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]

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

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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*}

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Mathematica [A]  time = 0.01, 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]

-1/8*Csch[(a + b*x)/2]^2/b - Log[Tanh[(a + b*x)/2]]/(2*b) - Sech[(a + b*x)/2]^2/(8*b)

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fricas [B]  time = 0.59, size = 387, normalized size = 11.38 \[ -\frac {2 \, \cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 2 \, \sinh \left (b x + a\right )^{3} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + 2 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]

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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giac [B]  time = 0.12, size = 84, normalized size = 2.47 \[ -\frac {\frac {4 \, {\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}}{{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 4} - \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right ) + \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.32, size = 27, normalized size = 0.79 \[ \frac {-\frac {\mathrm {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}+\arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]

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 = 0.33, size = 84, normalized size = 2.47 \[ \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac {e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

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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mupad [B]  time = 1.40, size = 86, normalized size = 2.53 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a + b*x)^3,x)

[Out]

atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b)/(-b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*
x) + 1)) - exp(a + b*x)/(b*(exp(2*a + 2*b*x) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]

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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