3.126 \(\int \coth ^n(x) \text {csch}^4(x) \, dx\)

Optimal. Leaf size=26 \[ \frac {\coth ^{n+1}(x)}{n+1}-\frac {\coth ^{n+3}(x)}{n+3} \]

[Out]

coth(x)^(1+n)/(1+n)-coth(x)^(3+n)/(3+n)

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2607, 14} \[ \frac {\coth ^{n+1}(x)}{n+1}-\frac {\coth ^{n+3}(x)}{n+3} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^n*Csch[x]^4,x]

[Out]

Coth[x]^(1 + n)/(1 + n) - Coth[x]^(3 + n)/(3 + n)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

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

Rubi steps

\begin {align*} \int \coth ^n(x) \text {csch}^4(x) \, dx &=-\left (i \operatorname {Subst}\left (\int (-i x)^n \left (1+x^2\right ) \, dx,x,i \coth (x)\right )\right )\\ &=-\left (i \operatorname {Subst}\left (\int \left ((-i x)^n-(-i x)^{2+n}\right ) \, dx,x,i \coth (x)\right )\right )\\ &=\frac {\coth ^{1+n}(x)}{1+n}-\frac {\coth ^{3+n}(x)}{3+n}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 30, normalized size = 1.15 \[ \frac {\text {csch}^2(x) (-n+\cosh (2 x)-2) \coth ^{n+1}(x)}{(n+1) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^n*Csch[x]^4,x]

[Out]

((-2 - n + Cosh[2*x])*Coth[x]^(1 + n)*Csch[x]^2)/((1 + n)*(3 + n))

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fricas [B]  time = 0.46, size = 114, normalized size = 4.38 \[ \frac {2 \, {\left ({\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} - {\left (2 \, n + 3\right )} \cosh \relax (x)\right )} \cosh \left (n \log \left (\frac {\cosh \relax (x)}{\sinh \relax (x)}\right )\right ) + {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} - {\left (2 \, n + 3\right )} \cosh \relax (x)\right )} \sinh \left (n \log \left (\frac {\cosh \relax (x)}{\sinh \relax (x)}\right )\right )\right )}}{{\left (n^{2} + 4 \, n + 3\right )} \sinh \relax (x)^{3} + 3 \, {\left ({\left (n^{2} + 4 \, n + 3\right )} \cosh \relax (x)^{2} - n^{2} - 4 \, n - 3\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^n*csch(x)^4,x, algorithm="fricas")

[Out]

2*((cosh(x)^3 + 3*cosh(x)*sinh(x)^2 - (2*n + 3)*cosh(x))*cosh(n*log(cosh(x)/sinh(x))) + (cosh(x)^3 + 3*cosh(x)
*sinh(x)^2 - (2*n + 3)*cosh(x))*sinh(n*log(cosh(x)/sinh(x))))/((n^2 + 4*n + 3)*sinh(x)^3 + 3*((n^2 + 4*n + 3)*
cosh(x)^2 - n^2 - 4*n - 3)*sinh(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \coth \relax (x)^{n} \operatorname {csch}\relax (x)^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^n*csch(x)^4,x, algorithm="giac")

[Out]

integrate(coth(x)^n*csch(x)^4, x)

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maple [C]  time = 0.54, size = 371, normalized size = 14.27 \[ -\frac {2 \left (-{\mathrm e}^{6 x}+2 n \,{\mathrm e}^{4 x}+3 \,{\mathrm e}^{4 x}+2 n \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{2 x}-1\right ) {\mathrm e}^{\frac {n \left (-i \pi \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right )^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-1}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right )^{2} \mathrm {csgn}\left (i \left (1+{\mathrm e}^{2 x}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-1}\right ) \mathrm {csgn}\left (i \left (1+{\mathrm e}^{2 x}\right )\right )+i \pi \,\mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right ) \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{\left ({\mathrm e}^{x}+1\right ) \left ({\mathrm e}^{x}-1\right )}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{x}-1}\right ) \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{\left ({\mathrm e}^{x}+1\right ) \left ({\mathrm e}^{x}-1\right )}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+1}\right )-i \pi \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{\left ({\mathrm e}^{x}+1\right ) \left ({\mathrm e}^{x}-1\right )}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (1+{\mathrm e}^{2 x}\right )}{\left ({\mathrm e}^{x}+1\right ) \left ({\mathrm e}^{x}-1\right )}\right )^{2} \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+1}\right )-2 \ln \left ({\mathrm e}^{x}+1\right )+2 \ln \left (1+{\mathrm e}^{2 x}\right )-2 \ln \left ({\mathrm e}^{x}-1\right )\right )}{2}}}{\left (n +1\right ) \left (n +3\right ) \left ({\mathrm e}^{2 x}-1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^n*csch(x)^4,x)

[Out]

