\(\int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx\) [52]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\text {arctanh}(\cosh (x))+\frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {4 i \cosh (x)}{3 (i+\sinh (x))} \]

[Out]

arctanh(cosh(x))+1/3*cosh(x)/(I+sinh(x))^2-4/3*I*cosh(x)/(I+sinh(x))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2845, 3057, 12, 3855} \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\text {arctanh}(\cosh (x))-\frac {4 i \cosh (x)}{3 (\sinh (x)+i)}+\frac {\cosh (x)}{3 (\sinh (x)+i)^2} \]

[In]

Int[Csch[x]/(I + Sinh[x])^2,x]

[Out]

ArcTanh[Cosh[x]] + Cosh[x]/(3*(I + Sinh[x])^2) - (((4*I)/3)*Cosh[x])/(I + Sinh[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {1}{3} \int \frac {\text {csch}(x) (3 i-\sinh (x))}{i+\sinh (x)} \, dx \\ & = \frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {4 i \cosh (x)}{3 (i+\sinh (x))}+\frac {1}{3} i \int 3 i \text {csch}(x) \, dx \\ & = \frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {4 i \cosh (x)}{3 (i+\sinh (x))}-\int \text {csch}(x) \, dx \\ & = \text {arctanh}(\cosh (x))+\frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {4 i \cosh (x)}{3 (i+\sinh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\log \left (\cosh \left (\frac {x}{2}\right )\right )-\log \left (\sinh \left (\frac {x}{2}\right )\right )-\frac {i}{3 i+3 \sinh (x)}-\frac {2 \sinh \left (\frac {x}{2}\right ) (5 i+4 \sinh (x))}{3 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^3} \]

[In]

Integrate[Csch[x]/(I + Sinh[x])^2,x]

[Out]

Log[Cosh[x/2]] - Log[Sinh[x/2]] - I/(3*I + 3*Sinh[x]) - (2*Sinh[x/2]*(5*I + 4*Sinh[x]))/(3*(Cosh[x/2] - I*Sinh
[x/2])^3)

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {2 \left (9 i {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x}-4\right )}{3 \left ({\mathrm e}^{x}+i\right )^{3}}+\ln \left ({\mathrm e}^{x}+1\right )-\ln \left ({\mathrm e}^{x}-1\right )\)	\(36\)
default	\(-\ln \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {4 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {4 i}{\tanh \left (\frac {x}{2}\right )+i}-\frac {2}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}\)	\(44\)
parallelrisch	\(\frac {\left (-3 \tanh \left (\frac {x}{2}\right )^{3}-9 i \tanh \left (\frac {x}{2}\right )^{2}+9 \tanh \left (\frac {x}{2}\right )+3 i\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )+4 \tanh \left (\frac {x}{2}\right )^{3}+6 i+6 \tanh \left (\frac {x}{2}\right )}{3 \tanh \left (\frac {x}{2}\right )^{3}+9 i \tanh \left (\frac {x}{2}\right )^{2}-9 \tanh \left (\frac {x}{2}\right )-3 i}\)	\(79\)

[In]

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

[Out]

-2/3*(9*I*exp(x)+3*exp(2*x)-4)/(exp(x)+I)^3+ln(exp(x)+1)-ln(exp(x)-1)

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. 78 vs. \(2 (24) = 48\).

Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.29 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\frac {3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} + 1\right ) - 3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} - 1\right ) - 6 \, e^{\left (2 \, x\right )} - 18 i \, e^{x} + 8}{3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \]

[In]

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

[Out]

1/3*(3*(e^(3*x) + 3*I*e^(2*x) - 3*e^x - I)*log(e^x + 1) - 3*(e^(3*x) + 3*I*e^(2*x) - 3*e^x - I)*log(e^x - 1) -
6*e^(2*x) - 18*I*e^x + 8)/(e^(3*x) + 3*I*e^(2*x) - 3*e^x - I)

Sympy [F]

\[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\int \frac {\operatorname {csch}{\left (x \right )}}{\left (\sinh {\left (x \right )} + i\right )^{2}}\, dx \]

[In]

integrate(csch(x)/(I+sinh(x))**2,x)

[Out]

Integral(csch(x)/(sinh(x) + I)**2, x)

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. 55 vs. \(2 (24) = 48\).

Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\frac {2 \, {\left (-9 i \, e^{\left (-x\right )} + 3 \, e^{\left (-2 \, x\right )} - 4\right )}}{3 \, {\left (3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i\right )}} + \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, e^{\left (2 \, x\right )} + 9 i \, e^{x} - 4\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} + \log \left (e^{x} + 1\right ) - \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\ln \left ({\mathrm {e}}^x+1\right )-\ln \left ({\mathrm {e}}^x-1\right )-\frac {2}{{\mathrm {e}}^x+1{}\mathrm {i}}-\frac {2{}\mathrm {i}}{{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^2}-\frac {4}{3\,{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^3} \]

[In]

int(1/(sinh(x)*(sinh(x) + 1i)^2),x)

[Out]

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