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(b*x+a)/b+sech(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 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]]

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]

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*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \[ \frac {\cosh (a+b x)}{b}+\frac {\text {sech}(a+b x)}{b} \]

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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fricas [A]  time = 0.70, size = 31, normalized size = 1.48 \[ \frac {\cosh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{2} + 3}{2 \, b \cosh \left (b x + a\right )} \]

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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giac [A]  time = 0.15, size = 41, normalized size = 1.95 \[ \frac {\frac {4}{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}} + e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{2 \, b} \]

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

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maple [A]  time = 0.11, size = 33, normalized size = 1.57 \[ \frac {\frac {\sinh ^{2}\left (b x +a \right )}{\cosh \left (b x +a \right )}+\frac {2}{\cosh \left (b x +a \right )}}{b} \]

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 = 0.42, size = 54, normalized size = 2.57 \[ \frac {e^{\left (-b x - a\right )}}{2 \, b} + \frac {5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1}{2 \, b {\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \]

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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mupad [B]  time = 0.07, size = 22, normalized size = 1.05 \[ \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2+1}{b\,\mathrm {cosh}\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

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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