### 3.145 $$\int \sinh (a+b x) \tanh ^3(c+b x) \, dx$$

Optimal. Leaf size=72 $\frac{\sinh (a-c) \text{sech}(b x+c)}{b}-\frac{3 \cosh (a-c) \tan ^{-1}(\sinh (b x+c))}{2 b}+\frac{\cosh (a-c) \tanh (b x+c) \text{sech}(b x+c)}{2 b}+\frac{\sinh (a+b x)}{b}$

[Out]

(-3*ArcTan[Sinh[c + b*x]]*Cosh[a - c])/(2*b) + (Sech[c + b*x]*Sinh[a - c])/b + Sinh[a + b*x]/b + (Cosh[a - c]*
Sech[c + b*x]*Tanh[c + b*x])/(2*b)

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Rubi [A]  time = 0.082597, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.467, Rules used = {5620, 5623, 2637, 3770, 2606, 8, 2611} $\frac{\sinh (a-c) \text{sech}(b x+c)}{b}-\frac{3 \cosh (a-c) \tan ^{-1}(\sinh (b x+c))}{2 b}+\frac{\cosh (a-c) \tanh (b x+c) \text{sech}(b x+c)}{2 b}+\frac{\sinh (a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]*Tanh[c + b*x]^3,x]

[Out]

(-3*ArcTan[Sinh[c + b*x]]*Cosh[a - c])/(2*b) + (Sech[c + b*x]*Sinh[a - c])/b + Sinh[a + b*x]/b + (Cosh[a - c]*
Sech[c + b*x]*Tanh[c + b*x])/(2*b)

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]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 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 8

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

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rubi steps

\begin{align*} \int \sinh (a+b x) \tanh ^3(c+b x) \, dx &=-\left (\cosh (a-c) \int \text{sech}(c+b x) \tanh ^2(c+b x) \, dx\right )+\int \cosh (a+b x) \tanh ^2(c+b x) \, dx\\ &=\frac{\cosh (a-c) \text{sech}(c+b x) \tanh (c+b x)}{2 b}-\frac{1}{2} \cosh (a-c) \int \text{sech}(c+b x) \, dx-\sinh (a-c) \int \text{sech}(c+b x) \tanh (c+b x) \, dx+\int \sinh (a+b x) \tanh (c+b x) \, dx\\ &=-\frac{\tan ^{-1}(\sinh (c+b x)) \cosh (a-c)}{2 b}+\frac{\cosh (a-c) \text{sech}(c+b x) \tanh (c+b x)}{2 b}-\cosh (a-c) \int \text{sech}(c+b x) \, dx+\frac{\sinh (a-c) \operatorname{Subst}(\int 1 \, dx,x,\text{sech}(c+b x))}{b}+\int \cosh (a+b x) \, dx\\ &=-\frac{3 \tan ^{-1}(\sinh (c+b x)) \cosh (a-c)}{2 b}+\frac{\text{sech}(c+b x) \sinh (a-c)}{b}+\frac{\sinh (a+b x)}{b}+\frac{\cosh (a-c) \text{sech}(c+b x) \tanh (c+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.334092, size = 70, normalized size = 0.97 $\frac{\text{sech}^2(b x+c) (2 \sinh (a-b x-2 c)+\sinh (a+3 b x+2 c)+5 \sinh (a+b x))-12 \cosh (a-c) \tan ^{-1}\left (\cosh (c) \tanh \left (\frac{b x}{2}\right )+\sinh (c)\right )}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]*Tanh[c + b*x]^3,x]

[Out]

(-12*ArcTan[Sinh[c] + Cosh[c]*Tanh[(b*x)/2]]*Cosh[a - c] + Sech[c + b*x]^2*(2*Sinh[a - 2*c - b*x] + 5*Sinh[a +
b*x] + Sinh[a + 2*c + 3*b*x]))/(4*b)

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Maple [C]  time = 0.093, size = 240, normalized size = 3.3 \begin{align*}{\frac{{{\rm e}^{bx+a}}}{2\,b}}-{\frac{{{\rm e}^{-bx-a}}}{2\,b}}+{\frac{{{\rm e}^{bx+a}} \left ( 3\,{{\rm e}^{2\,bx+4\,a+2\,c}}-{{\rm e}^{2\,bx+2\,a+4\,c}}+{{\rm e}^{4\,a}}-3\,{{\rm e}^{2\,a+2\,c}} \right ) }{2\,b \left ({{\rm e}^{2\,bx+2\,a+2\,c}}+{{\rm e}^{2\,a}} \right ) ^{2}}}+{\frac{{\frac{3\,i}{4}}\ln \left ({{\rm e}^{bx+a}}-i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,a}}}{b}}+{\frac{{\frac{3\,i}{4}}\ln \left ({{\rm e}^{bx+a}}-i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,c}}}{b}}-{\frac{{\frac{3\,i}{4}}\ln \left ({{\rm e}^{bx+a}}+i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,a}}}{b}}-{\frac{{\frac{3\,i}{4}}\ln \left ({{\rm e}^{bx+a}}+i{{\rm e}^{a-c}} \right ){{\rm e}^{-a-c}}{{\rm e}^{2\,c}}}{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)*tanh(b*x+c)^3,x)

