3.9 \(\int \coth ^3(a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac {\log (\sinh (a+b x))}{b}-\frac {\coth ^2(a+b x)}{2 b} \]

[Out]

-1/2*coth(b*x+a)^2/b+ln(sinh(b*x+a))/b

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Rubi [A]  time = 0.02, antiderivative size = 27, 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 = {3473, 3475} \[ \frac {\log (\sinh (a+b x))}{b}-\frac {\coth ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + b*x]^3,x]

[Out]

-Coth[a + b*x]^2/(2*b) + Log[Sinh[a + b*x]]/b

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

Rubi steps

\begin {align*} \int \coth ^3(a+b x) \, dx &=-\frac {\coth ^2(a+b x)}{2 b}+\int \coth (a+b x) \, dx\\ &=-\frac {\coth ^2(a+b x)}{2 b}+\frac {\log (\sinh (a+b x))}{b}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.26 \[ -\frac {\coth ^2(a+b x)-2 \log (\tanh (a+b x))-2 \log (\cosh (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + b*x]^3,x]

[Out]

-1/2*(Coth[a + b*x]^2 - 2*Log[Cosh[a + b*x]] - 2*Log[Tanh[a + b*x]])/b

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fricas [B]  time = 0.46, size = 346, normalized size = 12.81 \[ -\frac {b x \cosh \left (b x + a\right )^{4} + 4 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b x \sinh \left (b x + a\right )^{4} - 2 \, {\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b x \cosh \left (b x + a\right )^{2} - b x + 1\right )} \sinh \left (b x + a\right )^{2} + b x - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \, {\left (b x \cosh \left (b x + a\right )^{3} - {\left (b x - 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - 2*(b*x - 1)*cosh(b*x + a)^
2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x + 1)*sinh(b*x + a)^2 + b*x - (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(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(b*x*cosh(b*x + a)^
3 - (b*x - 1)*cosh(b*x + a))*sinh(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 [A]  time = 0.13, size = 49, normalized size = 1.81 \[ -\frac {b x + a + \frac {2 \, e^{\left (2 \, b x + 2 \, a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 43, normalized size = 1.59 \[ -\frac {\coth ^{2}\left (b x +a \right )}{2 b}-\frac {\ln \left (\coth \left (b x +a \right )-1\right )}{2 b}-\frac {\ln \left (\coth \left (b x +a \right )+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*coth(b*x+a)^2/b-1/2/b*ln(coth(b*x+a)-1)-1/2/b*ln(coth(b*x+a)+1)

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maxima [B]  time = 0.42, size = 79, normalized size = 2.93 \[ x + \frac {a}{b} + \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {2 \, e^{\left (-2 \, b x - 2 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 68, normalized size = 2.52 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-x-\frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^3,x)

[Out]

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

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sympy [A]  time = 2.49, size = 61, normalized size = 2.26 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\x \coth ^{3}{\relax (a )} & \text {for}\: b = 0 \\x - \frac {\log {\left (\tanh {\left (a + b x \right )} + 1 \right )}}{b} + \frac {\log {\left (\tanh {\left (a + b x \right )} \right )}}{b} - \frac {1}{2 b \tanh ^{2}{\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (x*coth(a)**3, Eq(b, 0)), (x - log(tanh(a +
b*x) + 1)/b + log(tanh(a + b*x))/b - 1/(2*b*tanh(a + b*x)**2), True))

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