\(\int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx\) [111]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {1}{8 a (1-\cosh (x))}-\frac {1}{8 a (1+\cosh (x))^2}+\frac {3}{4 a (1+\cosh (x))}+\frac {5 \log (1-\cosh (x))}{16 a}+\frac {11 \log (1+\cosh (x))}{16 a} \]

[Out]

1/8/a/(1-cosh(x))-1/8/a/(1+cosh(x))^2+3/4/a/(1+cosh(x))+5/16*ln(1-cosh(x))/a+11/16*ln(1+cosh(x))/a

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3964, 90} \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {1}{8 a (1-\cosh (x))}+\frac {3}{4 a (\cosh (x)+1)}-\frac {1}{8 a (\cosh (x)+1)^2}+\frac {5 \log (1-\cosh (x))}{16 a}+\frac {11 \log (\cosh (x)+1)}{16 a} \]

[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 90

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]))

Rule 3964

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]

Rubi steps \begin{align*} \text {integral}& = a^4 \text {Subst}\left (\int \frac {x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\cosh (x)\right ) \\ & = a^4 \text {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] (verified)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {\left (12-2 \coth ^2\left (\frac {x}{2}\right )+4 \cosh ^2\left (\frac {x}{2}\right ) \left (11 \log \left (\cosh \left (\frac {x}{2}\right )\right )+5 \log \left (\sinh \left (\frac {x}{2}\right )\right )\right )-\text {sech}^2\left (\frac {x}{2}\right )\right ) \text {sech}(x)}{16 a (1+\text {sech}(x))} \]

[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]))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {-\frac {\tanh \left (\frac {x}{2}\right )^{4}}{4}-\frac {5 \tanh \left (\frac {x}{2}\right )^{2}}{2}-8 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-8 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {1}{2 \tanh \left (\frac {x}{2}\right )^{2}}+5 \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{8 a}\)	\(56\)
risch	\(-\frac {x}{a}+\frac {{\mathrm e}^{x} \left (5 \,{\mathrm e}^{4 x}-6 \,{\mathrm e}^{3 x}-14 \,{\mathrm e}^{2 x}-6 \,{\mathrm e}^{x}+5\right )}{4 a \left ({\mathrm e}^{x}-1\right )^{2} \left ({\mathrm e}^{x}+1\right )^{4}}+\frac {11 \ln \left ({\mathrm e}^{x}+1\right )}{8 a}+\frac {5 \ln \left ({\mathrm e}^{x}-1\right )}{8 a}\)	\(71\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 773 vs. \(2 (56) = 112\).

Time = 0.27 (sec) , antiderivative size = 773, normalized size of antiderivative = 11.37 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\text {Too large to display} \]

[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)

Sympy [F]

\[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\coth ^{3}{\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.59 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} + \frac {5 \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} - 14 \, e^{\left (-3 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )}}{4 \, {\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac {11 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.38 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {11 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} - \frac {5 \, e^{\left (-x\right )} + 5 \, e^{x} - 6}{16 \, a {\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac {33 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 84 \, e^{\left (-x\right )} + 84 \, e^{x} + 52}{32 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.35 \[ \int \frac {\coth ^3(x)}{a+a \text {sech}(x)} \, dx=\frac {\ln \left (9\,{\mathrm {e}}^{2\,x}-9\right )}{a}-\frac {x}{a}-\frac {1}{2\,\left (a+4\,a\,{\mathrm {e}}^x+6\,a\,{\mathrm {e}}^{2\,x}+4\,a\,{\mathrm {e}}^{3\,x}+a\,{\mathrm {e}}^{4\,x}\right )}+\frac {1}{a+3\,a\,{\mathrm {e}}^x+3\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{3\,x}}-\frac {1}{4\,\left (a-2\,a\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}\right )}-\frac {2}{a+2\,a\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}}+\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{4\,\sqrt {-a^2}}+\frac {3}{2\,\left (a+a\,{\mathrm {e}}^x\right )}+\frac {1}{4\,\left (a-a\,{\mathrm {e}}^x\right )} \]

[In]

int(coth(x)^3/(a + a/cosh(x)),x)

[Out]

log(9*exp(2*x) - 9)/a - x/a - 1/(2*(a + 4*a*exp(x) + 6*a*exp(2*x) + 4*a*exp(3*x) + a*exp(4*x))) + 1/(a + 3*a*e
xp(x) + 3*a*exp(2*x) + a*exp(3*x)) - 1/(4*(a - 2*a*exp(x) + a*exp(2*x))) - 2/(a + 2*a*exp(x) + a*exp(2*x)) + (
3*atan((exp(x)*(-a^2)^(1/2))/a))/(4*(-a^2)^(1/2)) + 3/(2*(a + a*exp(x))) + 1/(4*(a - a*exp(x)))