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

Optimal. Leaf size=15 \[ \frac{\tanh ^4(a+b x)}{4 b} \]

[Out]

Tanh[a + b*x]^4/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0300007, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2607, 30} \[ \frac{\tanh ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tanh[a + b*x]^4/(4*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0048121, size = 15, normalized size = 1. \[ \frac{\tanh ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tanh[a + b*x]^4/(4*b)

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 42, normalized size = 2.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/4*sinh(b*x+a)^2/cosh(b*x+a)^4+1/4*sinh(b*x+a)^2/cosh(b*x+a)^2)

________________________________________________________________________________________

Maxima [A]  time = 1.07871, size = 18, normalized size = 1.2 \begin{align*} \frac{\tanh \left (b x + a\right )^{4}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*tanh(b*x + a)^4/b

________________________________________________________________________________________

Fricas [B]  time = 1.80813, size = 572, normalized size = 38.13 \begin{align*} -\frac{2 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} +{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + b \sinh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right )^{3} +{\left (10 \, b \cosh \left (b x + a\right )^{2} + 3 \, b\right )} \sinh \left (b x + a\right )^{3} + 5 \,{\left (2 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 10 \, b \cosh \left (b x + a\right ) +{\left (5 \, b \cosh \left (b x + a\right )^{4} + 9 \, b \cosh \left (b x + a\right )^{2} + 2 \, b\right )} \sinh \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a
) + cosh(b*x + a))/(b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)*sinh(b*x + a)^4 + b*sinh(b*x + a)^5 + 5*b*cosh(b*x +
a)^3 + (10*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^3 + 5*(2*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x +
a)^2 + 10*b*cosh(b*x + a) + (5*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 + 2*b)*sinh(b*x + a))

________________________________________________________________________________________

Sympy [A]  time = 2.38937, size = 44, normalized size = 2.93 \begin{align*} \begin{cases} - \frac{\tanh ^{2}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}}{4 b} - \frac{\operatorname{sech}^{2}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x \tanh ^{3}{\left (a \right )} \operatorname{sech}^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-tanh(a + b*x)**2*sech(a + b*x)**2/(4*b) - sech(a + b*x)**2/(4*b), Ne(b, 0)), (x*tanh(a)**3*sech(a)
**2, True))

________________________________________________________________________________________

Giac [B]  time = 1.25667, size = 50, normalized size = 3.33 \begin{align*} -\frac{2 \,{\left (e^{\left (6 \, b x + 6 \, a\right )} + e^{\left (2 \, b x + 2 \, a\right )}\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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