3.12 \(\int \coth ^6(a+b x) \, dx\)

Optimal. Leaf size=43 \[ -\frac {\coth ^5(a+b x)}{5 b}-\frac {\coth ^3(a+b x)}{3 b}-\frac {\coth (a+b x)}{b}+x \]

[Out]

x-coth(b*x+a)/b-1/3*coth(b*x+a)^3/b-1/5*coth(b*x+a)^5/b

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3473, 8} \[ -\frac {\coth ^5(a+b x)}{5 b}-\frac {\coth ^3(a+b x)}{3 b}-\frac {\coth (a+b x)}{b}+x \]

Antiderivative was successfully verified.

[In]

Int[Coth[a + b*x]^6,x]

[Out]

x - Coth[a + b*x]/b - Coth[a + b*x]^3/(3*b) - Coth[a + b*x]^5/(5*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps

\begin {align*} \int \coth ^6(a+b x) \, dx &=-\frac {\coth ^5(a+b x)}{5 b}+\int \coth ^4(a+b x) \, dx\\ &=-\frac {\coth ^3(a+b x)}{3 b}-\frac {\coth ^5(a+b x)}{5 b}+\int \coth ^2(a+b x) \, dx\\ &=-\frac {\coth (a+b x)}{b}-\frac {\coth ^3(a+b x)}{3 b}-\frac {\coth ^5(a+b x)}{5 b}+\int 1 \, dx\\ &=x-\frac {\coth (a+b x)}{b}-\frac {\coth ^3(a+b x)}{3 b}-\frac {\coth ^5(a+b x)}{5 b}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 31, normalized size = 0.72 \[ -\frac {\coth ^5(a+b x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\tanh ^2(a+b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + b*x]^6,x]

[Out]

-1/5*(Coth[a + b*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, Tanh[a + b*x]^2])/b

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fricas [B]  time = 0.53, size = 239, normalized size = 5.56 \[ \frac {{\left (15 \, b x + 23\right )} \sinh \left (b x + a\right )^{5} - 23 \, \cosh \left (b x + a\right )^{5} - 115 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 5 \, {\left (2 \, {\left (15 \, b x + 23\right )} \cosh \left (b x + a\right )^{2} - 15 \, b x - 23\right )} \sinh \left (b x + a\right )^{3} + 25 \, \cosh \left (b x + a\right )^{3} - 5 \, {\left (46 \, \cosh \left (b x + a\right )^{3} - 15 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 5 \, {\left ({\left (15 \, b x + 23\right )} \cosh \left (b x + a\right )^{4} - 3 \, {\left (15 \, b x + 23\right )} \cosh \left (b x + a\right )^{2} + 30 \, b x + 46\right )} \sinh \left (b x + a\right ) - 50 \, \cosh \left (b x + a\right )}{15 \, {\left (b \sinh \left (b x + a\right )^{5} + 5 \, {\left (2 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{3} + 5 \, {\left (b \cosh \left (b x + a\right )^{4} - 3 \, b \cosh \left (b x + a\right )^{2} + 2 \, b\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^6,x, algorithm="fricas")

[Out]

1/15*((15*b*x + 23)*sinh(b*x + a)^5 - 23*cosh(b*x + a)^5 - 115*cosh(b*x + a)*sinh(b*x + a)^4 + 5*(2*(15*b*x +
23)*cosh(b*x + a)^2 - 15*b*x - 23)*sinh(b*x + a)^3 + 25*cosh(b*x + a)^3 - 5*(46*cosh(b*x + a)^3 - 15*cosh(b*x
+ a))*sinh(b*x + a)^2 + 5*((15*b*x + 23)*cosh(b*x + a)^4 - 3*(15*b*x + 23)*cosh(b*x + a)^2 + 30*b*x + 46)*sinh
(b*x + a) - 50*cosh(b*x + a))/(b*sinh(b*x + a)^5 + 5*(2*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^3 + 5*(b*cosh(b*x
+ a)^4 - 3*b*cosh(b*x + a)^2 + 2*b)*sinh(b*x + a))

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giac [A]  time = 0.13, size = 74, normalized size = 1.72 \[ \frac {15 \, b x + 15 \, a - \frac {2 \, {\left (45 \, e^{\left (8 \, b x + 8 \, a\right )} - 90 \, e^{\left (6 \, b x + 6 \, a\right )} + 140 \, e^{\left (4 \, b x + 4 \, a\right )} - 70 \, e^{\left (2 \, b x + 2 \, a\right )} + 23\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{5}}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^6,x, algorithm="giac")

[Out]

1/15*(15*b*x + 15*a - 2*(45*e^(8*b*x + 8*a) - 90*e^(6*b*x + 6*a) + 140*e^(4*b*x + 4*a) - 70*e^(2*b*x + 2*a) +
23)/(e^(2*b*x + 2*a) - 1)^5)/b

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maple [A]  time = 0.01, size = 67, normalized size = 1.56 \[ -\frac {\coth ^{5}\left (b x +a \right )}{5 b}-\frac {\coth ^{3}\left (b x +a \right )}{3 b}-\frac {\coth \left (b x +a \right )}{b}-\frac {\ln \left (\coth \left (b x +a \right )-1\right )}{2 b}+\frac {\ln \left (\coth \left (b x +a \right )+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^6,x)

[Out]

-1/5*coth(b*x+a)^5/b-1/3*coth(b*x+a)^3/b-coth(b*x+a)/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 = 1.29, size = 115, normalized size = 2.67 \[ x + \frac {a}{b} - \frac {2 \, {\left (70 \, e^{\left (-2 \, b x - 2 \, a\right )} - 140 \, e^{\left (-4 \, b x - 4 \, a\right )} + 90 \, e^{\left (-6 \, b x - 6 \, a\right )} - 45 \, e^{\left (-8 \, b x - 8 \, a\right )} - 23\right )}}{15 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} - 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)^6,x, algorithm="maxima")

[Out]

x + a/b - 2/15*(70*e^(-2*b*x - 2*a) - 140*e^(-4*b*x - 4*a) + 90*e^(-6*b*x - 6*a) - 45*e^(-8*b*x - 8*a) - 23)/(
b*(5*e^(-2*b*x - 2*a) - 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) - 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) -
1))

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mupad [B]  time = 0.07, size = 34, normalized size = 0.79 \[ x-\frac {\frac {{\mathrm {coth}\left (a+b\,x\right )}^5}{5}+\frac {{\mathrm {coth}\left (a+b\,x\right )}^3}{3}+\mathrm {coth}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + b*x)^6,x)

[Out]

x - (coth(a + b*x) + coth(a + b*x)^3/3 + coth(a + b*x)^5/5)/b

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sympy [A]  time = 27.11, size = 63, normalized size = 1.47 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\x \coth ^{6}{\relax (a )} & \text {for}\: b = 0 \\x - \frac {1}{b \tanh {\left (a + b x \right )}} - \frac {1}{3 b \tanh ^{3}{\left (a + b x \right )}} - \frac {1}{5 b \tanh ^{5}{\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**6,x)

[Out]

Piecewise((zoo*x, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (x*coth(a)**6, Eq(b, 0)), (x - 1/(b*tanh(a
+ b*x)) - 1/(3*b*tanh(a + b*x)**3) - 1/(5*b*tanh(a + b*x)**5), True))

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