3.111 \(\int \frac {\coth ^5(x)}{i+\text {csch}(x)} \, dx\)

Optimal. Leaf size=30 \[ -\frac {1}{3} \text {csch}^3(x)+\frac {1}{2} i \text {csch}^2(x)-\text {csch}(x)-i \log (\sinh (x)) \]

[Out]

-csch(x)+1/2*I*csch(x)^2-1/3*csch(x)^3-I*ln(sinh(x))

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.00, 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, 75} \[ -\frac {1}{3} \text {csch}^3(x)+\frac {1}{2} i \text {csch}^2(x)-\text {csch}(x)-i \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^5/(I + Csch[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2 - Csch[x]^3/3 - I*Log[Sinh[x]]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.00 \[ -\frac {1}{3} \text {csch}^3(x)+\frac {1}{2} i \text {csch}^2(x)-\text {csch}(x)-i \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^5/(I + Csch[x]),x]

[Out]

-Csch[x] + (I/2)*Csch[x]^2 - Csch[x]^3/3 - I*Log[Sinh[x]]

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fricas [B]  time = 2.11, size = 97, normalized size = 3.23 \[ \frac {3 i \, x e^{\left (6 \, x\right )} + {\left (-9 i \, x + 6 i\right )} e^{\left (4 \, x\right )} + {\left (9 i \, x - 6 i\right )} e^{\left (2 \, x\right )} + {\left (-3 i \, e^{\left (6 \, x\right )} + 9 i \, e^{\left (4 \, x\right )} - 9 i \, e^{\left (2 \, x\right )} + 3 i\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) - 3 i \, x - 6 \, e^{\left (5 \, x\right )} + 4 \, e^{\left (3 \, x\right )} - 6 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} - 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*I*x*e^(6*x) + (-9*I*x + 6*I)*e^(4*x) + (9*I*x - 6*I)*e^(2*x) + (-3*I*e^(6*x) + 9*I*e^(4*x) - 9*I*e^(2*x
) + 3*I)*log(e^(2*x) - 1) - 3*I*x - 6*e^(5*x) + 4*e^(3*x) - 6*e^x)/(e^(6*x) - 3*e^(4*x) + 3*e^(2*x) - 1)

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giac [B]  time = 0.14, size = 68, normalized size = 2.27 \[ -\frac {11 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 12 i \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 12 \, e^{\left (-x\right )} - 12 \, e^{x} - 16 i}{6 \, {\left (-i \, e^{\left (-x\right )} + i \, e^{x}\right )}^{3}} - i \, \log \left (-i \, e^{\left (-x\right )} + i \, e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(11*(e^(-x) - e^x)^3 - 12*I*(e^(-x) - e^x)^2 + 12*e^(-x) - 12*e^x - 16*I)/(-I*e^(-x) + I*e^x)^3 - I*log(-
I*e^(-x) + I*e^x)

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maple [B]  time = 0.23, size = 78, normalized size = 2.60 \[ \frac {3 \tanh \left (\frac {x}{2}\right )}{8}+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24}+\frac {i \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}+i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {1}{24 \tanh \left (\frac {x}{2}\right )^{3}}-i \ln \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {i}{8 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {3}{8 \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5/(I+csch(x)),x)

[Out]

3/8*tanh(1/2*x)+1/24*tanh(1/2*x)^3+1/8*I*tanh(1/2*x)^2+I*ln(tanh(1/2*x)-1)+I*ln(tanh(1/2*x)+1)-1/24/tanh(1/2*x
)^3-I*ln(tanh(1/2*x))+1/8*I/tanh(1/2*x)^2-3/8/tanh(1/2*x)

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maxima [B]  time = 0.31, size = 75, normalized size = 2.50 \[ -i \, x + \frac {6 \, e^{\left (-x\right )} - 6 i \, e^{\left (-2 \, x\right )} - 4 \, e^{\left (-3 \, x\right )} + 6 i \, e^{\left (-4 \, x\right )} + 6 \, e^{\left (-5 \, x\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - i \, \log \left (e^{\left (-x\right )} + 1\right ) - i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x + 1/3*(6*e^(-x) - 6*I*e^(-2*x) - 4*e^(-3*x) + 6*I*e^(-4*x) + 6*e^(-5*x))/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6
*x) - 1) - I*log(e^(-x) + 1) - I*log(e^(-x) - 1)

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mupad [B]  time = 1.68, size = 81, normalized size = 2.70 \[ x\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^{2\,x}-1\right )\,1{}\mathrm {i}-\frac {8\,{\mathrm {e}}^x}{3\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {\frac {8\,{\mathrm {e}}^x}{3}-2{}\mathrm {i}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {2\,{\mathrm {e}}^x-2{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5/(1/sinh(x) + 1i),x)

[Out]

x*1i - log(exp(2*x) - 1)*1i - (8*exp(x))/(3*(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1)) - ((8*exp(x))/3 - 2i)/(e
xp(4*x) - 2*exp(2*x) + 1) - (2*exp(x) - 2i)/(exp(2*x) - 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{5}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**5/(I+csch(x)),x)

[Out]

Integral(coth(x)**5/(csch(x) + I), x)

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