3.111 \(\int \frac{\coth ^3(x)}{a+a \text{sech}(x)} \, dx\)

Optimal. Leaf size=68 \[ \frac{1}{8 a (1-\cosh (x))}+\frac{3}{4 a (\cosh (x)+1)}-\frac{1}{8 a (\cosh (x)+1)^2}+\frac{5 \log (1-\cosh (x))}{16 a}+\frac{11 \log (\cosh (x)+1)}{16 a} \]

[Out]

1/(8*a*(1 - Cosh[x])) - 1/(8*a*(1 + Cosh[x])^2) + 3/(4*a*(1 + Cosh[x])) + (5*Log[1 - Cosh[x]])/(16*a) + (11*Lo
g[1 + Cosh[x]])/(16*a)

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Rubi [A]  time = 0.0857017, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3879, 88} \[ \frac{1}{8 a (1-\cosh (x))}+\frac{3}{4 a (\cosh (x)+1)}-\frac{1}{8 a (\cosh (x)+1)^2}+\frac{5 \log (1-\cosh (x))}{16 a}+\frac{11 \log (\cosh (x)+1)}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + a*Sech[x]),x]

[Out]

1/(8*a*(1 - Cosh[x])) - 1/(8*a*(1 + Cosh[x])^2) + 3/(4*a*(1 + Cosh[x])) + (5*Log[1 - Cosh[x]])/(16*a) + (11*Lo
g[1 + Cosh[x]])/(16*a)

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\coth ^3(x)}{a+a \text{sech}(x)} \, dx &=a^4 \operatorname{Subst}\left (\int \frac{x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\cosh (x)\right )\\ &=a^4 \operatorname{Subst}\left (\int \left (\frac{1}{8 a^5 (-1+x)^2}+\frac{5}{16 a^5 (-1+x)}+\frac{1}{4 a^5 (1+x)^3}-\frac{3}{4 a^5 (1+x)^2}+\frac{11}{16 a^5 (1+x)}\right ) \, dx,x,\cosh (x)\right )\\ &=\frac{1}{8 a (1-\cosh (x))}-\frac{1}{8 a (1+\cosh (x))^2}+\frac{3}{4 a (1+\cosh (x))}+\frac{5 \log (1-\cosh (x))}{16 a}+\frac{11 \log (1+\cosh (x))}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.1474, size = 66, normalized size = 0.97 \[ \frac{\text{sech}(x) \left (-2 \coth ^2\left (\frac{x}{2}\right )-\text{sech}^2\left (\frac{x}{2}\right )+4 \cosh ^2\left (\frac{x}{2}\right ) \left (5 \log \left (\sinh \left (\frac{x}{2}\right )\right )+11 \log \left (\cosh \left (\frac{x}{2}\right )\right )\right )+12\right )}{16 a (\text{sech}(x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(a + a*Sech[x]),x]

[Out]

((12 - 2*Coth[x/2]^2 + 4*Cosh[x/2]^2*(11*Log[Cosh[x/2]] + 5*Log[Sinh[x/2]]) - Sech[x/2]^2)*Sech[x])/(16*a*(1 +
Sech[x]))

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Maple [A]  time = 0.036, size = 69, normalized size = 1. \begin{align*} -{\frac{1}{32\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}}-{\frac{5}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{16\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{5}{8\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(a+a*sech(x)),x)

[Out]

-1/32/a*tanh(1/2*x)^4-5/16/a*tanh(1/2*x)^2-1/a*ln(tanh(1/2*x)+1)-1/16/a/tanh(1/2*x)^2+5/8/a*ln(tanh(1/2*x))-1/
a*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.1447, size = 146, normalized size = 2.15 \begin{align*} \frac{x}{a} + \frac{5 \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} - 14 \, e^{\left (-3 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{4 \,{\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac{11 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} + \frac{5 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a + 1/4*(5*e^(-x) - 6*e^(-2*x) - 14*e^(-3*x) - 6*e^(-4*x) + 5*e^(-5*x))/(2*a*e^(-x) - a*e^(-2*x) - 4*a*e^(-3
*x) - a*e^(-4*x) + 2*a*e^(-5*x) + a*e^(-6*x) + a) + 11/8*log(e^(-x) + 1)/a + 5/8*log(e^(-x) - 1)/a

