Integrand size = 24, antiderivative size = 57 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \text {arctanh}(\cosh (c+d x))}{a d}-\frac {2 \coth (c+d x)}{a d}+\frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))} \] Output:
I*arctanh(cosh(d*x+c))/a/d-2*coth(d*x+c)/a/d+coth(d*x+c)/d/(a+I*a*sinh(d*x +c))
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {sech}(c+d x) \left (i-i \text {arctanh}\left (\sqrt {\cosh ^2(c+d x)}\right ) \sqrt {\cosh ^2(c+d x)}+\text {csch}(c+d x)+2 \sinh (c+d x)\right )}{a d} \] Input:
Integrate[Csch[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]
Output:
-((Sech[c + d*x]*(I - I*ArcTanh[Sqrt[Cosh[c + d*x]^2]]*Sqrt[Cosh[c + d*x]^ 2] + Csch[c + d*x] + 2*Sinh[c + d*x]))/(a*d))
Time = 0.47 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 25, 3247, 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.
\(\displaystyle \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\sin (i c+i d x)^2 (a+a \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\sin (i c+i d x)^2 (\sin (i c+i d x) a+a)}dx\) |
\(\Big \downarrow \) 3247 |
\(\displaystyle \frac {\int \text {csch}^2(c+d x) (2 a-i a \sinh (c+d x))dx}{a^2}+\frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\frac {2 a-a \sin (i c+i d x)}{\sin (i c+i d x)^2}dx}{a^2}+\frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {\int \frac {2 a-a \sin (i c+i d x)}{\sin (i c+i d x)^2}dx}{a^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {2 a \int -\text {csch}^2(c+d x)dx-a \int -i \text {csch}(c+d x)dx}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {-2 a \int \text {csch}^2(c+d x)dx-a \int -i \text {csch}(c+d x)dx}{a^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {-2 a \int \text {csch}^2(c+d x)dx+i a \int \text {csch}(c+d x)dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {i a \int i \csc (i c+i d x)dx-2 a \int -\csc (i c+i d x)^2dx}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {i a \int i \csc (i c+i d x)dx+2 a \int \csc (i c+i d x)^2dx}{a^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {2 a \int \csc (i c+i d x)^2dx-a \int \csc (i c+i d x)dx}{a^2}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {\frac {2 i a \int 1d(-i \coth (c+d x))}{d}-a \int \csc (i c+i d x)dx}{a^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {\frac {2 a \coth (c+d x)}{d}-a \int \csc (i c+i d x)dx}{a^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {\frac {2 a \coth (c+d x)}{d}-\frac {i a \text {arctanh}(\cosh (c+d x))}{d}}{a^2}\) |
Input:
Int[Csch[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]
Output:
-((((-I)*a*ArcTanh[Cosh[c + d*x]])/d + (2*a*Coth[c + d*x])/d)/a^2) + Coth[ c + d*x]/(d*(a + I*a*Sinh[c + d*x]))
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[((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[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x]))), x] + Simp[d/(a*(b*c - a*d)) Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[ c^2 - d^2, 0] && LtQ[n, 0] && (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 = 0.40 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\) | \(63\) |
default | \(\frac {-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}\) | \(63\) |
parallelrisch | \(\frac {2 \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\coth \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (-1-i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d a}\) | \(86\) |
risch | \(-\frac {2 i \left (-i {\mathrm e}^{d x +c}+{\mathrm e}^{2 d x +2 c}-2\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) \left ({\mathrm e}^{d x +c}-i\right ) d a}+\frac {i \ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}-\frac {i \ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}\) | \(91\) |
Input:
int(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/2/d/a*(-tanh(1/2*d*x+1/2*c)-4/(-I+tanh(1/2*d*x+1/2*c))-1/tanh(1/2*d*x+1/ 2*c)-2*I*ln(tanh(1/2*d*x+1/2*c)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (53) = 106\).
Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.56 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {{\left (i \, e^{\left (3 \, d x + 3 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 1\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) + {\left (-i \, e^{\left (3 \, d x + 3 \, c\right )} - e^{\left (2 \, d x + 2 \, c\right )} + i \, e^{\left (d x + c\right )} + 1\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 2 i \, e^{\left (2 \, d x + 2 \, c\right )} - 2 \, e^{\left (d x + c\right )} + 4 i}{a d e^{\left (3 \, d x + 3 \, c\right )} - i \, a d e^{\left (2 \, d x + 2 \, c\right )} - a d e^{\left (d x + c\right )} + i \, a d} \] Input:
integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
Output:
((I*e^(3*d*x + 3*c) + e^(2*d*x + 2*c) - I*e^(d*x + c) - 1)*log(e^(d*x + c) + 1) + (-I*e^(3*d*x + 3*c) - e^(2*d*x + 2*c) + I*e^(d*x + c) + 1)*log(e^( d*x + c) - 1) - 2*I*e^(2*d*x + 2*c) - 2*e^(d*x + c) + 4*I)/(a*d*e^(3*d*x + 3*c) - I*a*d*e^(2*d*x + 2*c) - a*d*e^(d*x + c) + I*a*d)
\[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \int \frac {\operatorname {csch}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(csch(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)
Output:
-I*Integral(csch(c + d*x)**2/(sinh(c + d*x) - I), x)/a
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (53) = 106\).
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.91 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {2 \, {\left (e^{\left (-d x - c\right )} - i \, e^{\left (-2 \, d x - 2 \, c\right )} + 2 i\right )}}{{\left (a e^{\left (-d x - c\right )} - i \, a e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-3 \, d x - 3 \, c\right )} + i \, a\right )} d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \] Input:
integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
Output:
-2*(e^(-d*x - c) - I*e^(-2*d*x - 2*c) + 2*I)/((a*e^(-d*x - c) - I*a*e^(-2* d*x - 2*c) - a*e^(-3*d*x - 3*c) + I*a)*d) + I*log(e^(-d*x - c) + 1)/(a*d) - I*log(e^(-d*x - c) - 1)/(a*d)
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {-\frac {i \, \log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac {i \, \log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2 \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 2\right )}}{a {\left (i \, e^{\left (3 \, d x + 3 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 1\right )}}}{d} \] Input:
integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
Output:
-(-I*log(e^(d*x + c) + 1)/a + I*log(e^(d*x + c) - 1)/a - 2*(e^(2*d*x + 2*c ) - I*e^(d*x + c) - 2)/(a*(I*e^(3*d*x + 3*c) + e^(2*d*x + 2*c) - I*e^(d*x + c) - 1)))/d
Time = 2.35 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.14 \[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d}-\frac {4{}\mathrm {i}}{a\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,2{}\mathrm {i}}{a\,d}}{{\mathrm {e}}^{c+d\,x}+{\mathrm {e}}^{2\,c+2\,d\,x}\,1{}\mathrm {i}-{\mathrm {e}}^{3\,c+3\,d\,x}-\mathrm {i}}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}-2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d} \] Input:
int(1/(sinh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)),x)
Output:
((2*exp(c + d*x))/(a*d) - 4i/(a*d) + (exp(2*c + 2*d*x)*2i)/(a*d))/(exp(c + d*x) + exp(2*c + 2*d*x)*1i - exp(3*c + 3*d*x) - 1i) - (log(exp(c + d*x)*2 i - 2i)*1i)/(a*d) + (log(exp(c + d*x)*2i + 2i)*1i)/(a*d)
\[ \int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\int \frac {\mathrm {csch}\left (d x +c \right )^{2}}{\sinh \left (d x +c \right ) i +1}d x}{a} \] Input:
int(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)
Output:
int(csch(c + d*x)**2/(sinh(c + d*x)*i + 1),x)/a