### 3.122 $$\int \coth ^2(a+b x) \text{csch}^3(a+b x) \, dx$$

Optimal. Leaf size=55 $\frac{\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac{\coth (a+b x) \text{csch}^3(a+b x)}{4 b}-\frac{\coth (a+b x) \text{csch}(a+b x)}{8 b}$

[Out]

ArcTanh[Cosh[a + b*x]]/(8*b) - (Coth[a + b*x]*Csch[a + b*x])/(8*b) - (Coth[a + b*x]*Csch[a + b*x]^3)/(4*b)

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Rubi [A]  time = 0.0596215, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {2611, 3768, 3770} $\frac{\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac{\coth (a+b x) \text{csch}^3(a+b x)}{4 b}-\frac{\coth (a+b x) \text{csch}(a+b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[a + b*x]^2*Csch[a + b*x]^3,x]

[Out]

ArcTanh[Cosh[a + b*x]]/(8*b) - (Coth[a + b*x]*Csch[a + b*x])/(8*b) - (Coth[a + b*x]*Csch[a + b*x]^3)/(4*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \coth ^2(a+b x) \text{csch}^3(a+b x) \, dx &=-\frac{\coth (a+b x) \text{csch}^3(a+b x)}{4 b}+\frac{1}{4} \int \text{csch}^3(a+b x) \, dx\\ &=-\frac{\coth (a+b x) \text{csch}(a+b x)}{8 b}-\frac{\coth (a+b x) \text{csch}^3(a+b x)}{4 b}-\frac{1}{8} \int \text{csch}(a+b x) \, dx\\ &=\frac{\tanh ^{-1}(\cosh (a+b x))}{8 b}-\frac{\coth (a+b x) \text{csch}(a+b x)}{8 b}-\frac{\coth (a+b x) \text{csch}^3(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.044009, size = 95, normalized size = 1.73 $-\frac{\text{csch}^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{\text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\text{sech}^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{\text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{32 b}-\frac{\log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[a + b*x]^2*Csch[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.022, size = 58, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{\cosh \left ( bx+a \right ) }{3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}}-{\frac{{\rm coth} \left (bx+a\right )}{3} \left ( -{\frac{ \left ({\rm csch} \left (bx+a\right ) \right ) ^{3}}{4}}+{\frac{3\,{\rm csch} \left (bx+a\right )}{8}} \right ) }+{\frac{{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{4}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^2*csch(b*x+a)^3,x)

[Out]

1/b*(-1/3/sinh(b*x+a)^4*cosh(b*x+a)-1/3*(-1/4*csch(b*x+a)^3+3/8*csch(b*x+a))*coth(b*x+a)+1/4*arctanh(exp(b*x+a
)))

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Maxima [B]  time = 1.09227, size = 174, normalized size = 3.16 \begin{align*} \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{8 \, b} - \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{8 \, b} + \frac{e^{\left (-b x - a\right )} + 7 \, e^{\left (-3 \, b x - 3 \, a\right )} + 7 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}}{4 \, b{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, 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(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

1/8*log(e^(-b*x - a) + 1)/b - 1/8*log(e^(-b*x - a) - 1)/b + 1/4*(e^(-b*x - a) + 7*e^(-3*b*x - 3*a) + 7*e^(-5*b
*x - 5*a) + e^(-7*b*x - 7*a))/(b*(4*e^(-2*b*x - 2*a) - 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) - e^(-8*b*x - 8
*a) - 1))

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Fricas [B]  time = 2.02739, size = 3066, normalized size = 55.75 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/8*(2*cosh(b*x + a)^7 + 14*cosh(b*x + a)*sinh(b*x + a)^6 + 2*sinh(b*x + a)^7 + 14*(3*cosh(b*x + a)^2 + 1)*si
nh(b*x + a)^5 + 14*cosh(b*x + a)^5 + 70*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^4 + 14*(5*cosh(b*x + a
)^4 + 10*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 14*cosh(b*x + a)^3 + 14*(3*cosh(b*x + a)^5 + 10*cosh(b*x + a)^
3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 +
4*(7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b
*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*
x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 15*cosh(b*x + a)^4 +
9*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 - 3*cosh(b*x + a)^5 + 3*cosh(
b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (cosh(b*x + a)^8 + 8*c
osh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6
+ 8*(7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*s
inh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a
)^3 + 4*(7*cosh(b*x + a)^6 - 15*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 +
8*(cosh(b*x + a)^7 - 3*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x +
a) + sinh(b*x + a) - 1) + 2*(7*cosh(b*x + a)^6 + 35*cosh(b*x + a)^4 + 21*cosh(b*x + a)^2 + 1)*sinh(b*x + a) +
2*cosh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 - 4*b*cosh(b*x +
a)^6 + 4*(7*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^6 + 8*(7*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)
^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 - 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)^4 + 8*(7*b*cosh
(b*x + a)^5 - 10*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 - 4*b*cosh(b*x + a)^2 + 4*(7*b*cosh(b*
x + a)^6 - 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a)^7 - 3*b*cosh(b
*x + a)^5 + 3*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 \coth ^{2}{\left (a + b x \right )} \operatorname{csch}^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**2*csch(b*x+a)**3,x)

[Out]

Integral(coth(a + b*x)**2*csch(a + b*x)**3, x)

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Giac [A]  time = 1.223, size = 108, normalized size = 1.96 \begin{align*} -\frac{\frac{2 \,{\left (e^{\left (7 \, b x + 7 \, a\right )} + 7 \, e^{\left (5 \, b x + 5 \, a\right )} + 7 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{4}} - \log \left (e^{\left (b x + a\right )} + 1\right ) + \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{8 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

-1/8*(2*(e^(7*b*x + 7*a) + 7*e^(5*b*x + 5*a) + 7*e^(3*b*x + 3*a) + e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^4 - log(
e^(b*x + a) + 1) + log(abs(e^(b*x + a) - 1)))/b