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(b*x+a))/b^2-x*sech(b*x+a)/b

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, 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 3770

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

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]

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 = 0.05, 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

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 116, normalized size = 4.83 \[ -\frac {2 \, {\left (b x \cosh \left (b x + a\right ) + b x \sinh \left (b x + a\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} + b^{2}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 70, normalized size = 2.92 \[ -\frac {2 \, {\left (\pi + b x e^{\left (b x + a\right )} + \pi e^{\left (2 \, b x + 2 \, a\right )} - \arctan \left (e^{\left (b x + a\right )}\right ) e^{\left (2 \, b x + 2 \, a\right )} - \arctan \left (e^{\left (b x + a\right )}\right )\right )}}{b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} \]

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)

________________________________________________________________________________________

maple [C]  time = 0.17, size = 59, normalized size = 2.46 \[ -\frac {2 x \,{\mathrm e}^{b x +a}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b^{2}}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b^{2}} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 37, normalized size = 1.54 \[ -\frac {2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + \frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b^{2}} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 49, normalized size = 2.04 \[ \frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^4}}{b^2}\right )}{\sqrt {b^4}}-\frac {2\,x\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________