### 3.90 $$\int \text{sech}^4(a+b x) \tanh ^2(a+b x) \, dx$$

Optimal. Leaf size=31 $\frac{\tanh ^3(a+b x)}{3 b}-\frac{\tanh ^5(a+b x)}{5 b}$

[Out]

Tanh[a + b*x]^3/(3*b) - Tanh[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.032713, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {2607, 14} $\frac{\tanh ^3(a+b x)}{3 b}-\frac{\tanh ^5(a+b x)}{5 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[a + b*x]^4*Tanh[a + b*x]^2,x]

[Out]

Tanh[a + b*x]^3/(3*b) - Tanh[a + b*x]^5/(5*b)

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \text{sech}^4(a+b x) \tanh ^2(a+b x) \, dx &=\frac{i \operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{i \operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac{\tanh ^3(a+b x)}{3 b}-\frac{\tanh ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0407975, size = 56, normalized size = 1.81 $\frac{2 \tanh (a+b x)}{15 b}-\frac{\tanh (a+b x) \text{sech}^4(a+b x)}{5 b}+\frac{\tanh (a+b x) \text{sech}^2(a+b x)}{15 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[a + b*x]^4*Tanh[a + b*x]^2,x]

[Out]

(2*Tanh[a + b*x])/(15*b) + (Sech[a + b*x]^2*Tanh[a + b*x])/(15*b) - (Sech[a + b*x]^4*Tanh[a + b*x])/(5*b)

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Maple [A]  time = 0.019, size = 52, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ( -{\frac{\sinh \left ( bx+a \right ) }{4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{5}}}+{\frac{\tanh \left ( bx+a \right ) }{4} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (bx+a\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (bx+a\right ) \right ) ^{2}}{15}} \right ) } \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^4*tanh(b*x+a)^2,x)

[Out]

1/b*(-1/4*sinh(b*x+a)/cosh(b*x+a)^5+1/4*(8/15+1/5*sech(b*x+a)^4+4/15*sech(b*x+a)^2)*tanh(b*x+a))

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Maxima [B]  time = 1.12916, size = 373, normalized size = 12.03 \begin{align*} \frac{4 \, e^{\left (-2 \, b x - 2 \, a\right )}}{3 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} - \frac{4 \, e^{\left (-4 \, b x - 4 \, a\right )}}{3 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac{4 \, e^{\left (-6 \, b x - 6 \, a\right )}}{b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac{4}{15 \, b{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

4/3*e^(-2*b*x - 2*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) +
e^(-10*b*x - 10*a) + 1)) - 4/3*e^(-4*b*x - 4*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x -
6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1)) + 4*e^(-6*b*x - 6*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b
*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1)) + 4/15/(b*(5*e^(-2*b*x - 2*a)
+ 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1))

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Fricas [B]  time = 1.85278, size = 836, normalized size = 26.97 \begin{align*} -\frac{8 \,{\left (8 \, \cosh \left (b x + a\right )^{3} + 24 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 7 \, \sinh \left (b x + a\right )^{3} +{\left (21 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )\right )}}{15 \,{\left (b \cosh \left (b x + a\right )^{7} + 7 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + b \sinh \left (b x + a\right )^{7} + 5 \, b \cosh \left (b x + a\right )^{5} +{\left (21 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )^{5} + 5 \,{\left (7 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 11 \, b \cosh \left (b x + a\right )^{3} +{\left (35 \, b \cosh \left (b x + a\right )^{4} + 50 \, b \cosh \left (b x + a\right )^{2} + 9 \, b\right )} \sinh \left (b x + a\right )^{3} +{\left (21 \, b \cosh \left (b x + a\right )^{5} + 50 \, b \cosh \left (b x + a\right )^{3} + 33 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 15 \, b \cosh \left (b x + a\right ) +{\left (7 \, b \cosh \left (b x + a\right )^{6} + 25 \, b \cosh \left (b x + a\right )^{4} + 27 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

-8/15*(8*cosh(b*x + a)^3 + 24*cosh(b*x + a)*sinh(b*x + a)^2 + 7*sinh(b*x + a)^3 + (21*cosh(b*x + a)^2 - 5)*sin
h(b*x + a))/(b*cosh(b*x + a)^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 + 5*b*cosh(b*x + a)^5 +
(21*b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + a))*sinh(b*x + a)^4 +
11*b*cosh(b*x + a)^3 + (35*b*cosh(b*x + a)^4 + 50*b*cosh(b*x + a)^2 + 9*b)*sinh(b*x + a)^3 + (21*b*cosh(b*x +
a)^5 + 50*b*cosh(b*x + a)^3 + 33*b*cosh(b*x + a))*sinh(b*x + a)^2 + 15*b*cosh(b*x + a) + (7*b*cosh(b*x + a)^6
+ 25*b*cosh(b*x + a)^4 + 27*b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a))

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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}^{4}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**4*tanh(b*x+a)**2,x)

[Out]

Integral(tanh(a + b*x)**2*sech(a + b*x)**4, x)

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Giac [A]  time = 1.25431, size = 72, normalized size = 2.32 \begin{align*} -\frac{4 \,{\left (15 \, e^{\left (6 \, b x + 6 \, a\right )} - 5 \, e^{\left (4 \, b x + 4 \, a\right )} + 5 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{15 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="giac")

[Out]

-4/15*(15*e^(6*b*x + 6*a) - 5*e^(4*b*x + 4*a) + 5*e^(2*b*x + 2*a) + 1)/(b*(e^(2*b*x + 2*a) + 1)^5)