3.83 \(\int \text {sech}^{1+n}(a+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=16 \[ -\frac {\text {sech}^n(a+b x)}{b n} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2622, 30} \[ -\frac {\text {sech}^n(a+b x)}{b n} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^(1 + n)*Sinh[a + b*x],x]

[Out]

-(Sech[a + b*x]^n/(b*n))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \text {sech}^{1+n}(a+b x) \sinh (a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int x^{-1+n} \, dx,x,\text {sech}(a+b x)\right )}{b}\\ &=-\frac {\text {sech}^n(a+b x)}{b n}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 16, normalized size = 1.00 \[ -\frac {\text {sech}^n(a+b x)}{b n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^(1 + n)*Sinh[a + b*x],x]

[Out]

-(Sech[a + b*x]^n/(b*n))

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fricas [B]  time = 0.40, size = 115, normalized size = 7.19 \[ -\frac {\cosh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right ) + \sinh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^n*tanh(b*x+a),x, algorithm="fricas")

[Out]

-(cosh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a
)^2 + 1))) + sinh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + s
inh(b*x + a)^2 + 1))))/(b*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {sech}\left (b x + a\right )^{n} \tanh \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^n*tanh(b*x+a),x, algorithm="giac")

[Out]

integrate(sech(b*x + a)^n*tanh(b*x + a), x)

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maple [A]  time = 0.10, size = 17, normalized size = 1.06 \[ -\frac {\mathrm {sech}\left (b x +a \right )^{n}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^n*tanh(b*x+a),x)

[Out]

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

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maxima [B]  time = 0.83, size = 36, normalized size = 2.25 \[ -\frac {2^{n} e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )\right )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^n*tanh(b*x+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.48, size = 31, normalized size = 1.94 \[ -\frac {{\left (\frac {2\,{\mathrm {e}}^{a+b\,x}}{{\mathrm {e}}^{2\,a+2\,b\,x}+1}\right )}^n}{b\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + b*x)*(1/cosh(a + b*x))^n,x)

[Out]

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

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sympy [A]  time = 0.49, size = 39, normalized size = 2.44 \[ \begin {cases} x \tanh {\relax (a )} & \text {for}\: b = 0 \wedge n = 0 \\x \tanh {\relax (a )} \operatorname {sech}^{n}{\relax (a )} & \text {for}\: b = 0 \\x - \frac {\log {\left (\tanh {\left (a + b x \right )} + 1 \right )}}{b} & \text {for}\: n = 0 \\- \frac {\operatorname {sech}^{n}{\left (a + b x \right )}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**n*tanh(b*x+a),x)

[Out]

Piecewise((x*tanh(a), Eq(b, 0) & Eq(n, 0)), (x*tanh(a)*sech(a)**n, Eq(b, 0)), (x - log(tanh(a + b*x) + 1)/b, E
q(n, 0)), (-sech(a + b*x)**n/(b*n), True))

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