3.664 \(\int (-\coth (x)+\text {csch}(x))^5 \, dx\)

Optimal. Leaf size=24 \[ -\frac {4}{\cosh (x)+1}+\frac {2}{(\cosh (x)+1)^2}-\log (\cosh (x)+1) \]

[Out]

2/(1+cosh(x))^2-4/(1+cosh(x))-ln(1+cosh(x))

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4392, 2667, 43} \[ -\frac {4}{\cosh (x)+1}+\frac {2}{(\cosh (x)+1)^2}-\log (\cosh (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[(-Coth[x] + Csch[x])^5,x]

[Out]

2/(1 + Cosh[x])^2 - 4/(1 + Cosh[x]) - Log[1 + Cosh[x]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

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

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Mathematica [B]  time = 0.09, size = 55, normalized size = 2.29 \[ \frac {1}{2} \text {sech}^4\left (\frac {x}{2}\right )-2 \text {sech}^2\left (\frac {x}{2}\right )+6 \log \left (\sinh \left (\frac {x}{2}\right )\right )-\log (\sinh (x))-5 \log \left (\tanh \left (\frac {x}{2}\right )\right )-6 \log \left (\cosh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-Coth[x] + Csch[x])^5,x]

[Out]

-6*Log[Cosh[x/2]] + 6*Log[Sinh[x/2]] - Log[Sinh[x]] - 5*Log[Tanh[x/2]] - 2*Sech[x/2]^2 + Sech[x/2]^4/2

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fricas [B]  time = 0.40, size = 265, normalized size = 11.04 \[ \frac {x \cosh \relax (x)^{4} + x \sinh \relax (x)^{4} + 4 \, {\left (x - 2\right )} \cosh \relax (x)^{3} + 4 \, {\left (x \cosh \relax (x) + x - 2\right )} \sinh \relax (x)^{3} + 2 \, {\left (3 \, x - 4\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, x \cosh \relax (x)^{2} + 6 \, {\left (x - 2\right )} \cosh \relax (x) + 3 \, x - 4\right )} \sinh \relax (x)^{2} + 4 \, {\left (x - 2\right )} \cosh \relax (x) - 2 \, {\left (\cosh \relax (x)^{4} + 4 \, {\left (\cosh \relax (x) + 1\right )} \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} + 6 \, {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) + 1\right )} \sinh \relax (x)^{2} + 6 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x)^{2} + 3 \, \cosh \relax (x) + 1\right )} \sinh \relax (x) + 4 \, \cosh \relax (x) + 1\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + 4 \, {\left (x \cosh \relax (x)^{3} + 3 \, {\left (x - 2\right )} \cosh \relax (x)^{2} + {\left (3 \, x - 4\right )} \cosh \relax (x) + x - 2\right )} \sinh \relax (x) + x}{\cosh \relax (x)^{4} + 4 \, {\left (\cosh \relax (x) + 1\right )} \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} + 6 \, {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) + 1\right )} \sinh \relax (x)^{2} + 6 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x)^{2} + 3 \, \cosh \relax (x) + 1\right )} \sinh \relax (x) + 4 \, \cosh \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-coth(x)+csch(x))^5,x, algorithm="fricas")

[Out]

(x*cosh(x)^4 + x*sinh(x)^4 + 4*(x - 2)*cosh(x)^3 + 4*(x*cosh(x) + x - 2)*sinh(x)^3 + 2*(3*x - 4)*cosh(x)^2 + 2
*(3*x*cosh(x)^2 + 6*(x - 2)*cosh(x) + 3*x - 4)*sinh(x)^2 + 4*(x - 2)*cosh(x) - 2*(cosh(x)^4 + 4*(cosh(x) + 1)*
sinh(x)^3 + sinh(x)^4 + 4*cosh(x)^3 + 6*(cosh(x)^2 + 2*cosh(x) + 1)*sinh(x)^2 + 6*cosh(x)^2 + 4*(cosh(x)^3 + 3
*cosh(x)^2 + 3*cosh(x) + 1)*sinh(x) + 4*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) + 4*(x*cosh(x)^3 + 3*(x - 2)*c
osh(x)^2 + (3*x - 4)*cosh(x) + x - 2)*sinh(x) + x)/(cosh(x)^4 + 4*(cosh(x) + 1)*sinh(x)^3 + sinh(x)^4 + 4*cosh
(x)^3 + 6*(cosh(x)^2 + 2*cosh(x) + 1)*sinh(x)^2 + 6*cosh(x)^2 + 4*(cosh(x)^3 + 3*cosh(x)^2 + 3*cosh(x) + 1)*si
nh(x) + 4*cosh(x) + 1)

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giac [A]  time = 0.13, size = 28, normalized size = 1.17 \[ x - \frac {8 \, {\left (e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x}\right )}}{{\left (e^{x} + 1\right )}^{4}} - 2 \, \log \left (e^{x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-coth(x)+csch(x))^5,x, algorithm="giac")

