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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, 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 = 0.02, size = 34, normalized size = 1.00 \[ \frac {\tan ^{-1}(\sinh (a+b x))}{2 b}-\frac {\tanh (a+b x) \text {sech}(a+b x)}{2 b} \]

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)

________________________________________________________________________________________

fricas [B]  time = 0.93, size = 269, normalized size = 7.91 \[ -\frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.15, size = 76, normalized size = 2.24 \[ \frac {\pi - \frac {4 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4} + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{4 \, b} \]

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

________________________________________________________________________________________

maple [A]  time = 0.34, size = 49, normalized size = 1.44 \[ -\frac {\sinh \left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}+\frac {\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}+\frac {\arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.42, size = 66, normalized size = 1.94 \[ -\frac {\arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \]

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))

________________________________________________________________________________________

mupad [B]  time = 1.47, size = 82, normalized size = 2.41 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}+\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b)/(b^2)^(1/2) + (2*exp(a + b*x))/(b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x)
+ 1)) - exp(a + b*x)/(b*(exp(2*a + 2*b*x) + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]

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)

________________________________________________________________________________________