### 3.351 $$\int x \text{sech}^2(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=30 $\frac{\tanh (a+b x)}{2 b^2}-\frac{x \text{sech}^2(a+b x)}{2 b}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0304105, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {5418, 3767, 8} $\frac{\tanh (a+b x)}{2 b^2}-\frac{x \text{sech}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.072022, size = 30, normalized size = 1. $\frac{\tanh (a+b x)}{2 b^2}-\frac{x \text{sech}^2(a+b x)}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.028, size = 43, normalized size = 1.4 \begin{align*} -{\frac{2\,bx{{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,bx+2\,a}}+1}{{b}^{2} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.06165, size = 177, normalized size = 5.9 \begin{align*} -\frac{2 \, b x e^{\left (4 \, b x + 4 \, a\right )} +{\left (4 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}{2 \,{\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} + \frac{2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} - 1}{2 \,{\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.75086, size = 270, normalized size = 9. \begin{align*} -\frac{2 \,{\left (b x \sinh \left (b x + a\right ) +{\left (b x + 1\right )} \cosh \left (b x + a\right )\right )}}{b^{2} \cosh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b^{2} \sinh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) +{\left (3 \, b^{2} \cosh \left (b x + a\right )^{2} + b^{2}\right )} \sinh \left (b x + a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*x*sinh(b*x + a) + (b*x + 1)*cosh(b*x + a))/(b^2*cosh(b*x + a)^3 + 3*b^2*cosh(b*x + a)*sinh(b*x + a)^2 +
b^2*sinh(b*x + a)^3 + 3*b^2*cosh(b*x + a) + (3*b^2*cosh(b*x + a)^2 + b^2)*sinh(b*x + a))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sinh{\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(x*sech(b*x+a)**3*sinh(b*x+a),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.25154, size = 248, normalized size = 8.27 \begin{align*} -\frac{4 \, b x e^{\left (2 \, b x + 2 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) - 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + e^{\left (4 \, b x + 4 \, a\right )} \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2}{2 \,{\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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