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Fricas [B]  time = 2.4752, size = 2475, normalized size = 36.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*x*cosh(x)^6 + 8*x*sinh(x)^6 + 2*(8*x - 5)*cosh(x)^5 + 2*(24*x*cosh(x) + 8*x - 5)*sinh(x)^5 - 4*(2*x -
3)*cosh(x)^4 + 2*(60*x*cosh(x)^2 + 5*(8*x - 5)*cosh(x) - 4*x + 6)*sinh(x)^4 - 4*(8*x - 7)*cosh(x)^3 + 4*(40*x*
cosh(x)^3 + 5*(8*x - 5)*cosh(x)^2 - 4*(2*x - 3)*cosh(x) - 8*x + 7)*sinh(x)^3 - 4*(2*x - 3)*cosh(x)^2 + 4*(30*x
*cosh(x)^4 + 5*(8*x - 5)*cosh(x)^3 - 6*(2*x - 3)*cosh(x)^2 - 3*(8*x - 7)*cosh(x) - 2*x + 3)*sinh(x)^2 + 2*(8*x
- 5)*cosh(x) - 11*(cosh(x)^6 + 2*(3*cosh(x) + 1)*sinh(x)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cos
h(x) - 1)*sinh(x)^4 - cosh(x)^4 + 4*(5*cosh(x)^3 + 5*cosh(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*co
sh(x)^4 + 20*cosh(x)^3 - 6*cosh(x)^2 - 12*cosh(x) - 1)*sinh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 -
2*cosh(x)^3 - 6*cosh(x)^2 - cosh(x) + 1)*sinh(x) + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) - 5*(cosh(x)^6 +
2*(3*cosh(x) + 1)*sinh(x)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh(x)^4
+ 4*(5*cosh(x)^3 + 5*cosh(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6*cosh(
x)^2 - 12*cosh(x) - 1)*sinh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2 - cosh
(x) + 1)*sinh(x) + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) - 1) + 2*(24*x*cosh(x)^5 + 5*(8*x - 5)*cosh(x)^4 - 8*(
2*x - 3)*cosh(x)^3 - 6*(8*x - 7)*cosh(x)^2 - 4*(2*x - 3)*cosh(x) + 8*x - 5)*sinh(x) + 8*x)/(a*cosh(x)^6 + a*si
nh(x)^6 + 2*a*cosh(x)^5 + 2*(3*a*cosh(x) + a)*sinh(x)^5 - a*cosh(x)^4 + (15*a*cosh(x)^2 + 10*a*cosh(x) - a)*si
nh(x)^4 - 4*a*cosh(x)^3 + 4*(5*a*cosh(x)^3 + 5*a*cosh(x)^2 - a*cosh(x) - a)*sinh(x)^3 - a*cosh(x)^2 + (15*a*co
sh(x)^4 + 20*a*cosh(x)^3 - 6*a*cosh(x)^2 - 12*a*cosh(x) - a)*sinh(x)^2 + 2*a*cosh(x) + 2*(3*a*cosh(x)^5 + 5*a*
cosh(x)^4 - 2*a*cosh(x)^3 - 6*a*cosh(x)^2 - a*cosh(x) + a)*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\coth ^{3}{\left (x \right )}}{\operatorname{sech}{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**3/(a+a*sech(x)),x)

[Out]

Integral(coth(x)**3/(sech(x) + 1), x)/a

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Giac [A]  time = 1.15991, size = 127, normalized size = 1.87 \begin{align*} \frac{11 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} + \frac{5 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} - \frac{5 \, e^{\left (-x\right )} + 5 \, e^{x} - 6}{16 \, a{\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac{33 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 84 \, e^{\left (-x\right )} + 84 \, e^{x} + 52}{32 \, a{\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/16*log(e^(-x) + e^x + 2)/a + 5/16*log(e^(-x) + e^x - 2)/a - 1/16*(5*e^(-x) + 5*e^x - 6)/(a*(e^(-x) + e^x -
2)) - 1/32*(33*(e^(-x) + e^x)^2 + 84*e^(-x) + 84*e^x + 52)/(a*(e^(-x) + e^x + 2)^2)