### 3.33 $$\int \text{csch}^2(a+b x) \text{sech}^5(a+b x) \, dx$$

Optimal. Leaf size=70 $-\frac{15 \text{csch}(a+b x)}{8 b}-\frac{15 \tan ^{-1}(\sinh (a+b x))}{8 b}+\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}+\frac{5 \text{csch}(a+b x) \text{sech}^2(a+b x)}{8 b}$

[Out]

(-15*ArcTan[Sinh[a + b*x]])/(8*b) - (15*Csch[a + b*x])/(8*b) + (5*Csch[a + b*x]*Sech[a + b*x]^2)/(8*b) + (Csch
[a + b*x]*Sech[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0427769, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {2621, 288, 321, 207} $-\frac{15 \text{csch}(a+b x)}{8 b}-\frac{15 \tan ^{-1}(\sinh (a+b x))}{8 b}+\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}+\frac{5 \text{csch}(a+b x) \text{sech}^2(a+b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]^2*Sech[a + b*x]^5,x]

[Out]

(-15*ArcTan[Sinh[a + b*x]])/(8*b) - (15*Csch[a + b*x])/(8*b) + (5*Csch[a + b*x]*Sech[a + b*x]^2)/(8*b) + (Csch
[a + b*x]*Sech[a + b*x]^4)/(4*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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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}^2(a+b x) \text{sech}^5(a+b x) \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,-i \text{csch}(a+b x)\right )}{b}\\ &=\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,-i \text{csch}(a+b x)\right )}{4 b}\\ &=\frac{5 \text{csch}(a+b x) \text{sech}^2(a+b x)}{8 b}+\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}-\frac{(15 i) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{8 b}\\ &=-\frac{15 \text{csch}(a+b x)}{8 b}+\frac{5 \text{csch}(a+b x) \text{sech}^2(a+b x)}{8 b}+\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}-\frac{(15 i) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{8 b}\\ &=-\frac{15 \tan ^{-1}(\sinh (a+b x))}{8 b}-\frac{15 \text{csch}(a+b x)}{8 b}+\frac{5 \text{csch}(a+b x) \text{sech}^2(a+b x)}{8 b}+\frac{\text{csch}(a+b x) \text{sech}^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [C]  time = 0.0151623, size = 29, normalized size = 0.41 $-\frac{\text{csch}(a+b x) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\sinh ^2(a+b x)\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[a + b*x]^2*Sech[a + b*x]^5,x]

[Out]

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

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Maple [A]  time = 0.019, size = 71, normalized size = 1. \begin{align*} -{\frac{1}{b\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}}-{\frac{5\, \left ({\rm sech} \left (bx+a\right ) \right ) ^{3}\tanh \left ( bx+a \right ) }{4\,b}}-{\frac{15\,{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{8\,b}}-{\frac{15\,\arctan \left ({{\rm e}^{bx+a}} \right ) }{4\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^2*sech(b*x+a)^5,x)

[Out]

-1/b/sinh(b*x+a)/cosh(b*x+a)^4-5/4*sech(b*x+a)^3*tanh(b*x+a)/b-15/8*sech(b*x+a)*tanh(b*x+a)/b-15/4*arctan(exp(
b*x+a))/b

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Maxima [B]  time = 1.77856, size = 184, normalized size = 2.63 \begin{align*} \frac{15 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{4 \, b} - \frac{15 \, e^{\left (-b x - a\right )} + 40 \, e^{\left (-3 \, b x - 3 \, a\right )} + 18 \, e^{\left (-5 \, b x - 5 \, a\right )} + 40 \, e^{\left (-7 \, b x - 7 \, a\right )} + 15 \, e^{\left (-9 \, b x - 9 \, a\right )}}{4 \, b{\left (3 \, e^{\left (-2 \, b x - 2 \, a\right )} + 2 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, e^{\left (-6 \, b x - 6 \, a\right )} - 3 \, e^{\left (-8 \, b x - 8 \, a\right )} - e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^2*sech(b*x+a)^5,x, algorithm="maxima")

[Out]

15/4*arctan(e^(-b*x - a))/b - 1/4*(15*e^(-b*x - a) + 40*e^(-3*b*x - 3*a) + 18*e^(-5*b*x - 5*a) + 40*e^(-7*b*x
- 7*a) + 15*e^(-9*b*x - 9*a))/(b*(3*e^(-2*b*x - 2*a) + 2*e^(-4*b*x - 4*a) - 2*e^(-6*b*x - 6*a) - 3*e^(-8*b*x -
8*a) - e^(-10*b*x - 10*a) + 1))

