3.144 \(\int \sinh (a+b x) \tanh ^2(c+b x) \, dx\)

Optimal. Leaf size=45 \[ -\frac {\sinh (a-c) \tan ^{-1}(\sinh (b x+c))}{b}+\frac {\cosh (a-c) \text {sech}(b x+c)}{b}+\frac {\cosh (a+b x)}{b} \]

[Out]

cosh(b*x+a)/b+cosh(a-c)*sech(b*x+c)/b-arctan(sinh(b*x+c))*sinh(a-c)/b

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5620, 5623, 2638, 3770, 2606, 8} \[ -\frac {\sinh (a-c) \tan ^{-1}(\sinh (b x+c))}{b}+\frac {\cosh (a-c) \text {sech}(b x+c)}{b}+\frac {\cosh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]*Tanh[c + b*x]^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2638

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

Rule 3770

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

Rule 5620

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

Rule 5623

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

Rubi steps

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

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Mathematica [B]  time = 0.11, size = 102, normalized size = 2.27 \[ \frac {\cosh (a-c) \text {sech}(b x+c)}{b}-\frac {2 \sinh (a-c) \tan ^{-1}\left (\frac {(\cosh (c)-\sinh (c)) \left (\sinh (c) \cosh \left (\frac {b x}{2}\right )+\cosh (c) \sinh \left (\frac {b x}{2}\right )\right )}{\cosh (c) \cosh \left (\frac {b x}{2}\right )-\sinh (c) \cosh \left (\frac {b x}{2}\right )}\right )}{b}+\frac {\sinh (a) \sinh (b x)}{b}+\frac {\cosh (a) \cosh (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[a]*Cosh[b*x])/b + (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 + (Sinh[a]*Sinh[b*x]
)/b

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fricas [B]  time = 0.51, size = 902, normalized size = 20.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(cosh(b*x + c)^4*cosh(-a + c)^2 + (cosh(-a + c)^2 - 2*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2)*sinh(b*x
+ c)^4 + 4*(cosh(b*x + c)*cosh(-a + c)^2 - 2*cosh(b*x + c)*cosh(-a + c)*sinh(-a + c) + cosh(b*x + c)*sinh(-a
+ c)^2)*sinh(b*x + c)^3 + 3*(cosh(-a + c)^2 + 1)*cosh(b*x + c)^2 + 3*(2*cosh(b*x + c)^2*cosh(-a + c)^2 + (2*co
sh(b*x + c)^2 + 1)*sinh(-a + c)^2 + cosh(-a + c)^2 - 2*(2*cosh(b*x + c)^2*cosh(-a + c) + cosh(-a + c))*sinh(-a
+ c) + 1)*sinh(b*x + c)^2 + (cosh(b*x + c)^4 + 3*cosh(b*x + c)^2)*sinh(-a + c)^2 - 2*((cosh(-a + c)^2 - 1)*co
sh(b*x + c)^3 + (cosh(-a + c)^2 - 2*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2 - 1)*sinh(b*x + c)^3 - 3*(2*cos
h(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)^3 + cosh(b*x + c))*sinh(-a + c)^2 + (cosh(-a + c)^2 - 1)*cosh(b*x + c) + (3*(cosh
(-a + c)^2 - 1)*cosh(b*x + c)^2 + (3*cosh(b*x + c)^2 + 1)*sinh(-a + c)^2 + cosh(-a + c)^2 - 2*(3*cosh(b*x + c)
^2*cosh(-a + c) + cosh(-a + c))*sinh(-a + c) - 1)*sinh(b*x + c) - 2*(cosh(b*x + c)^3*cosh(-a + c) + cosh(b*x +
c)*cosh(-a + c))*sinh(-a + c))*arctan(cosh(b*x + c) + sinh(b*x + c)) + 2*(2*cosh(b*x + c)^3*cosh(-a + c)^2 +
(2*cosh(b*x + c)^3 + 3*cosh(b*x + c))*sinh(-a + c)^2 + 3*(cosh(-a + c)^2 + 1)*cosh(b*x + c) - 2*(2*cosh(b*x +
c)^3*cosh(-a + c) + 3*cosh(b*x + c)*cosh(-a + c))*sinh(-a + c))*sinh(b*x + c) - 2*(cosh(b*x + c)^4*cosh(-a + c
) + 3*cosh(b*x + c)^2*cosh(-a + c))*sinh(-a + c) + 1)/(b*cosh(b*x + c)^3*cosh(-a + c) + (b*cosh(-a + c) - b*si
nh(-a + c))*sinh(b*x + c)^3 + b*cosh(b*x + c)*cosh(-a + c) + 3*(b*cosh(b*x + c)*cosh(-a + c) - b*cosh(b*x + c)
*sinh(-a + c))*sinh(b*x + c)^2 + (3*b*cosh(b*x + c)^2*cosh(-a + c) + b*cosh(-a + c) - (3*b*cosh(b*x + c)^2 + b
)*sinh(-a + c))*sinh(b*x + c) - (b*cosh(b*x + c)^3 + b*cosh(b*x + c))*sinh(-a + c))

