\(\int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx\) [54]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 58 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {7}{2} \text {arctanh}(\cosh (x))+\frac {16}{3} i \coth (x)+\frac {7}{2} \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {8 i \coth (x) \text {csch}(x)}{3 (i+\sinh (x))} \]

[Out]

-7/2*arctanh(cosh(x))+16/3*I*coth(x)+7/2*coth(x)*csch(x)+1/3*coth(x)*csch(x)/(I+sinh(x))^2-8/3*I*coth(x)*csch(
x)/(I+sinh(x))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2845, 3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {7}{2} \text {arctanh}(\cosh (x))+\frac {16}{3} i \coth (x)+\frac {7}{2} \coth (x) \text {csch}(x)-\frac {8 i \coth (x) \text {csch}(x)}{3 (\sinh (x)+i)}+\frac {\coth (x) \text {csch}(x)}{3 (\sinh (x)+i)^2} \]

[In]

Int[Csch[x]^3/(I + Sinh[x])^2,x]

[Out]

(-7*ArcTanh[Cosh[x]])/2 + ((16*I)/3)*Coth[x] + (7*Coth[x]*Csch[x])/2 + (Coth[x]*Csch[x])/(3*(I + Sinh[x])^2) -
 (((8*I)/3)*Coth[x]*Csch[x])/(I + Sinh[x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
 1] && IntegerQ[2*n]

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 {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {1}{3} \int \frac {\text {csch}^3(x) (5 i-3 \sinh (x))}{i+\sinh (x)} \, dx \\ & = \frac {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {8 i \coth (x) \text {csch}(x)}{3 (i+\sinh (x))}+\frac {1}{3} \int \text {csch}^3(x) (-21-16 i \sinh (x)) \, dx \\ & = \frac {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {8 i \coth (x) \text {csch}(x)}{3 (i+\sinh (x))}-\frac {16}{3} i \int \text {csch}^2(x) \, dx-7 \int \text {csch}^3(x) \, dx \\ & = \frac {7}{2} \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {8 i \coth (x) \text {csch}(x)}{3 (i+\sinh (x))}+\frac {7}{2} \int \text {csch}(x) \, dx-\frac {16}{3} \text {Subst}(\int 1 \, dx,x,-i \coth (x)) \\ & = -\frac {7}{2} \text {arctanh}(\cosh (x))+\frac {16}{3} i \coth (x)+\frac {7}{2} \coth (x) \text {csch}(x)+\frac {\coth (x) \text {csch}(x)}{3 (i+\sinh (x))^2}-\frac {8 i \coth (x) \text {csch}(x)}{3 (i+\sinh (x))} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(140\) vs. \(2(58)=116\).

Time = 1.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.41 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=\frac {1}{24} \left (24 i \coth \left (\frac {x}{2}\right )+3 \text {csch}^2\left (\frac {x}{2}\right )-84 \log \left (\cosh \left (\frac {x}{2}\right )\right )+84 \log \left (\sinh \left (\frac {x}{2}\right )\right )+3 \text {sech}^2\left (\frac {x}{2}\right )+\frac {8}{\left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^2}+\frac {160 i \sinh \left (\frac {x}{2}\right )}{\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )}+\frac {16 \sinh \left (\frac {x}{2}\right )}{\left (i \cosh \left (\frac {x}{2}\right )+\sinh \left (\frac {x}{2}\right )\right )^3}+24 i \tanh \left (\frac {x}{2}\right )\right ) \]

[In]

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

[Out]

((24*I)*Coth[x/2] + 3*Csch[x/2]^2 - 84*Log[Cosh[x/2]] + 84*Log[Sinh[x/2]] + 3*Sech[x/2]^2 + 8/(Cosh[x/2] - I*S
inh[x/2])^2 + ((160*I)*Sinh[x/2])/(Cosh[x/2] - I*Sinh[x/2]) + (16*Sinh[x/2])/(I*Cosh[x/2] + Sinh[x/2])^3 + (24
*I)*Tanh[x/2])/24

