\(\int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx\) [168]

Optimal result

Integrand size = 13, antiderivative size = 52 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=-\frac {3}{8} i \arctan (\sinh (x))+\frac {i}{8 (i-\sinh (x))}+\frac {1}{8 (i+\sinh (x))^2}-\frac {i}{4 (i+\sinh (x))} \]

[Out]

-3/8*I*arctan(sinh(x))+1/8*I/(I-sinh(x))+1/8/(I+sinh(x))^2-1/4*I/(I+sinh(x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2746, 46, 209} \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=-\frac {3}{8} i \arctan (\sinh (x))+\frac {i}{8 (-\sinh (x)+i)}-\frac {i}{4 (\sinh (x)+i)}+\frac {1}{8 (\sinh (x)+i)^2} \]

[In]

Int[Sech[x]^3/(I + Sinh[x]),x]

[Out]

((-3*I)/8)*ArcTan[Sinh[x]] + (I/8)/(I - Sinh[x]) + 1/(8*(I + Sinh[x])^2) - (I/4)/(I + Sinh[x])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{(i-x)^2 (i+x)^3} \, dx,x,\sinh (x)\right ) \\ & = \text {Subst}\left (\int \left (\frac {i}{8 (-i+x)^2}-\frac {1}{4 (i+x)^3}+\frac {i}{4 (i+x)^2}-\frac {3 i}{8 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right ) \\ & = \frac {i}{8 (i-\sinh (x))}+\frac {1}{8 (i+\sinh (x))^2}-\frac {i}{4 (i+\sinh (x))}-\frac {3}{8} i \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (x)\right ) \\ & = -\frac {3}{8} i \arctan (\sinh (x))+\frac {i}{8 (i-\sinh (x))}+\frac {1}{8 (i+\sinh (x))^2}-\frac {i}{4 (i+\sinh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=-\frac {i \text {sech}^2(x) \left (2+3 i \arctan (\sinh (x))+3 (i+\arctan (\sinh (x))) \sinh (x)+(3+3 i \arctan (\sinh (x))) \sinh ^2(x)+3 \arctan (\sinh (x)) \sinh ^3(x)\right )}{8 (i+\sinh (x))} \]

[In]

Integrate[Sech[x]^3/(I + Sinh[x]),x]

[Out]

((-1/8*I)*Sech[x]^2*(2 + (3*I)*ArcTan[Sinh[x]] + 3*(I + ArcTan[Sinh[x]])*Sinh[x] + (3 + (3*I)*ArcTan[Sinh[x]])
*Sinh[x]^2 + 3*ArcTan[Sinh[x]]*Sinh[x]^3))/(I + Sinh[x])

Maple [A] (verified)

Time = 177.48 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.21

method	result	size
risch	\(-\frac {i {\mathrm e}^{x} \left (6 i {\mathrm e}^{3 x}+3 \,{\mathrm e}^{4 x}-6 i {\mathrm e}^{x}+2 \,{\mathrm e}^{2 x}+3\right )}{4 \left ({\mathrm e}^{x}+i\right )^{4} \left ({\mathrm e}^{x}-i\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{x}-i\right )}{8}+\frac {3 \ln \left ({\mathrm e}^{x}+i\right )}{8}\)	\(63\)
default	\(-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}+\frac {i}{\tanh \left (\frac {x}{2}\right )+i}-\frac {i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}+\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )}{8}+\frac {i}{-4 i+4 \tanh \left (\frac {x}{2}\right )}-\frac {1}{4 \left (-i+\tanh \left (\frac {x}{2}\right )\right )^{2}}-\frac {3 \ln \left (-i+\tanh \left (\frac {x}{2}\right )\right )}{8}\)	\(91\)

[In]

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

[Out]

-1/4*I*exp(x)*(6*I*exp(x)^3+3*exp(x)^4-6*I*exp(x)+2*exp(x)^2+3)/(exp(x)+I)^4/(exp(x)-I)^2-3/8*ln(exp(x)-I)+3/8
*ln(exp(x)+I)

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. 143 vs. \(2 (30) = 60\).

