### 3.39 $$\int \text{csch}^4(a+b x) \text{sech}(a+b x) \, dx$$

Optimal. Leaf size=37 $-\frac{\text{csch}^3(a+b x)}{3 b}+\frac{\text{csch}(a+b x)}{b}+\frac{\tan ^{-1}(\sinh (a+b x))}{b}$

[Out]

ArcTan[Sinh[a + b*x]]/b + Csch[a + b*x]/b - Csch[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0270247, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2621, 302, 207} $-\frac{\text{csch}^3(a+b x)}{3 b}+\frac{\text{csch}(a+b x)}{b}+\frac{\tan ^{-1}(\sinh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]^4*Sech[a + b*x],x]

[Out]

ArcTan[Sinh[a + b*x]]/b + Csch[a + b*x]/b - Csch[a + b*x]^3/(3*b)

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \text{csch}^4(a+b x) \text{sech}(a+b x) \, dx &=\frac{i \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{b}\\ &=\frac{i \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,-i \text{csch}(a+b x)\right )}{b}\\ &=\frac{\text{csch}(a+b x)}{b}-\frac{\text{csch}^3(a+b x)}{3 b}+\frac{i \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{b}\\ &=\frac{\tan ^{-1}(\sinh (a+b x))}{b}+\frac{\text{csch}(a+b x)}{b}-\frac{\text{csch}^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [C]  time = 0.0169181, size = 33, normalized size = 0.89 $-\frac{\text{csch}^3(a+b x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\sinh ^2(a+b x)\right )}{3 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[a + b*x]^4*Sech[a + b*x],x]

[Out]

-(Csch[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Sinh[a + b*x]^2])/(3*b)

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Maple [A]  time = 0.017, size = 39, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,b \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{1}{b\sinh \left ( bx+a \right ) }}+2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^4*sech(b*x+a),x)

[Out]

-1/3/b/sinh(b*x+a)^3+1/b/sinh(b*x+a)+2*arctan(exp(b*x+a))/b

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Maxima [B]  time = 1.78521, size = 122, normalized size = 3.3 \begin{align*} -\frac{2 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac{2 \,{\left (3 \, e^{\left (-b x - a\right )} - 10 \, e^{\left (-3 \, b x - 3 \, a\right )} + 3 \, e^{\left (-5 \, b x - 5 \, a\right )}\right )}}{3 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} - 3 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^4*sech(b*x+a),x, algorithm="maxima")

[Out]

-2*arctan(e^(-b*x - a))/b - 2/3*(3*e^(-b*x - a) - 10*e^(-3*b*x - 3*a) + 3*e^(-5*b*x - 5*a))/(b*(3*e^(-2*b*x -
2*a) - 3*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a) - 1))

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Fricas [B]  time = 2.33626, size = 1423, normalized size = 38.46 \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)^4*sech(b*x+a),x, algorithm="fricas")

[Out]

2/3*(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 + 3*sinh(b*x + a)^5 + 10*(3*cosh(b*x + a)^2 - 1)*sin
h(b*x + a)^3 - 10*cosh(b*x + a)^3 + 30*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^6
+ 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(b*x +
a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1
)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 - 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)
- 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + 3*(5*cosh(b*x + a)^4 - 10*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + 3*
cosh(b*x + a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 - 3*b*cosh(b*x + a)^
4 + 3*(5*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^3
+ 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b*x + a)^4 - 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a)
^5 - 2*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) - b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}^{4}{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**4*sech(b*x+a),x)

[Out]

Integral(csch(a + b*x)**4*sech(a + b*x), x)

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Giac [B]  time = 1.25549, size = 111, normalized size = 3. \begin{align*} \frac{\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{2 \, b} + \frac{2 \,{\left (3 \,{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} - 4\right )}}{3 \, b{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^4*sech(b*x+a),x, algorithm="giac")

[Out]

1/2*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b + 2/3*(3*(e^(b*x + a) - e^(-b*x - a))^2 - 4)/(b*
(e^(b*x + a) - e^(-b*x - a))^3)