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

Optimal. Leaf size=24 \[ \frac{a \cosh (c+d x)}{d}-\frac{b \text{sech}(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0326643, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4133, 14} \[ \frac{a \cosh (c+d x)}{d}-\frac{b \text{sech}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)*Sinh[c + d*x],x]

[Out]

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

Rule 4133

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*(ff*x)^n)^p)/(ff*x)^(n*
p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]

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

Mathematica [A]  time = 0.021398, size = 35, normalized size = 1.46 \[ \frac{a \sinh (c) \sinh (d x)}{d}+\frac{a \cosh (c) \cosh (d x)}{d}-\frac{b \text{sech}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 26, normalized size = 1.1 \begin{align*} -{\frac{1}{d} \left ( b{\rm sech} \left (dx+c\right )-{\frac{a}{{\rm sech} \left (dx+c\right )}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(b*sech(d*x+c)-1/sech(d*x+c)*a)

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Maxima [A]  time = 1.00482, size = 49, normalized size = 2.04 \begin{align*} \frac{a \cosh \left (d x + c\right )}{d} - \frac{2 \, b}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.63862, size = 99, normalized size = 4.12 \begin{align*} \frac{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + a - 2 \, b}{2 \, d \cosh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14407, size = 63, normalized size = 2.62 \begin{align*} \frac{a{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{2 \, d} - \frac{2 \, b}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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