3.61 \(\int (1+\coth (x))^5 \, dx\)

Optimal. Leaf size=41 \[ 16 x-\frac{1}{4} (\coth (x)+1)^4-\frac{2}{3} (\coth (x)+1)^3-2 (\coth (x)+1)^2-8 \coth (x)+16 \log (\sinh (x)) \]

[Out]

16*x - 8*Coth[x] - 2*(1 + Coth[x])^2 - (2*(1 + Coth[x])^3)/3 - (1 + Coth[x])^4/4 + 16*Log[Sinh[x]]

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Rubi [A]  time = 0.0394656, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3478, 3477, 3475} \[ 16 x-\frac{1}{4} (\coth (x)+1)^4-\frac{2}{3} (\coth (x)+1)^3-2 (\coth (x)+1)^2-8 \coth (x)+16 \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x])^5,x]

[Out]

16*x - 8*Coth[x] - 2*(1 + Coth[x])^2 - (2*(1 + Coth[x])^3)/3 - (1 + Coth[x])^4/4 + 16*Log[Sinh[x]]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

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 (1+\coth (x))^5 \, dx &=-\frac{1}{4} (1+\coth (x))^4+2 \int (1+\coth (x))^4 \, dx\\ &=-\frac{2}{3} (1+\coth (x))^3-\frac{1}{4} (1+\coth (x))^4+4 \int (1+\coth (x))^3 \, dx\\ &=-2 (1+\coth (x))^2-\frac{2}{3} (1+\coth (x))^3-\frac{1}{4} (1+\coth (x))^4+8 \int (1+\coth (x))^2 \, dx\\ &=16 x-8 \coth (x)-2 (1+\coth (x))^2-\frac{2}{3} (1+\coth (x))^3-\frac{1}{4} (1+\coth (x))^4+16 \int \coth (x) \, dx\\ &=16 x-8 \coth (x)-2 (1+\coth (x))^2-\frac{2}{3} (1+\coth (x))^3-\frac{1}{4} (1+\coth (x))^4+16 \log (\sinh (x))\\ \end{align*}

Mathematica [C]  time = 0.238116, size = 94, normalized size = 2.29 \[ \frac{\sinh (x) (\coth (x)+1)^5 \left (-20 \sinh (x) \cosh ^3(x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\tanh ^2(x)\right )-120 \sinh ^3(x) \cosh (x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\tanh ^2(x)\right )-3 \cosh ^4(x)-66 \sinh ^2(x) \cosh ^2(x)+12 \sinh ^4(x) (x+16 \log (\tanh (x))+16 \log (\cosh (x)))\right )}{12 (\sinh (x)+\cosh (x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x])^5,x]

[Out]

((1 + Coth[x])^5*Sinh[x]*(-3*Cosh[x]^4 - 20*Cosh[x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[x]^2]*Sinh[x] - 66
*Cosh[x]^2*Sinh[x]^2 - 120*Cosh[x]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[x]^2]*Sinh[x]^3 + 12*(x + 16*Log[Cosh[
x]] + 16*Log[Tanh[x]])*Sinh[x]^4))/(12*(Cosh[x] + Sinh[x])^5)

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Maple [A]  time = 0.007, size = 31, normalized size = 0.8 \begin{align*} -{\frac{ \left ({\rm coth} \left (x\right ) \right ) ^{4}}{4}}-{\frac{5\, \left ({\rm coth} \left (x\right ) \right ) ^{3}}{3}}-{\frac{11\, \left ({\rm coth} \left (x\right ) \right ) ^{2}}{2}}-15\,{\rm coth} \left (x\right )-16\,\ln \left ({\rm coth} \left (x\right )-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+coth(x))^5,x)

[Out]

-1/4*coth(x)^4-5/3*coth(x)^3-11/2*coth(x)^2-15*coth(x)-16*ln(coth(x)-1)

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Maxima [B]  time = 1.0613, size = 189, normalized size = 4.61 \begin{align*} 27 \, x - \frac{20 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} - 2\right )}}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} + \frac{4 \,{\left (e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \frac{20 \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + \frac{20}{e^{\left (-2 \, x\right )} - 1} + 11 \, \log \left (e^{\left (-x\right )} + 1\right ) + 11 \, \log \left (e^{\left (-x\right )} - 1\right ) + 5 \, \log \left (\sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^5,x, algorithm="maxima")

[Out]

27*x - 20/3*(3*e^(-2*x) - 3*e^(-4*x) - 2)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1) + 4*(e^(-2*x) - e^(-4*x) +
e^(-6*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1) + 20*e^(-2*x)/(2*e^(-2*x) - e^(-4*x) - 1) + 20
/(e^(-2*x) - 1) + 11*log(e^(-x) + 1) + 11*log(e^(-x) - 1) + 5*log(sinh(x))

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Fricas [B]  time = 2.04947, size = 1481, normalized size = 36.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^5,x, algorithm="fricas")

[Out]

-4/3*(48*cosh(x)^6 + 288*cosh(x)*sinh(x)^5 + 48*sinh(x)^6 + 36*(20*cosh(x)^2 - 3)*sinh(x)^4 - 108*cosh(x)^4 +
48*(20*cosh(x)^3 - 9*cosh(x))*sinh(x)^3 + 8*(90*cosh(x)^4 - 81*cosh(x)^2 + 11)*sinh(x)^2 + 88*cosh(x)^2 - 12*(
cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 - 1)*sinh(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3
*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x
)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cos
h(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*sinh(x) + 1)*log(2*sinh(x)/(cosh(x) - sinh(x))) + 16*(18*cosh(x)
^5 - 27*cosh(x)^3 + 11*cosh(x))*sinh(x) - 25)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 -
1)*sinh(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*sinh(
x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*
cosh(x)^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*sinh(x) + 1)

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Sympy [A]  time = 2.58923, size = 48, normalized size = 1.17 \begin{align*} 32 x - 16 \log{\left (\tanh{\left (x \right )} + 1 \right )} + 16 \log{\left (\tanh{\left (x \right )} \right )} - \frac{15}{\tanh{\left (x \right )}} - \frac{11}{2 \tanh ^{2}{\left (x \right )}} - \frac{5}{3 \tanh ^{3}{\left (x \right )}} - \frac{1}{4 \tanh ^{4}{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))**5,x)

[Out]

32*x - 16*log(tanh(x) + 1) + 16*log(tanh(x)) - 15/tanh(x) - 11/(2*tanh(x)**2) - 5/(3*tanh(x)**3) - 1/(4*tanh(x
)**4)

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Giac [A]  time = 1.18271, size = 55, normalized size = 1.34 \begin{align*} -\frac{4 \,{\left (48 \, e^{\left (6 \, x\right )} - 108 \, e^{\left (4 \, x\right )} + 88 \, e^{\left (2 \, x\right )} - 25\right )}}{3 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{4}} + 16 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+coth(x))^5,x, algorithm="giac")

[Out]

-4/3*(48*e^(6*x) - 108*e^(4*x) + 88*e^(2*x) - 25)/(e^(2*x) - 1)^4 + 16*log(abs(e^(2*x) - 1))