\(\int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx\) [224]

Optimal result

Integrand size = 13, antiderivative size = 29 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=2 i \text {csch}(x)+\frac {\text {csch}^2(x)}{2}+2 \log (\sinh (x))-2 \log (i+\sinh (x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, 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 = {2786, 78} \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=\frac {\text {csch}^2(x)}{2}+2 i \text {csch}(x)+2 \log (\sinh (x))-2 \log (\sinh (x)+i) \]

[In]

Int[Coth[x]^3/(I + Sinh[x])^2,x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=2 i \text {csch}(x)+\frac {\text {csch}^2(x)}{2}+2 \log (\sinh (x))-2 \log (i+\sinh (x)) \]

[In]

Integrate[Coth[x]^3/(I + Sinh[x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 21.47 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55

method	result	size
risch	\(\frac {2 i {\mathrm e}^{x} \left (2 \,{\mathrm e}^{2 x}-2-i {\mathrm e}^{x}\right )}{\left ({\mathrm e}^{2 x}-1\right )^{2}}+2 \ln \left ({\mathrm e}^{2 x}-1\right )-4 \ln \left ({\mathrm e}^{x}+i\right )\)	\(45\)
default	\(-i \tanh \left (\frac {x}{2}\right )+\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8}-4 \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )+\frac {i}{\tanh \left (\frac {x}{2}\right )}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+2 \ln \left (\tanh \left (\frac {x}{2}\right )\right )\)	\(51\)

[In]

int(coth(x)^3/(I+sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

2*I*exp(x)*(2*exp(2*x)-2-I*exp(x))/(exp(2*x)-1)^2+2*ln(exp(2*x)-1)-4*ln(exp(x)+I)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (23) = 46\).

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=\frac {2 \, {\left ({\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) - 2 \, {\left (e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right )} \log \left (e^{x} + i\right ) + 2 i \, e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - 2 i \, e^{x}\right )}}{e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1} \]

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=\frac {4 i e^{3 x} + 2 e^{2 x} - 4 i e^{x}}{e^{4 x} - 2 e^{2 x} + 1} - 4 \log {\left (e^{x} + i \right )} + 2 \log {\left (e^{2 x} - 1 \right )} \]

[In]

integrate(coth(x)**3/(I+sinh(x))**2,x)

[Out]

(4*I*exp(3*x) + 2*exp(2*x) - 4*I*exp(x))/(exp(4*x) - 2*exp(2*x) + 1) - 4*log(exp(x) + I) + 2*log(exp(2*x) - 1)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (23) = 46\).

Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (2 i \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 2 i \, e^{\left (-3 \, x\right )}\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + 2 \, \log \left (e^{\left (-x\right )} + 1\right ) - 4 \, \log \left (e^{\left (-x\right )} - i\right ) + 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (-2 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + 2 i \, e^{x}\right )}}{{\left (e^{x} + 1\right )}^{2} {\left (e^{x} - 1\right )}^{2}} + 2 \, \log \left (e^{x} + 1\right ) - 4 \, \log \left (e^{x} + i\right ) + 2 \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

integrate(coth(x)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {\coth ^3(x)}{(i+\sinh (x))^2} \, dx=\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}+2\,\ln \left (-{\mathrm {e}}^{2\,x}\,6{}\mathrm {i}+6{}\mathrm {i}\right )-4\,\ln \left (144\,{\mathrm {e}}^x+144{}\mathrm {i}\right )+\frac {2+{\mathrm {e}}^x\,4{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1} \]

[In]

int(coth(x)^3/(sinh(x) + 1i)^2,x)

[Out]

2*log(6i - exp(2*x)*6i) - 4*log(144*exp(x) + 144i) + 2/(exp(4*x) - 2*exp(2*x) + 1) + (exp(x)*4i + 2)/(exp(2*x)
- 1)