### 3.617 $$\int (a \text{sech}(x)+b \tanh (x))^2 \, dx$$

Optimal. Leaf size=29 $-a b \cosh (x)-\text{sech}(x) (b-a \sinh (x)) (a+b \sinh (x))+b^2 x$

[Out]

b^2*x - a*b*Cosh[x] - Sech[x]*(b - a*Sinh[x])*(a + b*Sinh[x])

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Rubi [A]  time = 0.0627443, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {4391, 2691, 2638} $-a b \cosh (x)-\text{sech}(x) (b-a \sinh (x)) (a+b \sinh (x))+b^2 x$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Sech[x] + b*Tanh[x])^2,x]

[Out]

b^2*x - a*b*Cosh[x] - Sech[x]*(b - a*Sinh[x])*(a + b*Sinh[x])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2691

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(b + a*Sin[e + f*x]))/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1
)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[
2*m, 2*p] || IntegerQ[m])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0476505, size = 26, normalized size = 0.9 $\left (a^2-b^2\right ) \tanh (x)-2 a b \text{sech}(x)+b^2 \tanh ^{-1}(\tanh (x))$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Sech[x] + b*Tanh[x])^2,x]

[Out]

b^2*ArcTanh[Tanh[x]] - 2*a*b*Sech[x] + (a^2 - b^2)*Tanh[x]

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Maple [A]  time = 0.014, size = 36, normalized size = 1.2 \begin{align*}{a}^{2}\tanh \left ( x \right ) +2\,ab \left ({\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{\cosh \left ( x \right ) }}-\cosh \left ( x \right ) \right ) +{b}^{2} \left ( x-\tanh \left ( x \right ) \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05664, size = 58, normalized size = 2. \begin{align*} b^{2}{\left (x - \frac{2}{e^{\left (-2 \, x\right )} + 1}\right )} - \frac{4 \, a b}{e^{\left (-x\right )} + e^{x}} + \frac{2 \, a^{2}}{e^{\left (-2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(x - 2/(e^(-2*x) + 1)) - 4*a*b/(e^(-x) + e^x) + 2*a^2/(e^(-2*x) + 1)

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Fricas [A]  time = 2.26537, size = 95, normalized size = 3.28 \begin{align*} -\frac{2 \, a b -{\left (b^{2} x - a^{2} + b^{2}\right )} \cosh \left (x\right ) -{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )}{\cosh \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sech(x)+b*tanh(x))**2,x)

[Out]

Integral((a*sech(x) + b*tanh(x))**2, x)

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Giac [A]  time = 1.13688, size = 42, normalized size = 1.45 \begin{align*} b^{2} x - \frac{2 \,{\left (2 \, a b e^{x} + a^{2} - b^{2}\right )}}{e^{\left (2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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