### 3.95 $$\int \text{sech}^3(a+b x) \tanh ^2(a+b x) \, dx$$

Optimal. Leaf size=55 $\frac{\tan ^{-1}(\sinh (a+b x))}{8 b}-\frac{\tanh (a+b x) \text{sech}^3(a+b x)}{4 b}+\frac{\tanh (a+b x) \text{sech}(a+b x)}{8 b}$

[Out]

ArcTan[Sinh[a + b*x]]/(8*b) + (Sech[a + b*x]*Tanh[a + b*x])/(8*b) - (Sech[a + b*x]^3*Tanh[a + b*x])/(4*b)

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Rubi [A]  time = 0.0451404, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {2611, 3768, 3770} $\frac{\tan ^{-1}(\sinh (a+b x))}{8 b}-\frac{\tanh (a+b x) \text{sech}^3(a+b x)}{4 b}+\frac{\tanh (a+b x) \text{sech}(a+b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]^3*Tanh[a + b*x]^2,x]

[Out]

ArcTan[Sinh[a + b*x]]/(8*b) + (Sech[a + b*x]*Tanh[a + b*x])/(8*b) - (Sech[a + b*x]^3*Tanh[a + b*x])/(4*b)

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]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \text{sech}^3(a+b x) \tanh ^2(a+b x) \, dx &=-\frac{\text{sech}^3(a+b x) \tanh (a+b x)}{4 b}+\frac{1}{4} \int \text{sech}^3(a+b x) \, dx\\ &=\frac{\text{sech}(a+b x) \tanh (a+b x)}{8 b}-\frac{\text{sech}^3(a+b x) \tanh (a+b x)}{4 b}+\frac{1}{8} \int \text{sech}(a+b x) \, dx\\ &=\frac{\tan ^{-1}(\sinh (a+b x))}{8 b}+\frac{\text{sech}(a+b x) \tanh (a+b x)}{8 b}-\frac{\text{sech}^3(a+b x) \tanh (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0201913, size = 55, normalized size = 1. $\frac{\tan ^{-1}(\sinh (a+b x))}{8 b}-\frac{\tanh (a+b x) \text{sech}^3(a+b x)}{4 b}+\frac{\tanh (a+b x) \text{sech}(a+b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[a + b*x]^3*Tanh[a + b*x]^2,x]

[Out]

ArcTan[Sinh[a + b*x]]/(8*b) + (Sech[a + b*x]*Tanh[a + b*x])/(8*b) - (Sech[a + b*x]^3*Tanh[a + b*x])/(4*b)

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Maple [A]  time = 0.017, size = 69, normalized size = 1.3 \begin{align*} -{\frac{\sinh \left ( bx+a \right ) }{3\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}}+{\frac{ \left ({\rm sech} \left (bx+a\right ) \right ) ^{3}\tanh \left ( bx+a \right ) }{12\,b}}+{\frac{{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{8\,b}}+{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{4\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^3*tanh(b*x+a)^2,x)

[Out]

-1/3/b*sinh(b*x+a)/cosh(b*x+a)^4+1/12*sech(b*x+a)^3*tanh(b*x+a)/b+1/8*sech(b*x+a)*tanh(b*x+a)/b+1/4*arctan(exp
(b*x+a))/b

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Maxima [B]  time = 1.8487, size = 149, normalized size = 2.71 \begin{align*} -\frac{\arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} + \frac{e^{\left (-b x - a\right )} - 7 \, e^{\left (-3 \, b x - 3 \, a\right )} + 7 \, e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*arctan(e^(-b*x - a))/b + 1/4*(e^(-b*x - a) - 7*e^(-3*b*x - 3*a) + 7*e^(-5*b*x - 5*a) - e^(-7*b*x - 7*a))/
(b*(4*e^(-2*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))

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Fricas [B]  time = 1.87445, size = 2236, normalized size = 40.65 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(cosh(b*x + a)^7 + 7*cosh(b*x + a)*sinh(b*x + a)^6 + sinh(b*x + a)^7 + 7*(3*cosh(b*x + a)^2 - 1)*sinh(b*x
+ a)^5 - 7*cosh(b*x + a)^5 + 35*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^4 + 7*(5*cosh(b*x + a)^4 - 10*
cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 7*cosh(b*x + a)^3 + 7*(3*cosh(b*x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(
b*x + a))*sinh(b*x + a)^2 + (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b
*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^5 +
2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 +
10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 + 15*cosh(b*x + a)^4 + 9*cosh(b*x
+ a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 + 3*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3
+ cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (7*cosh(b*x + a)^6 - 35*cosh(b*x +
a)^4 + 21*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x
+ a)^7 + b*sinh(b*x + a)^8 + 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^6 + 8*(7*b*cosh(
b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 + 30*b*cosh(b*
x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x
+ a)^3 + 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*x + a)^6 + 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 + b)*sinh(b
*x + a)^2 + 8*(b*cosh(b*x + a)^7 + 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)
+ b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh ^{2}{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**3*tanh(b*x+a)**2,x)

[Out]

Integral(tanh(a + b*x)**2*sech(a + b*x)**3, x)

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Giac [A]  time = 1.21273, size = 90, normalized size = 1.64 \begin{align*} \frac{\frac{e^{\left (7 \, b x + 7 \, a\right )} - 7 \, e^{\left (5 \, b x + 5 \, a\right )} + 7 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} + \arctan \left (e^{\left (b x + a\right )}\right )}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*tanh(b*x+a)^2,x, algorithm="giac")

[Out]

1/4*((e^(7*b*x + 7*a) - 7*e^(5*b*x + 5*a) + 7*e^(3*b*x + 3*a) - e^(b*x + a))/(e^(2*b*x + 2*a) + 1)^4 + arctan(
e^(b*x + a)))/b