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]

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

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Rubi [A]  time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 8

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

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 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}^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*}

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Mathematica [A]  time = 0.08, size = 30, normalized size = 1.00 \[ \frac {\tanh (a+b x)}{2 b^2}-\frac {x \text {sech}^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x*Sech[a + b*x]^2)/b + Tanh[a + b*x]/(2*b^2)

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fricas [B]  time = 0.43, size = 105, normalized size = 3.50 \[ -\frac {2 \, {\left (b x \sinh \left (b x + a\right ) + {\left (b x + 1\right )} \cosh \left (b x + a\right )\right )}}{b^{2} \cosh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b^{2} \sinh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) + {\left (3 \, b^{2} \cosh \left (b x + a\right )^{2} + b^{2}\right )} \sinh \left (b x + a\right )} \]

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))

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giac [B]  time = 0.15, size = 184, normalized size = 6.13 \[ -\frac {4 \, b x e^{\left (2 \, b x + 2 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) - 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + e^{\left (4 \, b x + 4 \, a\right )} \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + \log \left (-e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + 2}{2 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} \]

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)

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maple [A]  time = 0.16, size = 43, normalized size = 1.43 \[ -\frac {2 b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}+1}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}} \]

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

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maxima [B]  time = 0.35, size = 131, normalized size = 4.37 \[ -\frac {2 \, b x e^{\left (4 \, b x + 4 \, a\right )} + {\left (4 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}{2 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} + \frac {2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} - 1}{2 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} \]

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)

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mupad [B]  time = 1.46, size = 36, normalized size = 1.20 \[ -\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (2\,b\,x+1\right )+1}{b^2\,{\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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