### 3.27 $$\int \text{csch}(a+b x) \text{sech}^4(a+b x) \, dx$$

Optimal. Leaf size=38 $\frac{\text{sech}^3(a+b x)}{3 b}+\frac{\text{sech}(a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

[Out]

-(ArcTanh[Cosh[a + b*x]]/b) + Sech[a + b*x]/b + Sech[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0292979, antiderivative size = 38, 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 = {2622, 302, 207} $\frac{\text{sech}^3(a+b x)}{3 b}+\frac{\text{sech}(a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]*Sech[a + b*x]^4,x]

[Out]

-(ArcTanh[Cosh[a + b*x]]/b) + Sech[a + b*x]/b + Sech[a + b*x]^3/(3*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 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}(a+b x) \text{sech}^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\text{sech}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\text{sech}(a+b x)\right )}{b}\\ &=\frac{\text{sech}(a+b x)}{b}+\frac{\text{sech}^3(a+b x)}{3 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\text{sech}(a+b x)\right )}{b}\\ &=-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}+\frac{\text{sech}(a+b x)}{b}+\frac{\text{sech}^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0272027, size = 41, normalized size = 1.08 $\frac{\text{sech}^3(a+b x)}{3 b}+\frac{\text{sech}(a+b x)}{b}+\frac{\log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Log[Tanh[(a + b*x)/2]]/b + Sech[a + b*x]/b + Sech[a + b*x]^3/(3*b)

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Maple [A]  time = 0.016, size = 33, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ({\frac{1}{3\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+ \left ( \cosh \left ( bx+a \right ) \right ) ^{-1}-2\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/3/cosh(b*x+a)^3+1/cosh(b*x+a)-2*arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.0422, size = 146, normalized size = 3.84 \begin{align*} -\frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\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)*sech(b*x+a)^4,x, algorithm="maxima")

[Out]

-log(e^(-b*x - a) + 1)/b + log(e^(-b*x - a) - 1)/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 = 2.0867, size = 1958, normalized size = 51.53 \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)*sech(b*x+a)^4,x, algorithm="fricas")

[Out]

1/3*(6*cosh(b*x + a)^5 + 30*cosh(b*x + a)*sinh(b*x + a)^4 + 6*sinh(b*x + a)^5 + 20*(3*cosh(b*x + a)^2 + 1)*sin
h(b*x + a)^3 + 20*cosh(b*x + a)^3 + 60*(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)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 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*(co
sh(b*x + a)^5 + 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) +
6*(5*cosh(b*x + a)^4 + 10*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + 6*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}{\left (a + b x \right )} \operatorname{sech}^{4}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.18892, size = 124, normalized size = 3.26 \begin{align*} -\frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right )}{2 \, b} + \frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\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)*sech(b*x+a)^4,x, algorithm="giac")

[Out]

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