\(\int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx\) [89]

Optimal result

Integrand size = 11, antiderivative size = 60 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\frac {5 x}{16}+\frac {1}{32 (1-\coth (x))^2}+\frac {1}{8 (1-\coth (x))}-\frac {1}{24 (1+\coth (x))^3}-\frac {3}{32 (1+\coth (x))^2}-\frac {3}{16 (1+\coth (x))} \]

[Out]

5/16*x+1/32/(1-coth(x))^2+1/8/(1-coth(x))-1/24/(1+coth(x))^3-3/32/(1+coth(x))^2-3/16/(1+coth(x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3568, 46, 213} \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\frac {5 x}{16}+\frac {1}{8 (1-\coth (x))}-\frac {3}{16 (\coth (x)+1)}+\frac {1}{32 (1-\coth (x))^2}-\frac {3}{32 (\coth (x)+1)^2}-\frac {1}{24 (\coth (x)+1)^3} \]

[In]

Int[Sinh[x]^4/(1 + Coth[x]),x]

[Out]

(5*x)/16 + 1/(32*(1 - Coth[x])^2) + 1/(8*(1 - Coth[x])) - 1/(24*(1 + Coth[x])^3) - 3/(32*(1 + Coth[x])^2) - 3/
(16*(1 + Coth[x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{(1-x)^3 (1+x)^4} \, dx,x,\coth (x)\right ) \\ & = \text {Subst}\left (\int \left (-\frac {1}{16 (-1+x)^3}+\frac {1}{8 (-1+x)^2}+\frac {1}{8 (1+x)^4}+\frac {3}{16 (1+x)^3}+\frac {3}{16 (1+x)^2}-\frac {5}{16 \left (-1+x^2\right )}\right ) \, dx,x,\coth (x)\right ) \\ & = \frac {1}{32 (1-\coth (x))^2}+\frac {1}{8 (1-\coth (x))}-\frac {1}{24 (1+\coth (x))^3}-\frac {3}{32 (1+\coth (x))^2}-\frac {3}{16 (1+\coth (x))}-\frac {5}{16} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\coth (x)\right ) \\ & = \frac {5 x}{16}+\frac {1}{32 (1-\coth (x))^2}+\frac {1}{8 (1-\coth (x))}-\frac {1}{24 (1+\coth (x))^3}-\frac {3}{32 (1+\coth (x))^2}-\frac {3}{16 (1+\coth (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\frac {1}{192} (60 x+15 \cosh (2 x)-6 \cosh (4 x)+\cosh (6 x)-45 \sinh (2 x)+9 \sinh (4 x)-\sinh (6 x)) \]

[In]

Integrate[Sinh[x]^4/(1 + Coth[x]),x]

[Out]

(60*x + 15*Cosh[2*x] - 6*Cosh[4*x] + Cosh[6*x] - 45*Sinh[2*x] + 9*Sinh[4*x] - Sinh[6*x])/192

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58

method	result	size
risch	\(\frac {5 x}{16}+\frac {{\mathrm e}^{4 x}}{128}-\frac {5 \,{\mathrm e}^{2 x}}{64}+\frac {5 \,{\mathrm e}^{-2 x}}{32}-\frac {5 \,{\mathrm e}^{-4 x}}{128}+\frac {{\mathrm e}^{-6 x}}{192}\)	\(35\)
parallelrisch	\(\frac {19}{96}-\frac {\cosh \left (4 x \right )}{32}+\frac {\cosh \left (6 x \right )}{192}+\frac {5 \cosh \left (2 x \right )}{64}+\frac {3 \sinh \left (4 x \right )}{64}-\frac {15 \sinh \left (2 x \right )}{64}-\frac {\sinh \left (6 x \right )}{192}-\frac {5 \ln \left (1-\tanh \left (x \right )\right )}{32}+\frac {5 \ln \left (1+\tanh \left (x \right )\right )}{32}\)	\(55\)
default	\(\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{6}}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {5}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {5}{12 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{16}+\frac {1}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{16}\)	\(110\)

[In]

int(sinh(x)^4/(1+coth(x)),x,method=_RETURNVERBOSE)

[Out]

5/16*x+1/128*exp(4*x)-5/64*exp(2*x)+5/32*exp(-2*x)-5/128*exp(-4*x)+1/192*exp(-6*x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (44) = 88\).

Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.55 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\frac {5 \, \cosh \left (x\right )^{5} + 25 \, \cosh \left (x\right ) \sinh \left (x\right )^{4} + \sinh \left (x\right )^{5} + 5 \, {\left (2 \, \cosh \left (x\right )^{2} - 3\right )} \sinh \left (x\right )^{3} - 45 \, \cosh \left (x\right )^{3} + 5 \, {\left (10 \, \cosh \left (x\right )^{3} - 27 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 60 \, {\left (2 \, x + 1\right )} \cosh \left (x\right ) + 5 \, {\left (\cosh \left (x\right )^{4} - 9 \, \cosh \left (x\right )^{2} + 24 \, x - 12\right )} \sinh \left (x\right )}{384 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]

[In]

integrate(sinh(x)^4/(1+coth(x)),x, algorithm="fricas")

[Out]

1/384*(5*cosh(x)^5 + 25*cosh(x)*sinh(x)^4 + sinh(x)^5 + 5*(2*cosh(x)^2 - 3)*sinh(x)^3 - 45*cosh(x)^3 + 5*(10*c
osh(x)^3 - 27*cosh(x))*sinh(x)^2 + 60*(2*x + 1)*cosh(x) + 5*(cosh(x)^4 - 9*cosh(x)^2 + 24*x - 12)*sinh(x))/(co
sh(x) + sinh(x))

Sympy [F]

\[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\int \frac {\sinh ^{4}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \]

[In]

integrate(sinh(x)**4/(1+coth(x)),x)

[Out]

Integral(sinh(x)**4/(coth(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=-\frac {1}{128} \, {\left (10 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (4 \, x\right )} + \frac {5}{16} \, x + \frac {5}{32} \, e^{\left (-2 \, x\right )} - \frac {5}{128} \, e^{\left (-4 \, x\right )} + \frac {1}{192} \, e^{\left (-6 \, x\right )} \]

[In]

integrate(sinh(x)^4/(1+coth(x)),x, algorithm="maxima")

[Out]

-1/128*(10*e^(-2*x) - 1)*e^(4*x) + 5/16*x + 5/32*e^(-2*x) - 5/128*e^(-4*x) + 1/192*e^(-6*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=-\frac {1}{384} \, {\left (110 \, e^{\left (6 \, x\right )} - 60 \, e^{\left (4 \, x\right )} + 15 \, e^{\left (2 \, x\right )} - 2\right )} e^{\left (-6 \, x\right )} + \frac {5}{16} \, x + \frac {1}{128} \, e^{\left (4 \, x\right )} - \frac {5}{64} \, e^{\left (2 \, x\right )} \]

[In]

integrate(sinh(x)^4/(1+coth(x)),x, algorithm="giac")

[Out]

-1/384*(110*e^(6*x) - 60*e^(4*x) + 15*e^(2*x) - 2)*e^(-6*x) + 5/16*x + 1/128*e^(4*x) - 5/64*e^(2*x)

Mupad [B] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.57 \[ \int \frac {\sinh ^4(x)}{1+\coth (x)} \, dx=\frac {5\,x}{16}+\frac {5\,{\mathrm {e}}^{-2\,x}}{32}-\frac {5\,{\mathrm {e}}^{2\,x}}{64}-\frac {5\,{\mathrm {e}}^{-4\,x}}{128}+\frac {{\mathrm {e}}^{4\,x}}{128}+\frac {{\mathrm {e}}^{-6\,x}}{192} \]

[In]

int(sinh(x)^4/(coth(x) + 1),x)

[Out]

(5*x)/16 + (5*exp(-2*x))/32 - (5*exp(2*x))/64 - (5*exp(-4*x))/128 + exp(4*x)/128 + exp(-6*x)/192