Maple [A] (verified)

Time = 4.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.24

method result size
risch \(\frac {-98 \,{\mathrm e}^{4 x}+63 i {\mathrm e}^{5 x}+97 \,{\mathrm e}^{2 x}-126 i {\mathrm e}^{3 x}+21 \,{\mathrm e}^{6 x}-32+75 i {\mathrm e}^{x}}{3 \left ({\mathrm e}^{2 x}-1\right )^{2} \left ({\mathrm e}^{x}+i\right )^{3}}-\frac {7 \ln \left ({\mathrm e}^{x}+1\right )}{2}+\frac {7 \ln \left ({\mathrm e}^{x}-1\right )}{2}\) \(72\)
default \(i \tanh \left (\frac {x}{2}\right )-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8}+\frac {i}{\tanh \left (\frac {x}{2}\right )}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {7 \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2}+\frac {8 i}{\tanh \left (\frac {x}{2}\right )+i}-\frac {4 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}+\frac {2}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}\) \(76\)
parallelrisch \(\frac {\left (84 \tanh \left (\frac {x}{2}\right )^{3}+252 i \tanh \left (\frac {x}{2}\right )^{2}-252 \tanh \left (\frac {x}{2}\right )-84 i\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )+15 i \tanh \left (\frac {x}{2}\right )^{4}-3 \tanh \left (\frac {x}{2}\right )^{5}-3 i \coth \left (\frac {x}{2}\right )^{2}-112 \tanh \left (\frac {x}{2}\right )^{3}-190 i+15 \coth \left (\frac {x}{2}\right )-234 \tanh \left (\frac {x}{2}\right )}{24 \tanh \left (\frac {x}{2}\right )^{3}+72 i \tanh \left (\frac {x}{2}\right )^{2}-72 \tanh \left (\frac {x}{2}\right )-24 i}\) \(111\)

[In]

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

[Out]

1/3*(-98*exp(x)^4+63*I*exp(x)^5+97*exp(x)^2-126*I*exp(x)^3+21*exp(x)^6-32+75*I*exp(x))/(exp(x)^2-1)^2/(exp(x)+
I)^3-7/2*ln(exp(x)+1)+7/2*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. 174 vs. \(2 (40) = 80\).

Time = 0.27 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.00 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {21 \, {\left (e^{\left (7 \, x\right )} + 3 i \, e^{\left (6 \, x\right )} - 5 \, e^{\left (5 \, x\right )} - 7 i \, e^{\left (4 \, x\right )} + 7 \, e^{\left (3 \, x\right )} + 5 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} + 1\right ) - 21 \, {\left (e^{\left (7 \, x\right )} + 3 i \, e^{\left (6 \, x\right )} - 5 \, e^{\left (5 \, x\right )} - 7 i \, e^{\left (4 \, x\right )} + 7 \, e^{\left (3 \, x\right )} + 5 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )} \log \left (e^{x} - 1\right ) - 42 \, e^{\left (6 \, x\right )} - 126 i \, e^{\left (5 \, x\right )} + 196 \, e^{\left (4 \, x\right )} + 252 i \, e^{\left (3 \, x\right )} - 194 \, e^{\left (2 \, x\right )} - 150 i \, e^{x} + 64}{6 \, {\left (e^{\left (7 \, x\right )} + 3 i \, e^{\left (6 \, x\right )} - 5 \, e^{\left (5 \, x\right )} - 7 i \, e^{\left (4 \, x\right )} + 7 \, e^{\left (3 \, x\right )} + 5 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \]

[In]

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

[Out]