[Out]

1/2*exp(b*x+a)/b-1/2*exp(-b*x-a)/b+1/2*exp(b*x+a)*(3*exp(2*b*x+4*a+2*c)-exp(2*b*x+2*a+4*c)+exp(4*a)-3*exp(2*a+
2*c))/b/(exp(2*b*x+2*a+2*c)+exp(2*a))^2+3/4*I/b*ln(exp(b*x+a)-I*exp(a-c))*exp(-a-c)*exp(2*a)+3/4*I/b*ln(exp(b*
x+a)-I*exp(a-c))*exp(-a-c)*exp(2*c)-3/4*I/b*ln(exp(b*x+a)+I*exp(a-c))*exp(-a-c)*exp(2*a)-3/4*I/b*ln(exp(b*x+a)
+I*exp(a-c))*exp(-a-c)*exp(2*c)

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Maxima [B]  time = 1.78294, size = 201, normalized size = 2.79 \begin{align*} \frac{3 \,{\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 )}}{2 \, b} - \frac{e^{\left (-b x - a\right )}}{2 \, b} + \frac{{\left (5 \, e^{\left (2 \, a + 2 \, c\right )} - e^{\left (4 \, c\right )}\right )} e^{\left (-2 \, b x - 2 \, a\right )} +{\left (2 \, e^{\left (4 \, a\right )} - 3 \, e^{\left (2 \, a + 2 \, c\right )}\right )} e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (4 \, c\right )}}{2 \, b{\left (e^{\left (-b x - a + 4 \, c\right )} + 2 \, e^{\left (-3 \, b x - a + 2 \, c\right )} + e^{\left (-5 \, b x - a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(e^(2*a) + e^(2*c))*arctan(e^(-b*x - c))*e^(-a - c)/b - 1/2*e^(-b*x - a)/b + 1/2*((5*e^(2*a + 2*c) - e^(4*
c))*e^(-2*b*x - 2*a) + (2*e^(4*a) - 3*e^(2*a + 2*c))*e^(-4*b*x - 4*a) + e^(4*c))/(b*(e^(-b*x - a + 4*c) + 2*e^
(-3*b*x - a + 2*c) + e^(-5*b*x - a)))

