3.175 $$\int \frac{1}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=45 $\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)}+\frac{x}{a+b}$

[Out]

x/(a + b) + (Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)*d)

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Rubi [A]  time = 0.0740487, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {3660, 3675, 205} $\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a+b)}+\frac{x}{a+b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Tanh[c + d*x]^2)^(-1),x]

[Out]

x/(a + b) + (Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a + b)*d)

Rule 3660

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> Simp[x/(a - b), x] - Dist[b/(a - b), Int[Sec[e
+ f*x]^2/(a + b*Tan[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a, b]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{a+b \tanh ^2(c+d x)} \, dx &=\frac{x}{a+b}+\frac{b \int \frac{\text{sech}^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx}{a+b}\\ &=\frac{x}{a+b}+\frac{b \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b) d}\\ &=\frac{x}{a+b}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.0770452, size = 65, normalized size = 1.44 $\frac{\frac{2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a}}-\log (1-\tanh (c+d x))+\log (\tanh (c+d x)+1)}{2 a d+2 b d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*Tanh[c + d*x]^2)^(-1),x]

[Out]

((2*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/Sqrt[a] - Log[1 - Tanh[c + d*x]] + Log[1 + Tanh[c + d*x]]
)/(2*a*d + 2*b*d)

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Maple [B]  time = 0.017, size = 76, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{d \left ( 2\,b+2\,a \right ) }}+{\frac{b}{d \left ( a+b \right ) }\arctan \left ({b\tanh \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,b+2\,a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(2*b+2*a)*ln(tanh(d*x+c)+1)+1/d*b/(a+b)/(a*b)^(1/2)*arctan(tanh(d*x+c)*b/(a*b)^(1/2))-1/d/(2*b+2*a)*ln(tan
h(d*x+c)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.95923, size = 1249, normalized size = 27.76 \begin{align*} \left [\frac{2 \, d x + \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - 6 \, a b + b^{2} + 4 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} +{\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 4 \,{\left ({\left (a^{2} + a b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} + a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{2} + a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} - a b\right )} \sqrt{-\frac{b}{a}}}{{\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \,{\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} +{\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b}\right )}{2 \,{\left (a + b\right )} d}, \frac{d x + \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )} \sqrt{\frac{b}{a}}}{2 \, b}\right )}{{\left (a + b\right )} d}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*d*x + sqrt(-b/a)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d
*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cos
h(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2
- b^2)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x +
c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 - a*b)*sqrt(-b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*s
inh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*s
inh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)))/((a + b)*d), (d*
x + sqrt(b/a)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x +
c)^2 + a - b)*sqrt(b/a)/b))/((a + b)*d)]

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Sympy [A]  time = 11.581, size = 280, normalized size = 6.22 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\tanh ^{2}{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{x}{a} & \text{for}\: b = 0 \\\frac{x - \frac{1}{d \tanh{\left (c + d x \right )}}}{b} & \text{for}\: a = 0 \\- \frac{d x \tanh ^{2}{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} + \frac{d x}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} + \frac{\tanh{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} & \text{for}\: a = - b \\\frac{x}{a + b \tanh ^{2}{\left (c \right )}} & \text{for}\: d = 0 \\\frac{2 i \sqrt{a} d x \sqrt{\frac{1}{b}}}{2 i a^{\frac{3}{2}} d \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b d \sqrt{\frac{1}{b}}} + \frac{\log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \tanh{\left (c + d x \right )} \right )}}{2 i a^{\frac{3}{2}} d \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b d \sqrt{\frac{1}{b}}} - \frac{\log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \tanh{\left (c + d x \right )} \right )}}{2 i a^{\frac{3}{2}} d \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b d \sqrt{\frac{1}{b}}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/tanh(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/a, Eq(b, 0)), ((x - 1/(d*tanh(c + d*x)))/b, E
q(a, 0)), (-d*x*tanh(c + d*x)**2/(2*b*d*tanh(c + d*x)**2 - 2*b*d) + d*x/(2*b*d*tanh(c + d*x)**2 - 2*b*d) + tan
h(c + d*x)/(2*b*d*tanh(c + d*x)**2 - 2*b*d), Eq(a, -b)), (x/(a + b*tanh(c)**2), Eq(d, 0)), (2*I*sqrt(a)*d*x*sq
rt(1/b)/(2*I*a**(3/2)*d*sqrt(1/b) + 2*I*sqrt(a)*b*d*sqrt(1/b)) + log(-I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))/(2*
I*a**(3/2)*d*sqrt(1/b) + 2*I*sqrt(a)*b*d*sqrt(1/b)) - log(I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))/(2*I*a**(3/2)*d
*sqrt(1/b) + 2*I*sqrt(a)*b*d*sqrt(1/b)), True))

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Giac [A]  time = 1.16146, size = 90, normalized size = 2. \begin{align*} \frac{b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{\sqrt{a b}{\left (a d + b d\right )}} + \frac{d x + c}{a d + b d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*(a*d + b*d)) + (d*x + c)/(a
*d + b*d)