-1/6*(21*(e^(7*x) + 3*I*e^(6*x) - 5*e^(5*x) - 7*I*e^(4*x) + 7*e^(3*x) + 5*I*e^(2*x) - 3*e^x - I)*log(e^x + 1)
- 21*(e^(7*x) + 3*I*e^(6*x) - 5*e^(5*x) - 7*I*e^(4*x) + 7*e^(3*x) + 5*I*e^(2*x) - 3*e^x - I)*log(e^x - 1) - 42
*e^(6*x) - 126*I*e^(5*x) + 196*e^(4*x) + 252*I*e^(3*x) - 194*e^(2*x) - 150*I*e^x + 64)/(e^(7*x) + 3*I*e^(6*x)
- 5*e^(5*x) - 7*I*e^(4*x) + 7*e^(3*x) + 5*I*e^(2*x) - 3*e^x - I)

Sympy [F]

\[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=\int \frac {\operatorname {csch}^{3}{\left (x \right )}}{\left (\sinh {\left (x \right )} + i\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(csch(x)**3/(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. 105 vs. \(2 (40) = 80\).

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.81 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=-\frac {-75 i \, e^{\left (-x\right )} + 97 \, e^{\left (-2 \, x\right )} + 126 i \, e^{\left (-3 \, x\right )} - 98 \, e^{\left (-4 \, x\right )} - 63 i \, e^{\left (-5 \, x\right )} + 21 \, e^{\left (-6 \, x\right )} - 32}{3 \, {\left (3 \, e^{\left (-x\right )} + 5 i \, e^{\left (-2 \, x\right )} - 7 \, e^{\left (-3 \, x\right )} - 7 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )} + 3 i \, e^{\left (-6 \, x\right )} - e^{\left (-7 \, x\right )} - i\right )}} - \frac {7}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {7}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

-1/3*(-75*I*e^(-x) + 97*e^(-2*x) + 126*I*e^(-3*x) - 98*e^(-4*x) - 63*I*e^(-5*x) + 21*e^(-6*x) - 32)/(3*e^(-x)
+ 5*I*e^(-2*x) - 7*e^(-3*x) - 7*I*e^(-4*x) + 5*e^(-5*x) + 3*I*e^(-6*x) - e^(-7*x) - I) - 7/2*log(e^(-x) + 1) +
 7/2*log(e^(-x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=\frac {e^{\left (3 \, x\right )} + 4 i \, e^{\left (2 \, x\right )} + e^{x} - 4 i}{{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} + \frac {2 \, {\left (9 \, e^{\left (2 \, x\right )} + 21 i \, e^{x} - 10\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} - \frac {7}{2} \, \log \left (e^{x} + 1\right ) + \frac {7}{2} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

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

[Out]

(e^(3*x) + 4*I*e^(2*x) + e^x - 4*I)/(e^(2*x) - 1)^2 + 2/3*(9*e^(2*x) + 21*I*e^x - 10)/(e^x + I)^3 - 7/2*log(e^
x + 1) + 7/2*log(abs(e^x - 1))

Mupad [B] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.36 \[ \int \frac {\text {csch}^3(x)}{(i+\sinh (x))^2} \, dx=\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1}-\frac {7\,\ln \left ({\mathrm {e}}^x+1\right )}{2}-\frac {7\,\ln \left (\frac {1}{{\mathrm {e}}^x-1}\right )}{2}+\frac {2\,{\mathrm {e}}^x}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^2}+\frac {6}{{\mathrm {e}}^x+1{}\mathrm {i}}+\frac {2{}\mathrm {i}}{{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^2}+\frac {4}{3\,{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^3}+\frac {4{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1} \]

[In]

int(1/(sinh(x)^3*(sinh(x) + 1i)^2),x)

[Out]

exp(x)/(exp(2*x) - 1) - (7*log(exp(x) + 1))/2 - (7*log(1/(exp(x) - 1)))/2 + (2*exp(x))/(exp(2*x) - 1)^2 + 6/(e
xp(x) + 1i) + 2i/(exp(x) + 1i)^2 + 4/(3*(exp(x) + 1i)^3) + 4i/(exp(2*x) - 1)