3.344 \(\int x \text{sech}(a+b x) \tanh (a+b x) \, dx\)

Optimal. Leaf size=24 \[ \frac{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{x \text{sech}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0190941, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5418, 3770} \[ \frac{\tan ^{-1}(\sinh (a+b x))}{b^2}-\frac{x \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sech[a + b*x]*Tanh[a + b*x],x]

[Out]

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

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

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

Mathematica [A]  time = 0.0457742, size = 32, normalized size = 1.33 \[ \frac{2 \tan ^{-1}\left (\tanh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b^2}-\frac{x \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.039, size = 59, normalized size = 2.5 \begin{align*} -2\,{\frac{x{{\rm e}^{bx+a}}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}+{\frac{i\ln \left ({{\rm e}^{bx+a}}+i \right ) }{{b}^{2}}}-{\frac{i\ln \left ({{\rm e}^{bx+a}}-i \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*exp(b*x+a)/b/(1+exp(2*b*x+2*a))+I/b^2*ln(exp(b*x+a)+I)-I/b^2*ln(exp(b*x+a)-I)

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Maxima [A]  time = 1.78737, size = 50, normalized size = 2.08 \begin{align*} -\frac{2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + \frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.77433, size = 327, normalized size = 13.62 \begin{align*} -\frac{2 \,{\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right ) -{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*x*cosh(b*x + a) + b*x*sinh(b*x + a) - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2
+ 1)*arctan(cosh(b*x + a) + sinh(b*x + a)))/(b^2*cosh(b*x + a)^2 + 2*b^2*cosh(b*x + a)*sinh(b*x + a) + b^2*si
nh(b*x + a)^2 + b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.31225, size = 95, normalized size = 3.96 \begin{align*} -\frac{2 \,{\left (\pi + b x e^{\left (b x + a\right )} + \pi e^{\left (2 \, b x + 2 \, a\right )} - \arctan \left (e^{\left (b x + a\right )}\right ) e^{\left (2 \, b x + 2 \, a\right )} - \arctan \left (e^{\left (b x + a\right )}\right )\right )}}{b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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