3.373 $$\int \text{sech}(a+b x) \tanh ^2(a+b x) \, dx$$

Optimal. Leaf size=34 $\frac{\tan ^{-1}(\sinh (a+b x))}{2 b}-\frac{\tanh (a+b x) \text{sech}(a+b x)}{2 b}$

[Out]

ArcTan[Sinh[a + b*x]]/(2*b) - (Sech[a + b*x]*Tanh[a + b*x])/(2*b)

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Rubi [A]  time = 0.0237051, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {2611, 3770} $\frac{\tan ^{-1}(\sinh (a+b x))}{2 b}-\frac{\tanh (a+b x) \text{sech}(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

ArcTan[Sinh[a + b*x]]/(2*b) - (Sech[a + b*x]*Tanh[a + b*x])/(2*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0158222, size = 34, normalized size = 1. $\frac{\tan ^{-1}(\sinh (a+b x))}{2 b}-\frac{\tanh (a+b x) \text{sech}(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

ArcTan[Sinh[a + b*x]]/(2*b) - (Sech[a + b*x]*Tanh[a + b*x])/(2*b)

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Maple [A]  time = 0.014, size = 49, normalized size = 1.4 \begin{align*} -{\frac{\sinh \left ( bx+a \right ) }{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}+{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^3*sinh(b*x+a)^2,x)

[Out]

-1/b*sinh(b*x+a)/cosh(b*x+a)^2+1/2*sech(b*x+a)*tanh(b*x+a)/b+arctan(exp(b*x+a))/b

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Maxima [B]  time = 1.618, size = 89, normalized size = 2.62 \begin{align*} -\frac{\arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac{e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-arctan(e^(-b*x - a))/b - (e^(-b*x - a) - e^(-3*b*x - 3*a))/(b*(2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + 1))

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Fricas [B]  time = 1.99816, size = 759, normalized size = 22.32 \begin{align*} -\frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} -{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sin
h(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x +
a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (3*cosh(b*x + a)^2 - 1)*sinh
(b*x + a) - cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 2*b*co
sh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x
+ a) + b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{2}{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**3*sinh(b*x+a)**2,x)

[Out]

Integral(sinh(a + b*x)**2*sech(a + b*x)**3, x)

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Giac [B]  time = 1.16275, size = 108, normalized size = 3.18 \begin{align*} \frac{\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{4 \, b} - \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

1/4*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b - (e^(b*x + a) - e^(-b*x - a))/(((e^(b*x + a) -
e^(-b*x - a))^2 + 4)*b)