3.117 \(\int \coth ^n(a+b x) \text {csch}^2(a+b x) \, dx\)

Optimal. Leaf size=20 \[ -\frac {\coth ^{n+1}(a+b x)}{b (n+1)} \]

[Out]

-coth(b*x+a)^(1+n)/b/(1+n)

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Rubi [A]  time = 0.03, antiderivative size = 20, 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} \[ -\frac {\coth ^{n+1}(a+b x)}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + b*x]^n*Csch[a + b*x]^2,x]

[Out]

-(Coth[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 \coth ^n(a+b x) \text {csch}^2(a+b x) \, dx &=\frac {i \operatorname {Subst}\left (\int (-i x)^n \, dx,x,i \coth (a+b x)\right )}{b}\\ &=-\frac {\coth ^{1+n}(a+b x)}{b (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \[ -\frac {\coth ^{n+1}(a+b x)}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + b*x]^n*Csch[a + b*x]^2,x]

[Out]

-(Coth[a + b*x]^(1 + n)/(b*(1 + n)))

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fricas [B]  time = 0.44, size = 70, normalized size = 3.50 \[ -\frac {\cosh \left (b x + a\right ) \cosh \left (n \log \left (\frac {\cosh \left (b x + a\right )}{\sinh \left (b x + a\right )}\right )\right ) + \cosh \left (b x + a\right ) \sinh \left (n \log \left (\frac {\cosh \left (b x + a\right )}{\sinh \left (b x + a\right )}\right )\right )}{{\left (b n + b\right )} \sinh \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(coth(b*x + a)^n*csch(b*x + a)^2, x)

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maple [A]  time = 0.15, size = 21, normalized size = 1.05 \[ -\frac {\coth ^{n +1}\left (b x +a \right )}{b \left (n +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^n*csch(b*x+a)^2,x)

[Out]

-coth(b*x+a)^(n+1)/b/(n+1)

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maxima [A]  time = 0.78, size = 20, normalized size = 1.00 \[ -\frac {\coth \left (b x + a\right )^{n + 1}}{b {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-coth(b*x + a)^(n + 1)/(b*(n + 1))

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mupad [B]  time = 1.48, size = 43, normalized size = 2.15 \[ -\frac {\mathrm {coth}\left (a+b\,x\right )\,{\left (\frac {{\mathrm {e}}^{2\,a+2\,b\,x}+1}{{\mathrm {e}}^{2\,a+2\,b\,x}-1}\right )}^n}{b\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^n/sinh(a + b*x)^2,x)

[Out]

-(coth(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 \[ \int \coth ^{n}{\left (a + b x \right )} \operatorname {csch}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**n*csch(b*x+a)**2,x)

[Out]

Integral(coth(a + b*x)**n*csch(a + b*x)**2, x)

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