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Fricas [B]  time = 2.31111, size = 4772, normalized size = 66.28 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(cosh(b*x + c)^6*cosh(-a + c)^2 + (cosh(-a + c)^2 - 2*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2)*sinh(b*x
+ c)^6 + 6*(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)^5 + (5*cosh(-a + c)^2 - 2)*cosh(b*x + c)^4 + (15*cosh(b*x + c)^2*cosh(-a + c)^2 + 5*(3*c
osh(b*x + c)^2 + 1)*sinh(-a + c)^2 + 5*cosh(-a + c)^2 - 10*(3*cosh(b*x + c)^2*cosh(-a + c) + cosh(-a + c))*sin
h(-a + c) - 2)*sinh(b*x + c)^4 + 4*(5*cosh(b*x + c)^3*cosh(-a + c)^2 + 5*(cosh(b*x + c)^3 + cosh(b*x + c))*sin
h(-a + c)^2 + (5*cosh(-a + c)^2 - 2)*cosh(b*x + c) - 10*(cosh(b*x + c)^3*cosh(-a + c) + cosh(b*x + c)*cosh(-a
+ c))*sinh(-a + c))*sinh(b*x + c)^3 + (2*cosh(-a + c)^2 - 5)*cosh(b*x + c)^2 + (15*cosh(b*x + c)^4*cosh(-a + c
)^2 + 6*(5*cosh(-a + c)^2 - 2)*cosh(b*x + c)^2 + (15*cosh(b*x + c)^4 + 30*cosh(b*x + c)^2 + 2)*sinh(-a + c)^2
+ 2*cosh(-a + c)^2 - 2*(15*cosh(b*x + c)^4*cosh(-a + c) + 30*cosh(b*x + c)^2*cosh(-a + c) + 2*cosh(-a + c))*si
nh(-a + c) - 5)*sinh(b*x + c)^2 + (cosh(b*x + c)^6 + 5*cosh(b*x + c)^4 + 2*cosh(b*x + c)^2)*sinh(-a + c)^2 - 3
*((cosh(-a + c)^2 + 1)*cosh(b*x + c)^5 + (cosh(-a + c)^2 - 2*cosh(-a + c)*sinh(-a + c) + sinh(-a + c)^2 + 1)*s
inh(b*x + c)^5 - 5*(2*cosh(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)^4 + 2*(cosh(-a + c)^2 + 1)*cosh(b*x + c)^3 + 2*(5*(cosh(-a + c)^2 + 1)*cosh
(b*x + c)^2 + (5*cosh(b*x + c)^2 + 1)*sinh(-a + c)^2 + cosh(-a + c)^2 - 2*(5*cosh(b*x + c)^2*cosh(-a + c) + co
sh(-a + c))*sinh(-a + c) + 1)*sinh(b*x + c)^3 + 2*(5*(cosh(-a + c)^2 + 1)*cosh(b*x + c)^3 + (5*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*(5*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)^5 + 2*cosh(b*x + c)^3 + cosh(b*
x + c))*sinh(-a + c)^2 + (cosh(-a + c)^2 + 1)*cosh(b*x + c) + (5*(cosh(-a + c)^2 + 1)*cosh(b*x + c)^4 + 6*(cos
h(-a + c)^2 + 1)*cosh(b*x + c)^2 + (5*cosh(b*x + c)^4 + 6*cosh(b*x + c)^2 + 1)*sinh(-a + c)^2 + cosh(-a + c)^2
- 2*(5*cosh(b*x + c)^4*cosh(-a + c) + 6*cosh(b*x + c)^2*cosh(-a + c) + cosh(-a + c))*sinh(-a + c) + 1)*sinh(b
*x + c) - 2*(cosh(b*x + c)^5*cosh(-a + 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*(3*cosh(b*x + c)^5*cosh(-a + c)^2 + 2*(5*cosh(-a + c)^2 - 2
)*cosh(b*x + c)^3 + (3*cosh(b*x + c)^5 + 10*cosh(b*x + c)^3 + 2*cosh(b*x + c))*sinh(-a + c)^2 + (2*cosh(-a + c
)^2 - 5)*cosh(b*x + c) - 2*(3*cosh(b*x + c)^5*cosh(-a + c) + 10*cosh(b*x + c)^3*cosh(-a + c) + 2*cosh(b*x + c)
*cosh(-a + c))*sinh(-a + c))*sinh(b*x + c) - 2*(cosh(b*x + c)^6*cosh(-a + c) + 5*cosh(b*x + c)^4*cosh(-a + c)
+ 2*cosh(b*x + c)^2*cosh(-a + c))*sinh(-a + c) - 1)/(b*cosh(b*x + c)^5*cosh(-a + c) + (b*cosh(-a + c) - b*sinh
(-a + c))*sinh(b*x + c)^5 + 2*b*cosh(b*x + c)^3*cosh(-a + c) + 5*(b*cosh(b*x + c)*cosh(-a + c) - b*cosh(b*x +
c)*sinh(-a + c))*sinh(b*x + c)^4 + 2*(5*b*cosh(b*x + c)^2*cosh(-a + c) + b*cosh(-a + c) - (5*b*cosh(b*x + c)^2
+ b)*sinh(-a + c))*sinh(b*x + c)^3 + b*cosh(b*x + c)*cosh(-a + c) + 2*(5*b*cosh(b*x + c)^3*cosh(-a + c) + 3*b
*cosh(b*x + c)*cosh(-a + c) - (5*b*cosh(b*x + c)^3 + 3*b*cosh(b*x + c))*sinh(-a + c))*sinh(b*x + c)^2 + (5*b*c
osh(b*x + c)^4*cosh(-a + c) + 6*b*cosh(b*x + c)^2*cosh(-a + c) + b*cosh(-a + c) - (5*b*cosh(b*x + c)^4 + 6*b*c
osh(b*x + c)^2 + b)*sinh(-a + c))*sinh(b*x + c) - (b*cosh(b*x + c)^5 + 2*b*cosh(b*x + c)^3 + b*cosh(b*x + c))*
sinh(-a + c))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+c)**3,x)

[Out]

Integral(sinh(a + b*x)*tanh(b*x + c)**3, x)

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Giac [A]  time = 1.22407, size = 151, normalized size = 2.1 \begin{align*} -\frac{3 \,{\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 (3 \, e^{\left (3 \, b x + 2 \, a + 2 \, c\right )} - e^{\left (3 \, b x + 4 \, c\right )} + e^{\left (b x + 2 \, a\right )} - 3 \, e^{\left (b x + 2 \, c\right )}\right )} e^{\left (-a\right )}}{{\left (e^{\left (2 \, b x + 2 \, c\right )} + 1\right )}^{2}} - e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*(e^(2*a) + e^(2*c))*arctan(e^(b*x + c))*e^(-a - c) - (3*e^(3*b*x + 2*a + 2*c) - e^(3*b*x + 4*c) + e^(b
*x + 2*a) - 3*e^(b*x + 2*c))*e^(-a)/(e^(2*b*x + 2*c) + 1)^2 - e^(b*x + a) + e^(-b*x - a))/b