3.3 \(\int \text{csch}^3(a+b x) \, dx\)

Optimal. Leaf size=34 \[ \frac{\tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{\coth (a+b x) \text{csch}(a+b x)}{2 b} \]

[Out]

ArcTanh[Cosh[a + b*x]]/(2*b) - (Coth[a + b*x]*Csch[a + b*x])/(2*b)

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Rubi [A]  time = 0.0221119, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3770} \[ \frac{\tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{\coth (a+b x) \text{csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]^3,x]

[Out]

ArcTanh[Cosh[a + b*x]]/(2*b) - (Coth[a + b*x]*Csch[a + b*x])/(2*b)

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

Mathematica [A]  time = 0.0136658, size = 57, normalized size = 1.68 \[ -\frac{\text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.01, size = 27, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{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)^3,x)

[Out]

1/b*(-1/2*csch(b*x+a)*coth(b*x+a)+arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.00849, size = 113, normalized size = 3.32 \begin{align*} \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} - \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac{e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{b{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.63838, size = 1088, normalized size = 32. \begin{align*} -\frac{2 \, \cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 2 \, \sinh \left (b x + a\right )^{3} -{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) +{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + 2 \, \cosh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*cosh(b*x + a)^3 + 6*cosh(b*x + a)*sinh(b*x + a)^2 + 2*sinh(b*x + a)^3 - (cosh(b*x + a)^4 + 4*cosh(b*x
+ a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(co
sh(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)^4 +
4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a
)^2 + 4*(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*(3*cos
h(b*x + a)^2 + 1)*sinh(b*x + a) + 2*cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*
sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(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}^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(a + b*x)**3, x)

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Giac [B]  time = 1.1466, size = 122, normalized size = 3.59 \begin{align*} \frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right )}{4 \, b} - \frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{4 \, b} - \frac{e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{{\left ({\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 4\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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