3.150 \(\int \text{sech}^2(c+b x) \sinh (a+b x) \, dx\)

Optimal. Leaf size=35 \[ \frac{\sinh (a-c) \tan ^{-1}(\sinh (b x+c))}{b}-\frac{\cosh (a-c) \text{sech}(b x+c)}{b} \]

[Out]

-((Cosh[a - c]*Sech[c + b*x])/b) + (ArcTan[Sinh[c + b*x]]*Sinh[a - c])/b

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Rubi [A]  time = 0.0327126, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {5624, 2606, 8, 3770} \[ \frac{\sinh (a-c) \tan ^{-1}(\sinh (b x+c))}{b}-\frac{\cosh (a-c) \text{sech}(b x+c)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + b*x]^2*Sinh[a + b*x],x]

[Out]

-((Cosh[a - c]*Sech[c + b*x])/b) + (ArcTan[Sinh[c + b*x]]*Sinh[a - c])/b

Rule 5624

Int[Sech[w_]^(n_.)*Sinh[v_], x_Symbol] :> Dist[Cosh[v - w], Int[Tanh[w]*Sech[w]^(n - 1), x], x] + Dist[Sinh[v
- w], Int[Sech[w]^(n - 1), x], x] /; GtQ[n, 0] && NeQ[w, v] && FreeQ[v - w, x]

Rule 2606

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

Rule 8

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

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

Mathematica [B]  time = 0.0872576, size = 83, normalized size = 2.37 \[ \frac{2 \sinh (a-c) \tan ^{-1}\left (\frac{(\cosh (c)-\sinh (c)) \left (\sinh (c) \cosh \left (\frac{b x}{2}\right )+\cosh (c) \sinh \left (\frac{b x}{2}\right )\right )}{\cosh (c) \cosh \left (\frac{b x}{2}\right )-\sinh (c) \cosh \left (\frac{b x}{2}\right )}\right )}{b}-\frac{\cosh (a-c) \text{sech}(b x+c)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.08, size = 181, normalized size = 5.2 \begin{align*} -{\frac{{{\rm e}^{bx+a}} \left ({{\rm e}^{2\,a}}+{{\rm e}^{2\,c}} \right ) }{b \left ({{\rm e}^{2\,bx+2\,a+2\,c}}+{{\rm e}^{2\,a}} \right ) }}+{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}}+i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}} \left ({{\rm e}^{a}} \right ) ^{2}}{b}}-{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}}+i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}} \left ({{\rm e}^{c}} \right ) ^{2}}{b}}-{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}}-i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}} \left ({{\rm e}^{a}} \right ) ^{2}}{b}}+{\frac{{\frac{i}{2}}\ln \left ({{\rm e}^{bx+a}}-i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}} \left ({{\rm e}^{c}} \right ) ^{2}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.88148, size = 95, normalized size = 2.71 \begin{align*} -\frac{{\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )} \arctan \left (e^{\left (-b x - c\right )}\right ) e^{\left (-a - c\right )}}{b} - \frac{{\left (e^{\left (2 \, a\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (-b x - a\right )}}{b{\left (e^{\left (-2 \, b x\right )} + e^{\left (2 \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.07769, size = 1129, normalized size = 32.26 \begin{align*} \frac{2 \, \cosh \left (b x + c\right ) \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) - \cosh \left (b x + c\right ) \sinh \left (-a + c\right )^{2} +{\left ({\left (\cosh \left (-a + c\right )^{2} - 1\right )} \cosh \left (b x + c\right )^{2} +{\left (\cosh \left (-a + c\right )^{2} - 2 \, \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) + \sinh \left (-a + c\right )^{2} - 1\right )} \sinh \left (b x + c\right )^{2} +{\left (\cosh \left (b x + c\right )^{2} + 1\right )} \sinh \left (-a + c\right )^{2} + \cosh \left (-a + c\right )^{2} - 2 \,{\left (2 \, \cosh \left (b x + c\right ) \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) - \cosh \left (b x + c\right ) \sinh \left (-a + c\right )^{2} -{\left (\cosh \left (-a + c\right )^{2} - 1\right )} \cosh \left (b x + c\right )\right )} \sinh \left (b x + c\right ) - 2 \,{\left (\cosh \left (b x + c\right )^{2} \cosh \left (-a + c\right ) + \cosh \left (-a + c\right )\right )} \sinh \left (-a + c\right ) - 1\right )} \arctan \left (\cosh \left (b x + c\right ) + \sinh \left (b x + c\right )\right ) -{\left (\cosh \left (-a + c\right )^{2} + 1\right )} \cosh \left (b x + c\right ) -{\left (\cosh \left (-a + c\right )^{2} - 2 \, \cosh \left (-a + c\right ) \sinh \left (-a + c\right ) + \sinh \left (-a + c\right )^{2} + 1\right )} \sinh \left (b x + c\right )}{b \cosh \left (b x + c\right )^{2} \cosh \left (-a + c\right ) +{\left (b \cosh \left (-a + c\right ) - b \sinh \left (-a + c\right )\right )} \sinh \left (b x + c\right )^{2} + b \cosh \left (-a + c\right ) + 2 \,{\left (b \cosh \left (b x + c\right ) \cosh \left (-a + c\right ) - b \cosh \left (b x + c\right ) \sinh \left (-a + c\right )\right )} \sinh \left (b x + c\right ) -{\left (b \cosh \left (b x + c\right )^{2} + b\right )} \sinh \left (-a + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*cosh(b*x + c)*cosh(-a + c)*sinh(-a + c) - cosh(b*x + c)*sinh(-a + c)^2 + ((cosh(-a + c)^2 - 1)*cosh(b*x + c
)^2 + (cosh(-a + c)^2 - 2*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2 - 1)*sinh(b*x + c)^2 + (cosh(b*x + c)^2 +
1)*sinh(-a + c)^2 + cosh(-a + c)^2 - 2*(2*cosh(b*x + c)*cosh(-a + c)*sinh(-a + c) - cosh(b*x + c)*sinh(-a + c
)^2 - (cosh(-a + c)^2 - 1)*cosh(b*x + c))*sinh(b*x + c) - 2*(cosh(b*x + c)^2*cosh(-a + c) + cosh(-a + c))*sinh
(-a + c) - 1)*arctan(cosh(b*x + c) + sinh(b*x + c)) - (cosh(-a + c)^2 + 1)*cosh(b*x + c) - (cosh(-a + c)^2 - 2
*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2 + 1)*sinh(b*x + c))/(b*cosh(b*x + c)^2*cosh(-a + c) + (b*cosh(-a +
c) - b*sinh(-a + c))*sinh(b*x + c)^2 + b*cosh(-a + c) + 2*(b*cosh(b*x + c)*cosh(-a + c) - b*cosh(b*x + c)*sin
h(-a + c))*sinh(b*x + c) - (b*cosh(b*x + c)^2 + b)*sinh(-a + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18654, size = 92, normalized size = 2.63 \begin{align*} \frac{{\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )} \arctan \left (e^{\left (b x + c\right )}\right ) e^{\left (-a - c\right )} - \frac{{\left (e^{\left (b x + 2 \, a\right )} + e^{\left (b x + 2 \, c\right )}\right )} e^{\left (-a\right )}}{e^{\left (2 \, b x + 2 \, c\right )} + 1}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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