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giac [A]  time = 0.15, size = 88, normalized size = 1.96 \[ -\frac {2 \, {\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )} \arctan \left (e^{\left (b x + c\right )}\right ) e^{\left (-a - c\right )} - \frac {{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, c\right )} + 1\right )} e^{\left (-a\right )}}{e^{\left (3 \, b x + 2 \, c\right )} + e^{\left (b x\right )}} - e^{\left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.32, size = 205, normalized size = 4.56 \[ \frac {{\mathrm e}^{b x +a}}{2 b}+\frac {{\mathrm e}^{-b x -a}}{2 b}+\frac {{\mathrm e}^{b x +a} \left ({\mathrm e}^{2 a}+{\mathrm e}^{2 c}\right )}{b \left ({\mathrm e}^{2 b x +2 a +2 c}+{\mathrm e}^{2 a}\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 a}}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 c}}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 a}}{2 b}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -c}\right ) {\mathrm e}^{-a -c} {\mathrm e}^{2 c}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 105, normalized size = 2.33 \[ \frac {{\left (e^{\left (2 \, a\right )} - e^{\left (2 \, c\right )}\right )} \arctan \left (e^{\left (-b x - c\right )}\right ) e^{\left (-a - c\right )}}{b} + \frac {e^{\left (-b x - a\right )}}{2 \, b} + \frac {{\left (3 \, e^{\left (2 \, a\right )} + 2 \, e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (2 \, c\right )}}{2 \, b {\left (e^{\left (-b x - a + 2 \, c\right )} + e^{\left (-3 \, b x - a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.62, size = 173, normalized size = 3.84 \[ \frac {{\mathrm {e}}^{a+b\,x}}{2\,b}+\frac {{\mathrm {e}}^{-a-b\,x}}{2\,b}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-a}\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{b\,x}\,\left (\sqrt {b^2}-{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,c}\,\sqrt {b^2}\right )}{b\,\sqrt {{\mathrm {e}}^{-2\,a}\,{\mathrm {e}}^{2\,c}\,\left ({\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,c}-2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,c}+1\right )}}\right )\,\sqrt {{\mathrm {e}}^{2\,c-2\,a}\,\left ({\mathrm {e}}^{4\,a-4\,c}-2\,{\mathrm {e}}^{2\,a-2\,c}+1\right )}}{\sqrt {b^2}}+\frac {{\mathrm {e}}^{a+b\,x}\,\left ({\mathrm {e}}^{2\,a-2\,c}+1\right )}{b\,\left ({\mathrm {e}}^{2\,a-2\,c}+{\mathrm {e}}^{2\,a+2\,b\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)*tanh(c + b*x)^2,x)

[Out]

exp(a + b*x)/(2*b) + exp(- a - b*x)/(2*b) + (atan((exp(-a)*exp(2*c)*exp(b*x)*((b^2)^(1/2) - exp(2*a)*exp(-2*c)
*(b^2)^(1/2)))/(b*(exp(-2*a)*exp(2*c)*(exp(4*a)*exp(-4*c) - 2*exp(2*a)*exp(-2*c) + 1))^(1/2)))*(exp(2*c - 2*a)
*(exp(4*a - 4*c) - 2*exp(2*a - 2*c) + 1))^(1/2))/(b^2)^(1/2) + (exp(a + b*x)*(exp(2*a - 2*c) + 1))/(b*(exp(2*a
- 2*c) + exp(2*a + 2*b*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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