Integrand size = 11, antiderivative size = 34 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\text {arctanh}(\cosh (x))+\frac {\cosh (x)}{3 (i+\sinh (x))^2}-\frac {4 i \cosh (x)}{3 (i+\sinh (x))} \] Output:
arctanh(cosh(x))+1/3*cosh(x)/(I+sinh(x))^2-4/3*I*cosh(x)/(I+sinh(x))
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\text {arctanh}(\cosh (x))+\frac {\cosh (x) (5-4 i \sinh (x))}{3 (i+\sinh (x))^2} \] Input:
Integrate[Csch[x]/(I + Sinh[x])^2,x]
Output:
ArcTanh[Cosh[x]] + (Cosh[x]*(5 - (4*I)*Sinh[x]))/(3*(I + Sinh[x])^2)
Time = 0.40 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.091, Rules used = {3042, 26, 25, 3245, 26, 3042, 26, 3457, 27, 3042, 26, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {csch}(x)}{(\sinh (x)+i)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{(i-i \sin (i x))^2 \sin (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int -\frac {1}{(1-\sin (i x))^2 \sin (i x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i \int \frac {1}{(1-\sin (i x))^2 \sin (i x)}dx\) |
\(\Big \downarrow \) 3245 |
\(\displaystyle -i \left (\frac {1}{3} \int -\frac {i \text {csch}(x) (i \sinh (x)+3)}{1-i \sinh (x)}dx-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (-\frac {1}{3} i \int \frac {\text {csch}(x) (i \sinh (x)+3)}{1-i \sinh (x)}dx-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (-\frac {1}{3} i \int \frac {i (\sin (i x)+3)}{(1-\sin (i x)) \sin (i x)}dx-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{3} \int \frac {\sin (i x)+3}{(1-\sin (i x)) \sin (i x)}dx-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle -i \left (\frac {1}{3} \left (\int -3 i \text {csch}(x)dx-\frac {4 i \cosh (x)}{1-i \sinh (x)}\right )-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -i \left (\frac {1}{3} \left (-3 i \int \text {csch}(x)dx-\frac {4 i \cosh (x)}{1-i \sinh (x)}\right )-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{3} \left (-3 i \int i \csc (i x)dx-\frac {4 i \cosh (x)}{1-i \sinh (x)}\right )-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{3} \left (3 \int \csc (i x)dx-\frac {4 i \cosh (x)}{1-i \sinh (x)}\right )-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -i \left (\frac {1}{3} \left (3 i \text {arctanh}(\cosh (x))-\frac {4 i \cosh (x)}{1-i \sinh (x)}\right )-\frac {i \cosh (x)}{3 (1-i \sinh (x))^2}\right )\) |
Input:
Int[Csch[x]/(I + Sinh[x])^2,x]
Output:
(-I)*(((3*I)*ArcTanh[Cosh[x]] - ((4*I)*Cosh[x])/(1 - I*Sinh[x]))/3 - ((I/3 )*Cosh[x])/(1 - I*Sinh[x])^2)
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[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] + Simp[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] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[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] + Simp[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])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.48 (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 | \(\frac {4 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {4 i}{\tanh \left (\frac {x}{2}\right )+i}-\frac {2}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\ln \left (\tanh \left (\frac {x}{2}\right )\right )\) | \(44\) |
parallelrisch | \(\frac {\left (-3 \tanh \left (\frac {x}{2}\right )^{3}+9 \tanh \left (\frac {x}{2}\right )-9 i \tanh \left (\frac {x}{2}\right )^{2}+3 i\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )+6 i \tanh \left (\frac {x}{2}\right )^{2}+6 \tanh \left (\frac {x}{2}\right )^{3}+4 i}{-9 \tanh \left (\frac {x}{2}\right )+9 i \tanh \left (\frac {x}{2}\right )^{2}+3 \tanh \left (\frac {x}{2}\right )^{3}-3 i}\) | \(82\) |
Input:
int(csch(x)/(I+sinh(x))^2,x,method=_RETURNVERBOSE)
Output:
-2/3*(9*I*exp(x)+3*exp(2*x)-4)/(exp(x)+I)^3-ln(exp(x)-1)+ln(exp(x)+1)
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 = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.29 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\frac {3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} + 1\right ) - 3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} - 1\right ) - 6 \, e^{\left (2 \, x\right )} - 18 i \, e^{x} + 8}{3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \] Input:
integrate(csch(x)/(I+sinh(x))^2,x, algorithm="fricas")
Output:
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)
\[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\int \frac {\operatorname {csch}{\left (x \right )}}{\left (\sinh {\left (x \right )} + i\right )^{2}}\, dx \] Input:
integrate(csch(x)/(I+sinh(x))**2,x)
Output:
Integral(csch(x)/(sinh(x) + I)**2, x)
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 = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\frac {2 \, {\left (-9 i \, e^{\left (-x\right )} + 3 \, e^{\left (-2 \, x\right )} - 4\right )}}{3 \, {\left (3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i\right )}} + \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \] Input:
integrate(csch(x)/(I+sinh(x))^2,x, algorithm="maxima")
Output:
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)
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, e^{\left (2 \, x\right )} + 9 i \, e^{x} - 4\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} + \log \left (e^{x} + 1\right ) - \log \left ({\left | e^{x} - 1 \right |}\right ) \] Input:
integrate(csch(x)/(I+sinh(x))^2,x, algorithm="giac")
Output:
-2/3*(3*e^(2*x) + 9*I*e^x - 4)/(e^x + I)^3 + log(e^x + 1) - log(abs(e^x - 1))
Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\ln \left ({\mathrm {e}}^x+1\right )-\ln \left ({\mathrm {e}}^x-1\right )-\frac {2}{{\mathrm {e}}^x+1{}\mathrm {i}}-\frac {2{}\mathrm {i}}{{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^2}-\frac {4}{3\,{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^3} \] Input:
int(1/(sinh(x)*(sinh(x) + 1i)^2),x)
Output:
log(exp(x) + 1) - log(exp(x) - 1) - 2/(exp(x) + 1i) - 2i/(exp(x) + 1i)^2 - 4/(3*(exp(x) + 1i)^3)
\[ \int \frac {\text {csch}(x)}{(i+\sinh (x))^2} \, dx=\int \frac {\mathrm {csch}\left (x \right )}{\sinh \left (x \right )^{2}+2 \sinh \left (x \right ) i -1}d x \] Input:
int(csch(x)/(I+sinh(x))^2,x)
Output:
int(csch(x)/(sinh(x)**2 + 2*sinh(x)*i - 1),x)