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

Optimal. Leaf size=19 \[ \frac {\tanh ^{n+1}(a+b x)}{b (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 19, 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, 32} \[ \frac {\tanh ^{n+1}(a+b x)}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tanh[a + b*x]^(1 + n)/(b*(1 + n))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[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 ^n(a+b x) \, dx &=-\frac {i \operatorname {Subst}\left (\int (-i x)^n \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {\tanh ^{1+n}(a+b x)}{b (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.00 \[ \frac {\tanh ^{n+1}(a+b x)}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tanh[a + b*x]^(1 + n)/(b*(1 + n))

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fricas [B]  time = 0.42, size = 69, normalized size = 3.63 \[ \frac {\cosh \left (n \log \left (\frac {\sinh \left (b x + a\right )}{\cosh \left (b x + a\right )}\right )\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right ) \sinh \left (n \log \left (\frac {\sinh \left (b x + a\right )}{\cosh \left (b x + a\right )}\right )\right )}{{\left (b n + b\right )} \cosh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(n*log(sinh(b*x + a)/cosh(b*x + a)))*sinh(b*x + a) + sinh(b*x + a)*sinh(n*log(sinh(b*x + a)/cosh(b*x + a)
)))/((b*n + b)*cosh(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh \left (b x + a\right )^{n} \operatorname {sech}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 20, normalized size = 1.05 \[ \frac {\tanh ^{n +1}\left (b x +a \right )}{b \left (n +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 19, normalized size = 1.00 \[ \frac {\tanh \left (b x + a\right )^{n + 1}}{b {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tanh(b*x + a)^(n + 1)/(b*(n + 1))

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mupad [B]  time = 1.53, size = 42, normalized size = 2.21 \[ \frac {\mathrm {tanh}\left (a+b\,x\right )\,{\left (\frac {{\mathrm {e}}^{2\,a+2\,b\,x}-1}{{\mathrm {e}}^{2\,a+2\,b\,x}+1}\right )}^n}{b\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tanh(a + b*x)*((exp(2*a + 2*b*x) - 1)/(exp(2*a + 2*b*x) + 1))^n)/(b*(n + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh ^{n}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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