### 3.45 $$\int \text{csch}^5(a+b x) \text{sech}^2(a+b x) \, dx$$

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

[Out]

(-15*ArcTanh[Cosh[a + b*x]])/(8*b) + (15*Sech[a + b*x])/(8*b) + (5*Csch[a + b*x]^2*Sech[a + b*x])/(8*b) - (Csc
h[a + b*x]^4*Sech[a + b*x])/(4*b)

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Rubi [A]  time = 0.0501103, 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 = {2622, 288, 321, 207} $\frac{15 \text{sech}(a+b x)}{8 b}-\frac{15 \tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac{\text{csch}^4(a+b x) \text{sech}(a+b x)}{4 b}+\frac{5 \text{csch}^2(a+b x) \text{sech}(a+b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-15*ArcTanh[Cosh[a + b*x]])/(8*b) + (15*Sech[a + b*x])/(8*b) + (5*Csch[a + b*x]^2*Sech[a + b*x])/(8*b) - (Csc
h[a + b*x]^4*Sech[a + b*x])/(4*b)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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}^5(a+b x) \text{sech}^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\text{sech}(a+b x)\right )}{b}\\ &=-\frac{\text{csch}^4(a+b x) \text{sech}(a+b x)}{4 b}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\text{sech}(a+b x)\right )}{4 b}\\ &=\frac{5 \text{csch}^2(a+b x) \text{sech}(a+b x)}{8 b}-\frac{\text{csch}^4(a+b x) \text{sech}(a+b x)}{4 b}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\text{sech}(a+b x)\right )}{8 b}\\ &=\frac{15 \text{sech}(a+b x)}{8 b}+\frac{5 \text{csch}^2(a+b x) \text{sech}(a+b x)}{8 b}-\frac{\text{csch}^4(a+b x) \text{sech}(a+b x)}{4 b}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\text{sech}(a+b x)\right )}{8 b}\\ &=-\frac{15 \tanh ^{-1}(\cosh (a+b x))}{8 b}+\frac{15 \text{sech}(a+b x)}{8 b}+\frac{5 \text{csch}^2(a+b x) \text{sech}(a+b x)}{8 b}-\frac{\text{csch}^4(a+b x) \text{sech}(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0348478, size = 105, normalized size = 1.5 $-\frac{\text{csch}^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{7 \text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\text{sech}^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{7 \text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\text{sech}(a+b x)}{b}+\frac{15 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(7*Csch[(a + b*x)/2]^2)/(32*b) - Csch[(a + b*x)/2]^4/(64*b) + (15*Log[Tanh[(a + b*x)/2]])/(8*b) + (7*Sech[(a +
b*x)/2]^2)/(32*b) + Sech[(a + b*x)/2]^4/(64*b) + Sech[a + b*x]/b

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Maple [A]  time = 0.019, size = 61, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{4}\cosh \left ( bx+a \right ) }}+{\frac{5}{8\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{15}{8\,\cosh \left ( bx+a \right ) }}-{\frac{15\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{4}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/4/sinh(b*x+a)^4/cosh(b*x+a)+5/8/sinh(b*x+a)^2/cosh(b*x+a)+15/8/cosh(b*x+a)-15/4*arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.03293, size = 209, normalized size = 2.99 \begin{align*} -\frac{15 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{8 \, b} + \frac{15 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{8 \, 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)^5*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

-15/8*log(e^(-b*x - a) + 1)/b + 15/8*log(e^(-b*x - a) - 1)/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 = 2.5051, size = 4477, normalized size = 63.96 \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)^5*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(30*cosh(b*x + a)^9 + 270*cosh(b*x + a)*sinh(b*x + a)^8 + 30*sinh(b*x + a)^9 + 40*(27*cosh(b*x + a)^2 - 2)
*sinh(b*x + a)^7 - 80*cosh(b*x + a)^7 + 280*(9*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^6 + 12*(315*co
sh(b*x + a)^4 - 140*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^5 + 36*cosh(b*x + a)^5 + 20*(189*cosh(b*x + a)^5 - 140*
cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh(b*x + a)^4 + 40*(63*cosh(b*x + a)^6 - 70*cosh(b*x + a)^4 + 9*cosh(b*x
+ a)^2 - 2)*sinh(b*x + a)^3 - 80*cosh(b*x + a)^3 + 120*(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 + 1
0*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*cos
h(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)*log(cosh(b*x + a) +
sinh(b*x + a) + 1) + 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*cos
h(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*cos
h(b*x + a)^5 + 4*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1)
+ 10*(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) + 3
0*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)^10 - 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*c
osh(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*cosh(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*co
sh(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*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}^{5}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.16122, size = 185, normalized size = 2.64 \begin{align*} -\frac{15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right )}{16 \, b} + \frac{15 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\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)^5*sech(b*x+a)^2,x, algorithm="giac")

[Out]

-15/16*log(e^(b*x + a) + e^(-b*x - a) + 2)/b + 15/16*log(e^(b*x + a) + e^(-b*x - a) - 2)/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)))