### 3.99 $$\int \text{sech}^5(x) \tanh ^2(x) \, dx$$

Optimal. Leaf size=36 $\frac{1}{16} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tanh (x) \text{sech}^5(x)+\frac{1}{24} \tanh (x) \text{sech}^3(x)+\frac{1}{16} \tanh (x) \text{sech}(x)$

[Out]

ArcTan[Sinh[x]]/16 + (Sech[x]*Tanh[x])/16 + (Sech[x]^3*Tanh[x])/24 - (Sech[x]^5*Tanh[x])/6

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Rubi [A]  time = 0.0370977, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {2611, 3768, 3770} $\frac{1}{16} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tanh (x) \text{sech}^5(x)+\frac{1}{24} \tanh (x) \text{sech}^3(x)+\frac{1}{16} \tanh (x) \text{sech}(x)$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[x]^5*Tanh[x]^2,x]

[Out]

ArcTan[Sinh[x]]/16 + (Sech[x]*Tanh[x])/16 + (Sech[x]^3*Tanh[x])/24 - (Sech[x]^5*Tanh[x])/6

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}^5(x) \tanh ^2(x) \, dx &=-\frac{1}{6} \text{sech}^5(x) \tanh (x)+\frac{1}{6} \int \text{sech}^5(x) \, dx\\ &=\frac{1}{24} \text{sech}^3(x) \tanh (x)-\frac{1}{6} \text{sech}^5(x) \tanh (x)+\frac{1}{8} \int \text{sech}^3(x) \, dx\\ &=\frac{1}{16} \text{sech}(x) \tanh (x)+\frac{1}{24} \text{sech}^3(x) \tanh (x)-\frac{1}{6} \text{sech}^5(x) \tanh (x)+\frac{1}{16} \int \text{sech}(x) \, dx\\ &=\frac{1}{16} \tan ^{-1}(\sinh (x))+\frac{1}{16} \text{sech}(x) \tanh (x)+\frac{1}{24} \text{sech}^3(x) \tanh (x)-\frac{1}{6} \text{sech}^5(x) \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.0086348, size = 36, normalized size = 1. $\frac{1}{16} \tan ^{-1}(\sinh (x))-\frac{1}{6} \tanh (x) \text{sech}^5(x)+\frac{1}{24} \tanh (x) \text{sech}^3(x)+\frac{1}{16} \tanh (x) \text{sech}(x)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[x]^5*Tanh[x]^2,x]

[Out]

ArcTan[Sinh[x]]/16 + (Sech[x]*Tanh[x])/16 + (Sech[x]^3*Tanh[x])/24 - (Sech[x]^5*Tanh[x])/6

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Maple [A]  time = 0.012, size = 36, normalized size = 1. \begin{align*} -{\frac{\sinh \left ( x \right ) }{5\, \left ( \cosh \left ( x \right ) \right ) ^{6}}}+{\frac{\tanh \left ( x \right ) }{5} \left ({\frac{ \left ({\rm sech} \left (x\right ) \right ) ^{5}}{6}}+{\frac{5\, \left ({\rm sech} \left (x\right ) \right ) ^{3}}{24}}+{\frac{5\,{\rm sech} \left (x\right )}{16}} \right ) }+{\frac{\arctan \left ({{\rm e}^{x}} \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^5*tanh(x)^2,x)

[Out]

-1/5*sinh(x)/cosh(x)^6+1/5*(1/6*sech(x)^5+5/24*sech(x)^3+5/16*sech(x))*tanh(x)+1/8*arctan(exp(x))

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Maxima [B]  time = 1.56763, size = 115, normalized size = 3.19 \begin{align*} \frac{3 \, e^{\left (-x\right )} + 17 \, e^{\left (-3 \, x\right )} - 114 \, e^{\left (-5 \, x\right )} + 114 \, e^{\left (-7 \, x\right )} - 17 \, e^{\left (-9 \, x\right )} - 3 \, e^{\left (-11 \, x\right )}}{24 \,{\left (6 \, e^{\left (-2 \, x\right )} + 15 \, e^{\left (-4 \, x\right )} + 20 \, e^{\left (-6 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} + 6 \, e^{\left (-10 \, x\right )} + e^{\left (-12 \, x\right )} + 1\right )}} - \frac{1}{8} \, \arctan \left (e^{\left (-x\right )}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5*tanh(x)^2,x, algorithm="maxima")

[Out]

1/24*(3*e^(-x) + 17*e^(-3*x) - 114*e^(-5*x) + 114*e^(-7*x) - 17*e^(-9*x) - 3*e^(-11*x))/(6*e^(-2*x) + 15*e^(-4
*x) + 20*e^(-6*x) + 15*e^(-8*x) + 6*e^(-10*x) + e^(-12*x) + 1) - 1/8*arctan(e^(-x))

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Fricas [B]  time = 1.83163, size = 3189, normalized size = 88.58 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5*tanh(x)^2,x, algorithm="fricas")

[Out]

