### 3.387 $$\int \sinh (a+b x) \tanh ^2(a+b x) \, dx$$

Optimal. Leaf size=21 $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$

[Out]

Cosh[a + b*x]/b + Sech[a + b*x]/b

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Rubi [A]  time = 0.0242972, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {2590, 14} $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

Cosh[a + b*x]/b + Sech[a + b*x]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0285767, size = 21, normalized size = 1. $\frac{\cosh (a+b x)}{b}+\frac{\text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Cosh[a + b*x]/b + Sech[a + b*x]/b

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Maple [A]  time = 0.016, size = 32, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{\cosh \left ( bx+a \right ) }}+2\,\cosh \left ( bx+a \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^2*sinh(b*x+a)^3,x)

[Out]

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

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Maxima [B]  time = 1.03865, size = 73, normalized size = 3.48 \begin{align*} \frac{e^{\left (-b x - a\right )}}{2 \, b} + \frac{5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1}{2 \, b{\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/2*e^(-b*x - a)/b + 1/2*(5*e^(-2*b*x - 2*a) + 1)/(b*(e^(-b*x - a) + e^(-3*b*x - 3*a)))

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Fricas [A]  time = 1.94858, size = 85, normalized size = 4.05 \begin{align*} \frac{\cosh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{2} + 3}{2 \, b \cosh \left (b x + a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{3}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**2*sinh(b*x+a)**3,x)

[Out]

Integral(sinh(a + b*x)**3*sech(a + b*x)**2, x)

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Giac [B]  time = 1.20887, size = 61, normalized size = 2.9 \begin{align*} \frac{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{2 \, b} + \frac{2}{b{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

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