-2*(-exp(6*x)+2*n*exp(4*x)+3*exp(4*x)+2*n*exp(2*x)+3*exp(2*x)-1)/(n+1)/(n+3)/(exp(2*x)-1)^3*exp(1/2*n*(-I*Pi*c
sgn(I/(exp(x)-1)*(1+exp(2*x)))^3+I*Pi*csgn(I/(exp(x)-1)*(1+exp(2*x)))^2*csgn(I/(exp(x)-1))+I*Pi*csgn(I/(exp(x)
-1)*(1+exp(2*x)))^2*csgn(I*(1+exp(2*x)))-I*Pi*csgn(I/(exp(x)-1)*(1+exp(2*x)))*csgn(I/(exp(x)-1))*csgn(I*(1+exp
(2*x)))+I*Pi*csgn(I/(exp(x)-1)*(1+exp(2*x)))*csgn(I/(exp(x)+1)*(1+exp(2*x))/(exp(x)-1))^2-I*Pi*csgn(I/(exp(x)-
1)*(1+exp(2*x)))*csgn(I/(exp(x)+1)*(1+exp(2*x))/(exp(x)-1))*csgn(I/(exp(x)+1))-I*Pi*csgn(I/(exp(x)+1)*(1+exp(2
*x))/(exp(x)-1))^3+I*Pi*csgn(I/(exp(x)+1)*(1+exp(2*x))/(exp(x)-1))^2*csgn(I/(exp(x)+1))-2*ln(exp(x)+1)+2*ln(1+
exp(2*x))-2*ln(exp(x)-1)))

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maxima [B]  time = 0.45, size = 368, normalized size = 14.15 \[ -\frac {2 \, {\left (2 \, n + 3\right )} e^{\left (-n \log \left (e^{\left (-x\right )} + 1\right ) - n \log \left (-e^{\left (-x\right )} + 1\right ) + n \log \left (e^{\left (-2 \, x\right )} + 1\right ) - 2 \, x\right )}}{n^{2} - 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-4 \, x\right )} - {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-6 \, x\right )} + 4 \, n + 3} - \frac {2 \, {\left (2 \, n + 3\right )} e^{\left (-n \log \left (e^{\left (-x\right )} + 1\right ) - n \log \left (-e^{\left (-x\right )} + 1\right ) + n \log \left (e^{\left (-2 \, x\right )} + 1\right ) - 4 \, x\right )}}{n^{2} - 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-4 \, x\right )} - {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-6 \, x\right )} + 4 \, n + 3} + \frac {2 \, e^{\left (-n \log \left (e^{\left (-x\right )} + 1\right ) - n \log \left (-e^{\left (-x\right )} + 1\right ) + n \log \left (e^{\left (-2 \, x\right )} + 1\right ) - 6 \, x\right )}}{n^{2} - 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-4 \, x\right )} - {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-6 \, x\right )} + 4 \, n + 3} + \frac {2 \, e^{\left (-n \log \left (e^{\left (-x\right )} + 1\right ) - n \log \left (-e^{\left (-x\right )} + 1\right ) + n \log \left (e^{\left (-2 \, x\right )} + 1\right )\right )}}{n^{2} - 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-4 \, x\right )} - {\left (n^{2} + 4 \, n + 3\right )} e^{\left (-6 \, x\right )} + 4 \, n + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^n*csch(x)^4,x, algorithm="maxima")

[Out]

-2*(2*n + 3)*e^(-n*log(e^(-x) + 1) - n*log(-e^(-x) + 1) + n*log(e^(-2*x) + 1) - 2*x)/(n^2 - 3*(n^2 + 4*n + 3)*
e^(-2*x) + 3*(n^2 + 4*n + 3)*e^(-4*x) - (n^2 + 4*n + 3)*e^(-6*x) + 4*n + 3) - 2*(2*n + 3)*e^(-n*log(e^(-x) + 1
) - n*log(-e^(-x) + 1) + n*log(e^(-2*x) + 1) - 4*x)/(n^2 - 3*(n^2 + 4*n + 3)*e^(-2*x) + 3*(n^2 + 4*n + 3)*e^(-
4*x) - (n^2 + 4*n + 3)*e^(-6*x) + 4*n + 3) + 2*e^(-n*log(e^(-x) + 1) - n*log(-e^(-x) + 1) + n*log(e^(-2*x) + 1
) - 6*x)/(n^2 - 3*(n^2 + 4*n + 3)*e^(-2*x) + 3*(n^2 + 4*n + 3)*e^(-4*x) - (n^2 + 4*n + 3)*e^(-6*x) + 4*n + 3)
+ 2*e^(-n*log(e^(-x) + 1) - n*log(-e^(-x) + 1) + n*log(e^(-2*x) + 1))/(n^2 - 3*(n^2 + 4*n + 3)*e^(-2*x) + 3*(n
^2 + 4*n + 3)*e^(-4*x) - (n^2 + 4*n + 3)*e^(-6*x) + 4*n + 3)

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mupad [B]  time = 1.54, size = 87, normalized size = 3.35 \[ \frac {\left (\frac {4\,\mathrm {cosh}\left (3\,x\right )}{n^2+4\,n+3}-\frac {2\,\mathrm {cosh}\relax (x)\,\left (4\,n+6\right )}{n^2+4\,n+3}\right )\,{\left (\frac {{\mathrm {e}}^{2\,x}+1}{{\mathrm {e}}^{2\,x}-1}\right )}^n}{2\,\mathrm {sinh}\left (3\,x\right )-\frac {2\,\mathrm {sinh}\relax (x)\,\left (3\,n^2+12\,n+9\right )}{n^2+4\,n+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^n/sinh(x)^4,x)

[Out]

(((4*cosh(3*x))/(4*n + n^2 + 3) - (2*cosh(x)*(4*n + 6))/(4*n + n^2 + 3))*((exp(2*x) + 1)/(exp(2*x) - 1))^n)/(2
*sinh(3*x) - (2*sinh(x)*(12*n + 3*n^2 + 9))/(4*n + n^2 + 3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \coth ^{n}{\relax (x )} \operatorname {csch}^{4}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**n*csch(x)**4,x)

[Out]

Integral(coth(x)**n*csch(x)**4, x)

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