3.30 \(\int \text {csch}^2(a+b x) \text {sech}^2(a+b x) \, dx\)

Optimal. Leaf size=23 \[ -\frac {\tanh (a+b x)}{b}-\frac {\coth (a+b x)}{b} \]

[Out]

-coth(b*x+a)/b-tanh(b*x+a)/b

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 1.00, 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 = {2620, 14} \[ -\frac {\tanh (a+b x)}{b}-\frac {\coth (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Coth[a + b*x]/b) - Tanh[a + b*x]/b

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]]

Rule 2620

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 0.57 \[ -\frac {2 \coth (2 (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Coth[2*(a + b*x)])/b

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fricas [B]  time = 0.41, size = 81, normalized size = 3.52 \[ -\frac {4}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b*cosh(b*x
+ a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 - b)

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giac [A]  time = 0.15, size = 18, normalized size = 0.78 \[ -\frac {4}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(b*(e^(4*b*x + 4*a) - 1))

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maple [A]  time = 0.33, size = 32, normalized size = 1.39 \[ \frac {-\frac {1}{\sinh \left (b x +a \right ) \cosh \left (b x +a \right )}-2 \tanh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 18, normalized size = 0.78 \[ \frac {4}{b {\left (e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/(b*(e^(-4*b*x - 4*a) - 1))

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mupad [B]  time = 0.08, size = 18, normalized size = 0.78 \[ -\frac {4}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)^2*sinh(a + b*x)^2),x)

[Out]

-4/(b*(exp(4*a + 4*b*x) - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**2*sech(b*x+a)**2,x)

[Out]

Integral(csch(a + b*x)**2*sech(a + b*x)**2, x)

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