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.75 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=\frac {3 \, {\left (e^{\left (6 \, x\right )} + 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{x} + i\right ) - 3 \, {\left (e^{\left (6 \, x\right )} + 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{x} - i\right ) - 6 i \, e^{\left (5 \, x\right )} + 12 \, e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 6 i \, e^{x}}{8 \, {\left (e^{\left (6 \, x\right )} + 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )}} \]

[In]

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

[Out]

1/8*(3*(e^(6*x) + 2*I*e^(5*x) + e^(4*x) + 4*I*e^(3*x) - e^(2*x) + 2*I*e^x - 1)*log(e^x + I) - 3*(e^(6*x) + 2*I
*e^(5*x) + e^(4*x) + 4*I*e^(3*x) - e^(2*x) + 2*I*e^x - 1)*log(e^x - I) - 6*I*e^(5*x) + 12*e^(4*x) - 4*I*e^(3*x
) - 12*e^(2*x) - 6*I*e^x)/(e^(6*x) + 2*I*e^(5*x) + e^(4*x) + 4*I*e^(3*x) - e^(2*x) + 2*I*e^x - 1)

Sympy [F]

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

[In]

integrate(sech(x)**3/(I+sinh(x)),x)

[Out]

Integral(sech(x)**3/(sinh(x) + I), 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. 92 vs. \(2 (30) = 60\).

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.77 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=\frac {8 \, {\left (3 i \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} + 2 i \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 3 i \, e^{\left (-5 \, x\right )}\right )}}{-64 i \, e^{\left (-x\right )} - 32 \, e^{\left (-2 \, x\right )} - 128 i \, e^{\left (-3 \, x\right )} + 32 \, e^{\left (-4 \, x\right )} - 64 i \, e^{\left (-5 \, x\right )} + 32 \, e^{\left (-6 \, x\right )} - 32} - \frac {3}{8} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac {3}{8} \, \log \left (e^{\left (-x\right )} - i\right ) \]

[In]

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

[Out]

8*(3*I*e^(-x) - 6*e^(-2*x) + 2*I*e^(-3*x) + 6*e^(-4*x) + 3*I*e^(-5*x))/(-64*I*e^(-x) - 32*e^(-2*x) - 128*I*e^(
-3*x) + 32*e^(-4*x) - 64*I*e^(-5*x) + 32*e^(-6*x) - 32) - 3/8*log(e^(-x) + I) + 3/8*log(e^(-x) - I)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (30) = 60\).

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.77 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=\frac {3 \, e^{\left (-x\right )} - 3 \, e^{x} + 10 i}{16 \, {\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}} - \frac {9 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 52 i \, e^{\left (-x\right )} + 52 i \, e^{x} - 84}{32 \, {\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{2}} + \frac {3}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac {3}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \]

[In]

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

[Out]

1/16*(3*e^(-x) - 3*e^x + 10*I)/(e^(-x) - e^x + 2*I) - 1/32*(9*(e^(-x) - e^x)^2 - 52*I*e^(-x) + 52*I*e^x - 84)/
(e^(-x) - e^x - 2*I)^2 + 3/16*log(-e^(-x) + e^x + 2*I) - 3/16*log(-e^(-x) + e^x - 2*I)

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.21 \[ \int \frac {\text {sech}^3(x)}{i+\sinh (x)} \, dx=\frac {3\,\ln \left (-\frac {3}{4}+\frac {{\mathrm {e}}^x\,3{}\mathrm {i}}{4}\right )}{8}-\frac {3\,\ln \left (\frac {3}{4}+\frac {{\mathrm {e}}^x\,3{}\mathrm {i}}{4}\right )}{8}-\frac {1}{2\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1+{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,4{}\mathrm {i}\right )}-\frac {1}{4\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{4\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{2\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}} \]

[In]

int(1/(cosh(x)^3*(sinh(x) + 1i)),x)

[Out]

(3*log((exp(x)*3i)/4 - 3/4))/8 - (3*log((exp(x)*3i)/4 + 3/4))/8 - 1/(2*(exp(3*x)*4i - 6*exp(2*x) + exp(4*x) -
exp(x)*4i + 1)) - 1/(4*(exp(x)*2i - exp(2*x) + 1)) - 1i/(4*(exp(x) - 1i)) - 1i/(2*(exp(x) + 1i)) - 1i/(exp(2*x
)*3i + exp(3*x) - 3*exp(x) - 1i)