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Fricas [B]  time = 2.20505, size = 3312, normalized size = 47.31 \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)^2*sech(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/4*(15*cosh(b*x + a)^9 + 135*cosh(b*x + a)*sinh(b*x + a)^8 + 15*sinh(b*x + a)^9 + 20*(27*cosh(b*x + a)^2 + 2
)*sinh(b*x + a)^7 + 40*cosh(b*x + a)^7 + 140*(9*cosh(b*x + a)^3 + 2*cosh(b*x + a))*sinh(b*x + a)^6 + 6*(315*co
sh(b*x + a)^4 + 140*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^5 + 18*cosh(b*x + a)^5 + 10*(189*cosh(b*x + a)^5 + 140*
cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh(b*x + a)^4 + 20*(63*cosh(b*x + a)^6 + 70*cosh(b*x + a)^4 + 9*cosh(b*x
+ a)^2 + 2)*sinh(b*x + a)^3 + 40*cosh(b*x + a)^3 + 60*(9*cosh(b*x + a)^7 + 14*cosh(b*x + a)^5 + 3*cosh(b*x + a
)^3 + 2*cosh(b*x + a))*sinh(b*x + a)^2 + 15*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x +
a)^10 + 3*(15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^8 + 3*cosh(b*x + a)^8 + 24*(5*cosh(b*x + a)^3 + cosh(b*x + a)
)*sinh(b*x + a)^7 + 2*(105*cosh(b*x + a)^4 + 42*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 2*cosh(b*x + a)^6 + 12*
(21*cosh(b*x + a)^5 + 14*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^5 + 2*(105*cosh(b*x + a)^6 + 105*cosh(
b*x + a)^4 + 15*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 2*cosh(b*x + a)^4 + 8*(15*cosh(b*x + a)^7 + 21*cosh(b*x
+ a)^5 + 5*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + 3*(15*cosh(b*x + a)^8 + 28*cosh(b*x + a)^6 + 10
*cosh(b*x + a)^4 - 4*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 3*cosh(b*x + a)^2 + 2*(5*cosh(b*x + a)^9 + 12*cosh
(b*x + a)^7 + 6*cosh(b*x + a)^5 - 4*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a) - 1)*arctan(cosh(b*x + a)
+ sinh(b*x + a)) + 5*(27*cosh(b*x + a)^8 + 56*cosh(b*x + a)^6 + 18*cosh(b*x + a)^4 + 24*cosh(b*x + a)^2 + 3)*
sinh(b*x + a) + 15*cosh(b*x + a))/(b*cosh(b*x + a)^10 + 10*b*cosh(b*x + a)*sinh(b*x + a)^9 + b*sinh(b*x + a)^1
0 + 3*b*cosh(b*x + a)^8 + 3*(15*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^8 + 24*(5*b*cosh(b*x + a)^3 + b*cosh(b*x
+ a))*sinh(b*x + a)^7 + 2*b*cosh(b*x + a)^6 + 2*(105*b*cosh(b*x + a)^4 + 42*b*cosh(b*x + a)^2 + b)*sinh(b*x +
a)^6 + 12*(21*b*cosh(b*x + a)^5 + 14*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)^5 - 2*b*cosh(b*x + a)^
4 + 2*(105*b*cosh(b*x + a)^6 + 105*b*cosh(b*x + a)^4 + 15*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^4 + 8*(15*b*cos
h(b*x + a)^7 + 21*b*cosh(b*x + a)^5 + 5*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a)^3 - 3*b*cosh(b*x +
a)^2 + 3*(15*b*cosh(b*x + a)^8 + 28*b*cosh(b*x + a)^6 + 10*b*cosh(b*x + a)^4 - 4*b*cosh(b*x + a)^2 - b)*sinh(b
*x + a)^2 + 2*(5*b*cosh(b*x + a)^9 + 12*b*cosh(b*x + a)^7 + 6*b*cosh(b*x + a)^5 - 4*b*cosh(b*x + a)^3 - 3*b*co
sh(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}^{2}{\left (a + b x \right )} \operatorname{sech}^{5}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**2*sech(b*x+a)**5,x)

[Out]

Integral(csch(a + b*x)**2*sech(a + b*x)**5, x)

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Giac [B]  time = 1.17037, size = 174, normalized size = 2.49 \begin{align*} -\frac{15 \,{\left (\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 )\right )}}{16 \, b} - \frac{7 \,{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 36 \, e^{\left (b x + a\right )} - 36 \, e^{\left (-b x - a\right )}}{4 \,{\left ({\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 4\right )}^{2} b} - \frac{2}{b{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^2*sech(b*x+a)^5,x, algorithm="giac")

[Out]

-15/16*(pi + 2*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b - 1/4*(7*(e^(b*x + a) - e^(-b*x - a))^3 + 36*
e^(b*x + a) - 36*e^(-b*x - a))/(((e^(b*x + a) - e^(-b*x - a))^2 + 4)^2*b) - 2/(b*(e^(b*x + a) - e^(-b*x - a)))