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]

1/4*tanh(b*x+a)^4/b

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 30

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

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])

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*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ \frac {\tanh ^4(a+b x)}{4 b} \]

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)

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fricas [B]  time = 0.44, size = 208, normalized size = 13.87 \[ -\frac {2 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + b \sinh \left (b x + a\right )^{5} + 5 \, b \cosh \left (b x + a\right )^{3} + {\left (10 \, b \cosh \left (b x + a\right )^{2} + 3 \, b\right )} \sinh \left (b x + a\right )^{3} + 5 \, {\left (2 \, b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 10 \, b \cosh \left (b x + a\right ) + {\left (5 \, b \cosh \left (b x + a\right )^{4} + 9 \, b \cosh \left (b x + a\right )^{2} + 2 \, b\right )} \sinh \left (b x + a\right )} \]

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))

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giac [B]  time = 0.14, size = 37, normalized size = 2.47 \[ -\frac {2 \, {\left (e^{\left (6 \, b x + 6 \, a\right )} + e^{\left (2 \, b x + 2 \, a\right )}\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} \]

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)

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maple [B]  time = 0.14, size = 34, normalized size = 2.27 \[ \frac {-\frac {\sinh ^{2}\left (b x +a \right )}{2 \cosh \left (b x +a \right )^{4}}-\frac {1}{4 \cosh \left (b x +a \right )^{4}}}{b} \]

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

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maxima [A]  time = 0.31, size = 13, normalized size = 0.87 \[ \frac {\tanh \left (b x + a\right )^{4}}{4 \, b} \]

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

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mupad [B]  time = 0.11, size = 230, normalized size = 15.33 \[ \frac {\frac {1}{2\,b}-\frac {3\,{\mathrm {e}}^{2\,a+2\,b\,x}}{2\,b}+\frac {3\,{\mathrm {e}}^{4\,a+4\,b\,x}}{2\,b}-\frac {{\mathrm {e}}^{6\,a+6\,b\,x}}{2\,b}}{4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1}-\frac {\frac {1}{2\,b}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{b}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{2\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1}+\frac {\frac {1}{2\,b}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{2\,b}}{2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1}-\frac {1}{2\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/(2*b) - (3*exp(2*a + 2*b*x))/(2*b) + (3*exp(4*a + 4*b*x))/(2*b) - exp(6*a + 6*b*x)/(2*b))/(4*exp(2*a + 2*b*
x) + 6*exp(4*a + 4*b*x) + 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1) - (1/(2*b) - exp(2*a + 2*b*x)/b + exp(4*a
+ 4*b*x)/(2*b))/(3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1) + (1/(2*b) - exp(2*a + 2*b*x
)/(2*b))/(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1) - 1/(2*b*(exp(2*a + 2*b*x) + 1))

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sympy [A]  time = 1.53, size = 44, normalized size = 2.93 \[ \begin {cases} - \frac {\tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}}{4 b} - \frac {\operatorname {sech}^{2}{\left (a + b x \right )}}{4 b} & \text {for}\: b \neq 0 \\x \tanh ^{3}{\relax (a )} \operatorname {sech}^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

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))

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