### 3.525 $$\int \text{csch}^3(a+b x) \text{sech}^3(a+b x) \, dx$$

Optimal. Leaf size=43 $\frac{\tanh ^2(a+b x)}{2 b}-\frac{\coth ^2(a+b x)}{2 b}-\frac{2 \log (\tanh (a+b x))}{b}$

[Out]

-Coth[a + b*x]^2/(2*b) - (2*Log[Tanh[a + b*x]])/b + Tanh[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0443224, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {2620, 266, 43} $\frac{\tanh ^2(a+b x)}{2 b}-\frac{\coth ^2(a+b x)}{2 b}-\frac{2 \log (\tanh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-Coth[a + b*x]^2/(2*b) - (2*Log[Tanh[a + b*x]])/b + Tanh[a + b*x]^2/(2*b)

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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \text{csch}^3(a+b x) \text{sech}^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^3} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1+x)^2}{x^2} \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}+\frac{2}{x}\right ) \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=-\frac{\coth ^2(a+b x)}{2 b}-\frac{2 \log (\tanh (a+b x))}{b}+\frac{\tanh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.011343, size = 47, normalized size = 1.09 $8 \left (-\frac{\text{csch}^2(a+b x)}{16 b}-\frac{\text{sech}^2(a+b x)}{16 b}-\frac{\log (\tanh (a+b x))}{4 b}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

8*(-Csch[a + b*x]^2/(16*b) - Log[Tanh[a + b*x]]/(4*b) - Sech[a + b*x]^2/(16*b))

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Maple [A]  time = 0.023, size = 48, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,b \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}-{\frac{1}{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}-2\,{\frac{\ln \left ( \tanh \left ( bx+a \right ) \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.56957, size = 138, normalized size = 3.21 \begin{align*} -\frac{2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac{2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac{2 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac{4 \,{\left (e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}}{b{\left (2 \, e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(e^(-b*x - a) + 1)/b - 2*log(e^(-b*x - a) - 1)/b + 2*log(e^(-2*b*x - 2*a) + 1)/b + 4*(e^(-2*b*x - 2*a) +
e^(-6*b*x - 6*a))/(b*(2*e^(-4*b*x - 4*a) - e^(-8*b*x - 8*a) - 1))

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Fricas [B]  time = 1.84967, size = 2102, normalized size = 48.88 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*cosh(b*x + a)^6 + 40*cosh(b*x + a)^3*sinh(b*x + a)^3 + 30*cosh(b*x + a)^2*sinh(b*x + a)^4 + 12*cosh(b*x
+ a)*sinh(b*x + a)^5 + 2*sinh(b*x + a)^6 + 2*(15*cosh(b*x + a)^4 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 - (c
osh(b*x + a)^8 + 56*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*cosh(b*x + a)*sin
h(b*x + a)^7 + sinh(b*x + a)^8 + 2*(35*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 8*(7*cosh(b*
x + a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(cos
h(b*x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + (cos
h(b*x + a)^8 + 56*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*cosh(b*x + a)*sinh(
b*x + a)^7 + sinh(b*x + a)^8 + 2*(35*cosh(b*x + a)^4 - 1)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 8*(7*cosh(b*x
+ a)^5 - cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 3*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(cosh(
b*x + a)^7 - cosh(b*x + a)^3)*sinh(b*x + a) + 1)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(3*c
osh(b*x + a)^5 + cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^8 + 56*b*cosh(b*x + a)^3*sinh(b*x + a)^5 + 28*
b*cosh(b*x + a)^2*sinh(b*x + a)^6 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 - 2*b*cosh(b*x + a)^
4 + 2*(35*b*cosh(b*x + a)^4 - b)*sinh(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 - b*cosh(b*x + a))*sinh(b*x + a)^3 +
4*(7*b*cosh(b*x + a)^6 - 3*b*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 - b*cosh(b*x + a)^3)*sin
h(b*x + a) + b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}^{3}{\left (a + b x \right )} \operatorname{sech}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**3*sech(b*x+a)**3,x)

[Out]

Integral(csch(a + b*x)**3*sech(a + b*x)**3, x)

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Giac [B]  time = 1.13687, size = 136, normalized size = 3.16 \begin{align*} \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right )}{b} - \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{b} - \frac{4 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}}{{\left ({\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 4\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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