[Out]

x - 8*(e^(3*x) + e^(2*x) + e^x)/(e^x + 1)^4 - 2*log(e^x + 1)

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maple [B]  time = 0.35, size = 73, normalized size = 3.04 \[ -\ln \left (\sinh \relax (x )\right )+\frac {\left (\coth ^{2}\relax (x )\right )}{2}+\frac {\left (\coth ^{4}\relax (x )\right )}{4}-\frac {5 \left (\cosh ^{3}\relax (x )\right )}{\sinh \relax (x )^{4}}+\frac {5 \cosh \relax (x )}{3 \sinh \relax (x )^{4}}+\frac {8 \left (-\frac {\mathrm {csch}\relax (x )^{3}}{4}+\frac {3 \,\mathrm {csch}\relax (x )}{8}\right ) \coth \relax (x )}{3}-2 \arctanh \left ({\mathrm e}^{x}\right )+\frac {5 \left (\cosh ^{2}\relax (x )\right )}{\sinh \relax (x )^{4}}-\frac {5}{4 \sinh \relax (x )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-coth(x)+csch(x))^5,x)

[Out]

-ln(sinh(x))+1/2*coth(x)^2+1/4*coth(x)^4-5/sinh(x)^4*cosh(x)^3+5/3/sinh(x)^4*cosh(x)+8/3*(-1/4*csch(x)^3+3/8*c
sch(x))*coth(x)-2*arctanh(exp(x))+5/sinh(x)^4*cosh(x)^2-5/4/sinh(x)^4

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maxima [B]  time = 0.35, size = 238, normalized size = 9.92 \[ \frac {5}{2} \, \coth \relax (x)^{4} - x + \frac {5 \, {\left (5 \, e^{\left (-x\right )} + 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-7 \, x\right )}\right )}}{4 \, {\left (4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} - \frac {3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}}{4 \, {\left (4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} + \frac {5 \, {\left (e^{\left (-x\right )} + 7 \, e^{\left (-3 \, x\right )} + 7 \, e^{\left (-5 \, x\right )} + e^{\left (-7 \, x\right )}\right )}}{2 \, {\left (4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} - \frac {4 \, {\left (e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \frac {20}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{4}} - 2 \, \log \left (e^{\left (-x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-coth(x)+csch(x))^5,x, algorithm="maxima")

[Out]

5/2*coth(x)^4 - x + 5/4*(5*e^(-x) + 3*e^(-3*x) + 3*e^(-5*x) + 5*e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x
) - e^(-8*x) - 1) - 1/4*(3*e^(-x) - 11*e^(-3*x) - 11*e^(-5*x) + 3*e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6
*x) - e^(-8*x) - 1) + 5/2*(e^(-x) + 7*e^(-3*x) + 7*e^(-5*x) + e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x)
- e^(-8*x) - 1) - 4*(e^(-2*x) - e^(-4*x) + e^(-6*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1) + 2
0/(e^(-x) - e^x)^4 - 2*log(e^(-x) + 1)

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mupad [B]  time = 1.50, size = 77, normalized size = 3.21 \[ x-2\,\ln \left ({\mathrm {e}}^x+1\right )+\frac {16}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1}+\frac {8}{6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1}-\frac {8}{{\mathrm {e}}^x+1}-\frac {16}{3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(coth(x) - 1/sinh(x))^5,x)

[Out]

x - 2*log(exp(x) + 1) + 16/(exp(2*x) + 2*exp(x) + 1) + 8/(6*exp(2*x) + 4*exp(3*x) + exp(4*x) + 4*exp(x) + 1) -
8/(exp(x) + 1) - 16/(3*exp(2*x) + exp(3*x) + 3*exp(x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int 5 \coth {\relax (x )} \operatorname {csch}^{4}{\relax (x )}\, dx - \int \left (- 10 \coth ^{2}{\relax (x )} \operatorname {csch}^{3}{\relax (x )}\right )\, dx - \int 10 \coth ^{3}{\relax (x )} \operatorname {csch}^{2}{\relax (x )}\, dx - \int \left (- 5 \coth ^{4}{\relax (x )} \operatorname {csch}{\relax (x )}\right )\, dx - \int \coth ^{5}{\relax (x )}\, dx - \int \left (- \operatorname {csch}^{5}{\relax (x )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-coth(x)+csch(x))**5,x)

[Out]

-Integral(5*coth(x)*csch(x)**4, x) - Integral(-10*coth(x)**2*csch(x)**3, x) - Integral(10*coth(x)**3*csch(x)**
2, x) - Integral(-5*coth(x)**4*csch(x), x) - Integral(coth(x)**5, x) - Integral(-csch(x)**5, x)

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