1/b) + sech(c + d*x))/(2*a*d) - log(tanh(c + d*x) + 1)
/(a*d), True))

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Giac [B]  time = 1.39835, size = 76, normalized size = 3.3 \begin{align*} -\frac{\frac{2 \, d x}{a} - \frac{\log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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