1/24*(3*cosh(x)^11 + 33*cosh(x)*sinh(x)^10 + 3*sinh(x)^11 + (165*cosh(x)^2 + 17)*sinh(x)^9 + 17*cosh(x)^9 + 9*
(55*cosh(x)^3 + 17*cosh(x))*sinh(x)^8 + 6*(165*cosh(x)^4 + 102*cosh(x)^2 - 19)*sinh(x)^7 - 114*cosh(x)^7 + 42*
(33*cosh(x)^5 + 34*cosh(x)^3 - 19*cosh(x))*sinh(x)^6 + 6*(231*cosh(x)^6 + 357*cosh(x)^4 - 399*cosh(x)^2 + 19)*
sinh(x)^5 + 114*cosh(x)^5 + 6*(165*cosh(x)^7 + 357*cosh(x)^5 - 665*cosh(x)^3 + 95*cosh(x))*sinh(x)^4 + (495*co
sh(x)^8 + 1428*cosh(x)^6 - 3990*cosh(x)^4 + 1140*cosh(x)^2 - 17)*sinh(x)^3 - 17*cosh(x)^3 + 3*(55*cosh(x)^9 +
204*cosh(x)^7 - 798*cosh(x)^5 + 380*cosh(x)^3 - 17*cosh(x))*sinh(x)^2 + 3*(cosh(x)^12 + 12*cosh(x)*sinh(x)^11
+ sinh(x)^12 + 6*(11*cosh(x)^2 + 1)*sinh(x)^10 + 6*cosh(x)^10 + 20*(11*cosh(x)^3 + 3*cosh(x))*sinh(x)^9 + 15*(
33*cosh(x)^4 + 18*cosh(x)^2 + 1)*sinh(x)^8 + 15*cosh(x)^8 + 24*(33*cosh(x)^5 + 30*cosh(x)^3 + 5*cosh(x))*sinh(
x)^7 + 4*(231*cosh(x)^6 + 315*cosh(x)^4 + 105*cosh(x)^2 + 5)*sinh(x)^6 + 20*cosh(x)^6 + 24*(33*cosh(x)^7 + 63*
cosh(x)^5 + 35*cosh(x)^3 + 5*cosh(x))*sinh(x)^5 + 15*(33*cosh(x)^8 + 84*cosh(x)^6 + 70*cosh(x)^4 + 20*cosh(x)^
2 + 1)*sinh(x)^4 + 15*cosh(x)^4 + 20*(11*cosh(x)^9 + 36*cosh(x)^7 + 42*cosh(x)^5 + 20*cosh(x)^3 + 3*cosh(x))*s
inh(x)^3 + 6*(11*cosh(x)^10 + 45*cosh(x)^8 + 70*cosh(x)^6 + 50*cosh(x)^4 + 15*cosh(x)^2 + 1)*sinh(x)^2 + 6*cos
h(x)^2 + 12*(cosh(x)^11 + 5*cosh(x)^9 + 10*cosh(x)^7 + 10*cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*sinh(x) + 1)*arct
an(cosh(x) + sinh(x)) + 3*(11*cosh(x)^10 + 51*cosh(x)^8 - 266*cosh(x)^6 + 190*cosh(x)^4 - 17*cosh(x)^2 - 1)*si
nh(x) - 3*cosh(x))/(cosh(x)^12 + 12*cosh(x)*sinh(x)^11 + sinh(x)^12 + 6*(11*cosh(x)^2 + 1)*sinh(x)^10 + 6*cosh
(x)^10 + 20*(11*cosh(x)^3 + 3*cosh(x))*sinh(x)^9 + 15*(33*cosh(x)^4 + 18*cosh(x)^2 + 1)*sinh(x)^8 + 15*cosh(x)
^8 + 24*(33*cosh(x)^5 + 30*cosh(x)^3 + 5*cosh(x))*sinh(x)^7 + 4*(231*cosh(x)^6 + 315*cosh(x)^4 + 105*cosh(x)^2
+ 5)*sinh(x)^6 + 20*cosh(x)^6 + 24*(33*cosh(x)^7 + 63*cosh(x)^5 + 35*cosh(x)^3 + 5*cosh(x))*sinh(x)^5 + 15*(3
3*cosh(x)^8 + 84*cosh(x)^6 + 70*cosh(x)^4 + 20*cosh(x)^2 + 1)*sinh(x)^4 + 15*cosh(x)^4 + 20*(11*cosh(x)^9 + 36
*cosh(x)^7 + 42*cosh(x)^5 + 20*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 6*(11*cosh(x)^10 + 45*cosh(x)^8 + 70*cosh(x)
^6 + 50*cosh(x)^4 + 15*cosh(x)^2 + 1)*sinh(x)^2 + 6*cosh(x)^2 + 12*(cosh(x)^11 + 5*cosh(x)^9 + 10*cosh(x)^7 +
10*cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh ^{2}{\left (x \right )} \operatorname{sech}^{5}{\left (x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**5*tanh(x)**2,x)

[Out]

Integral(tanh(x)**2*sech(x)**5, x)

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Giac [B]  time = 1.20283, size = 99, normalized size = 2.75 \begin{align*} \frac{1}{32} \, \pi - \frac{3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{5} + 32 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 48 \, e^{\left (-x\right )} + 48 \, e^{x}}{24 \,{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{3}} + \frac{1}{16} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5*tanh(x)^2,x, algorithm="giac")

[Out]

1/32*pi - 1/24*(3*(e^(-x) - e^x)^5 + 32*(e^(-x) - e^x)^3 - 48*e^(-x) + 48*e^x)/((e^(-x) - e^x)^2 + 4)^3 + 1/16
*arctan(1/2*(e^(2*x) - 1)*e^(-x))