Integrand size = 13, antiderivative size = 42 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=2 i \text {arctanh}(\cosh (x))+\frac {10 \coth (x)}{3}+\frac {\coth (x)}{3 (i+\sinh (x))^2}-\frac {2 i \coth (x)}{i+\sinh (x)} \]
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(42)=84\).
Time = 1.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.10 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=\frac {1}{6} \left (3 \coth \left (\frac {x}{2}\right )+12 i \log \left (\cosh \left (\frac {x}{2}\right )\right )-12 i \log \left (\sinh \left (\frac {x}{2}\right )\right )+\frac {2}{i+\sinh (x)}-\frac {4 \sinh \left (\frac {x}{2}\right ) (8 i+7 \sinh (x))}{\left (i \cosh \left (\frac {x}{2}\right )+\sinh \left (\frac {x}{2}\right )\right )^3}+3 \tanh \left (\frac {x}{2}\right )\right ) \]
(3*Coth[x/2] + (12*I)*Log[Cosh[x/2]] - (12*I)*Log[Sinh[x/2]] + 2/(I + Sinh [x]) - (4*Sinh[x/2]*(8*I + 7*Sinh[x]))/(I*Cosh[x/2] + Sinh[x/2])^3 + 3*Tan h[x/2])/6
Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.538, Rules used = {3042, 25, 25, 3245, 27, 3042, 25, 3457, 25, 3042, 25, 3227, 25, 26, 3042, 25, 26, 4254, 24, 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.
\(\displaystyle \int \frac {\text {csch}^2(x)}{(\sinh (x)+i)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{(i-i \sin (i x))^2 \sin (i x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {1}{(1-\sin (i x))^2 \sin (i x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {1}{(1-\sin (i x))^2 \sin (i x)^2}dx\) |
\(\Big \downarrow \) 3245 |
\(\displaystyle \frac {1}{3} \int -\frac {2 \text {csch}^2(x) (i \sinh (x)+2)}{1-i \sinh (x)}dx-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{3} \int \frac {\text {csch}^2(x) (i \sinh (x)+2)}{1-i \sinh (x)}dx-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{3} \int -\frac {\sin (i x)+2}{(1-\sin (i x)) \sin (i x)^2}dx-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \int \frac {\sin (i x)+2}{(1-\sin (i x)) \sin (i x)^2}dx-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {2}{3} \left (\int -\text {csch}^2(x) (3 i \sinh (x)+5)dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \left (-\int \text {csch}^2(x) (3 i \sinh (x)+5)dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (-\int -\frac {3 \sin (i x)+5}{\sin (i x)^2}dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \left (\int \frac {3 \sin (i x)+5}{\sin (i x)^2}dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {2}{3} \left (5 \int -\text {csch}^2(x)dx+3 \int -i \text {csch}(x)dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \left (-5 \int \text {csch}^2(x)dx+3 \int -i \text {csch}(x)dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {2}{3} \left (-5 \int \text {csch}^2(x)dx-3 i \int \text {csch}(x)dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (-3 i \int i \csc (i x)dx-5 \int -\csc (i x)^2dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{3} \left (-3 i \int i \csc (i x)dx+5 \int \csc (i x)^2dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {2}{3} \left (3 \int \csc (i x)dx+5 \int \csc (i x)^2dx-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {2}{3} \left (3 \int \csc (i x)dx+5 i \int 1d(-i \coth (x))-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {2}{3} \left (3 \int \csc (i x)dx+5 \coth (x)-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {2}{3} \left (3 i \text {arctanh}(\cosh (x))+5 \coth (x)-\frac {3 \coth (x)}{1-i \sinh (x)}\right )-\frac {\coth (x)}{3 (1-i \sinh (x))^2}\) |
(2*((3*I)*ArcTanh[Cosh[x]] + 5*Coth[x] - (3*Coth[x])/(1 - I*Sinh[x])))/3 - Coth[x]/(3*(1 - I*Sinh[x])^2)
3.1.53.3.1 Defintions of rubi rules used
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38
method | result | size |
default | \(\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {2 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {4}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}+\frac {6}{\tanh \left (\frac {x}{2}\right )+i}-2 i \ln \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{2 \tanh \left (\frac {x}{2}\right )}\) | \(58\) |
risch | \(-\frac {4 i \left (9 i {\mathrm e}^{3 x}+3 \,{\mathrm e}^{4 x}-12 i {\mathrm e}^{x}-11 \,{\mathrm e}^{2 x}+5\right )}{3 \left ({\mathrm e}^{2 x}-1\right ) \left ({\mathrm e}^{x}+i\right )^{3}}+2 i \ln \left ({\mathrm e}^{x}+1\right )-2 i \ln \left ({\mathrm e}^{x}-1\right )\) | \(62\) |
parallelrisch | \(\frac {\left (-12 i \tanh \left (\frac {x}{2}\right )^{3}+36 \tanh \left (\frac {x}{2}\right )^{2}+36 i \tanh \left (\frac {x}{2}\right )-12\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )+19 i \tanh \left (\frac {x}{2}\right )^{3}+3 \tanh \left (\frac {x}{2}\right )^{4}-3 i \coth \left (\frac {x}{2}\right )+36 i \tanh \left (\frac {x}{2}\right )-31}{6 \tanh \left (\frac {x}{2}\right )^{3}+18 i \tanh \left (\frac {x}{2}\right )^{2}-18 \tanh \left (\frac {x}{2}\right )-6 i}\) | \(95\) |
1/2*tanh(1/2*x)-2*I/(tanh(1/2*x)+I)^2-4/3/(tanh(1/2*x)+I)^3+6/(tanh(1/2*x) +I)-2*I*ln(tanh(1/2*x))+1/2/tanh(1/2*x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (30) = 60\).
Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 3.10 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (-i \, e^{\left (5 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - 4 \, e^{\left (2 \, x\right )} - 3 i \, e^{x} + 1\right )} \log \left (e^{x} + 1\right ) + 3 \, {\left (i \, e^{\left (5 \, x\right )} - 3 \, e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} - 1\right )} \log \left (e^{x} - 1\right ) + 6 i \, e^{\left (4 \, x\right )} - 18 \, e^{\left (3 \, x\right )} - 22 i \, e^{\left (2 \, x\right )} + 24 \, e^{x} + 10 i\right )}}{3 \, {\left (e^{\left (5 \, x\right )} + 3 i \, e^{\left (4 \, x\right )} - 4 \, e^{\left (3 \, x\right )} - 4 i \, e^{\left (2 \, x\right )} + 3 \, e^{x} + i\right )}} \]
-2/3*(3*(-I*e^(5*x) + 3*e^(4*x) + 4*I*e^(3*x) - 4*e^(2*x) - 3*I*e^x + 1)*l og(e^x + 1) + 3*(I*e^(5*x) - 3*e^(4*x) - 4*I*e^(3*x) + 4*e^(2*x) + 3*I*e^x - 1)*log(e^x - 1) + 6*I*e^(4*x) - 18*e^(3*x) - 22*I*e^(2*x) + 24*e^x + 10 *I)/(e^(5*x) + 3*I*e^(4*x) - 4*e^(3*x) - 4*I*e^(2*x) + 3*e^x + I)
\[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=\int \frac {\operatorname {csch}^{2}{\left (x \right )}}{\left (\sinh {\left (x \right )} + i\right )^{2}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (30) = 60\).
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.88 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=\frac {4 \, {\left (12 \, e^{\left (-x\right )} + 11 i \, e^{\left (-2 \, x\right )} - 9 \, e^{\left (-3 \, x\right )} - 3 i \, e^{\left (-4 \, x\right )} - 5 i\right )}}{3 \, {\left (3 \, e^{\left (-x\right )} + 4 i \, e^{\left (-2 \, x\right )} - 4 \, e^{\left (-3 \, x\right )} - 3 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i\right )}} + 2 i \, \log \left (e^{\left (-x\right )} + 1\right ) - 2 i \, \log \left (e^{\left (-x\right )} - 1\right ) \]
4/3*(12*e^(-x) + 11*I*e^(-2*x) - 9*e^(-3*x) - 3*I*e^(-4*x) - 5*I)/(3*e^(-x ) + 4*I*e^(-2*x) - 4*e^(-3*x) - 3*I*e^(-4*x) + e^(-5*x) - I) + 2*I*log(e^( -x) + 1) - 2*I*log(e^(-x) - 1)
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=\frac {2}{e^{\left (2 \, x\right )} - 1} - \frac {2 \, {\left (6 i \, e^{\left (2 \, x\right )} - 15 \, e^{x} - 7 i\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} + 2 i \, \log \left (e^{x} + 1\right ) - 2 i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
2/(e^(2*x) - 1) - 2/3*(6*I*e^(2*x) - 15*e^x - 7*I)/(e^x + I)^3 + 2*I*log(e ^x + 1) - 2*I*log(abs(e^x - 1))
Time = 1.48 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.02 \[ \int \frac {\text {csch}^2(x)}{(i+\sinh (x))^2} \, dx=\frac {2}{{\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}}+\frac {2}{{\mathrm {e}}^{2\,x}-1}-\ln \left ({\mathrm {e}}^x\,4{}\mathrm {i}-4{}\mathrm {i}\right )\,2{}\mathrm {i}+\ln \left ({\mathrm {e}}^x\,4{}\mathrm {i}+4{}\mathrm {i}\right )\,2{}\mathrm {i}-\frac {4{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}}-\frac {4{}\mathrm {i}}{3\,\left ({\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}\right )} \]