3.11 \(\int \coth ^5(a+b x) \, dx\)

Optimal. Leaf size=42 \[ -\frac {\coth ^4(a+b x)}{4 b}-\frac {\coth ^2(a+b x)}{2 b}+\frac {\log (\sinh (a+b x))}{b} \]

[Out]

-1/2*coth(b*x+a)^2/b-1/4*coth(b*x+a)^4/b+ln(sinh(b*x+a))/b

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3473, 3475} \[ -\frac {\coth ^4(a+b x)}{4 b}-\frac {\coth ^2(a+b x)}{2 b}+\frac {\log (\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + b*x]^5,x]

[Out]

-Coth[a + b*x]^2/(2*b) - Coth[a + b*x]^4/(4*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 ^5(a+b x) \, dx &=-\frac {\coth ^4(a+b x)}{4 b}+\int \coth ^3(a+b x) \, dx\\ &=-\frac {\coth ^2(a+b x)}{2 b}-\frac {\coth ^4(a+b x)}{4 b}+\int \coth (a+b x) \, dx\\ &=-\frac {\coth ^2(a+b x)}{2 b}-\frac {\coth ^4(a+b x)}{4 b}+\frac {\log (\sinh (a+b x))}{b}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 44, normalized size = 1.05 \[ -\frac {\coth ^4(a+b x)+2 \coth ^2(a+b x)-4 \log (\tanh (a+b x))-4 \log (\cosh (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + b*x]^5,x]

[Out]

-1/4*(2*Coth[a + b*x]^2 + Coth[a + b*x]^4 - 4*Log[Cosh[a + b*x]] - 4*Log[Tanh[a + b*x]])/b

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fricas [B]  time = 0.55, size = 978, normalized size = 23.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^5,x, algorithm="fricas")

[Out]

-(b*x*cosh(b*x + a)^8 + 8*b*x*cosh(b*x + a)*sinh(b*x + a)^7 + b*x*sinh(b*x + a)^8 - 4*(b*x - 1)*cosh(b*x + a)^
6 + 4*(7*b*x*cosh(b*x + a)^2 - b*x + 1)*sinh(b*x + a)^6 + 8*(7*b*x*cosh(b*x + a)^3 - 3*(b*x - 1)*cosh(b*x + a)
)*sinh(b*x + a)^5 + 2*(3*b*x - 2)*cosh(b*x + a)^4 + 2*(35*b*x*cosh(b*x + a)^4 - 30*(b*x - 1)*cosh(b*x + a)^2 +
3*b*x - 2)*sinh(b*x + a)^4 + 8*(7*b*x*cosh(b*x + a)^5 - 10*(b*x - 1)*cosh(b*x + a)^3 + (3*b*x - 2)*cosh(b*x +
a))*sinh(b*x + a)^3 - 4*(b*x - 1)*cosh(b*x + a)^2 + 4*(7*b*x*cosh(b*x + a)^6 - 15*(b*x - 1)*cosh(b*x + a)^4 +
3*(3*b*x - 2)*cosh(b*x + a)^2 - b*x + 1)*sinh(b*x + a)^2 + b*x - (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*cos
h(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(2*sinh(b*x + a)/(cosh(b*x + a) -
sinh(b*x + a))) + 8*(b*x*cosh(b*x + a)^7 - 3*(b*x - 1)*cosh(b*x + a)^5 + (3*b*x - 2)*cosh(b*x + a)^3 - (b*x -
1)*cosh(b*x + a))*sinh(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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giac [A]  time = 0.16, size = 70, normalized size = 1.67 \[ -\frac {b x + a + \frac {4 \, {\left (e^{\left (6 \, b x + 6 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} + e^{\left (2 \, b x + 2 \, a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{4}} - \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^5,x, algorithm="giac")

[Out]

-(b*x + a + 4*(e^(6*b*x + 6*a) - e^(4*b*x + 4*a) + e^(2*b*x + 2*a))/(e^(2*b*x + 2*a) - 1)^4 - log(abs(e^(2*b*x
+ 2*a) - 1)))/b

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maple [A]  time = 0.01, size = 56, normalized size = 1.33 \[ -\frac {\coth ^{4}\left (b x +a \right )}{4 b}-\frac {\coth ^{2}\left (b x +a \right )}{2 b}-\frac {\ln \left (\coth \left (b x +a \right )-1\right )}{2 b}-\frac {\ln \left (\coth \left (b x +a \right )+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^5,x)

[Out]

-1/4*coth(b*x+a)^4/b-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 = 1.05, size = 122, normalized size = 2.90 \[ x + \frac {a}{b} + \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {4 \, {\left (e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}}{b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^5,x, algorithm="maxima")

[Out]

x + a/b + log(e^(-b*x - a) + 1)/b + log(e^(-b*x - a) - 1)/b + 4*(e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a) + e^(-6*b
*x - 6*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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mupad [B]  time = 0.99, size = 159, normalized size = 3.79 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-x-\frac {4}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {8}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {8}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {4}{b\,\left (6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^5,x)

[Out]

log(exp(2*a)*exp(2*b*x) - 1)/b - x - 4/(b*(exp(2*a + 2*b*x) - 1)) - 8/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x
) + 1)) - 8/(b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1)) - 4/(b*(6*exp(4*a + 4*b*x) -
4*exp(2*a + 2*b*x) - 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1))

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sympy [A]  time = 12.17, size = 75, normalized size = 1.79 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\x \coth ^{5}{\relax (a )} & \text {for}\: b = 0 \\x - \frac {\log {\left (\tanh {\left (a + b x \right )} + 1 \right )}}{b} + \frac {\log {\left (\tanh {\left (a + b x \right )} \right )}}{b} - \frac {1}{2 b \tanh ^{2}{\left (a + b x \right )}} - \frac {1}{4 b \tanh ^{4}{\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**5,x)

[Out]

Piecewise((zoo*x, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (x*coth(a)**5, 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) - 1/(4*b*tanh(a + b*x)**4), True))

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