3.179 \(\int \frac{\text{sech}^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=60 \[ \frac{1}{16 (-\sinh (x)+i)}-\frac{3}{16 (\sinh (x)+i)}-\frac{i}{8 (\sinh (x)+i)^2}+\frac{1}{12 (\sinh (x)+i)^3}-\frac{1}{4} \tan ^{-1}(\sinh (x)) \]

[Out]

-ArcTan[Sinh[x]]/4 + 1/(16*(I - Sinh[x])) + 1/(12*(I + Sinh[x])^3) - (I/8)/(I + Sinh[x])^2 - 3/(16*(I + Sinh[x
]))

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Rubi [A]  time = 0.0551326, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2667, 44, 203} \[ \frac{1}{16 (-\sinh (x)+i)}-\frac{3}{16 (\sinh (x)+i)}-\frac{i}{8 (\sinh (x)+i)^2}+\frac{1}{12 (\sinh (x)+i)^3}-\frac{1}{4} \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(I + Sinh[x])^2,x]

[Out]

-ArcTan[Sinh[x]]/4 + 1/(16*(I - Sinh[x])) + 1/(12*(I + Sinh[x])^3) - (I/8)/(I + Sinh[x])^2 - 3/(16*(I + Sinh[x
]))

Rule 2667

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])

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{(i+\sinh (x))^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{(i-x)^2 (i+x)^4} \, dx,x,\sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{16 (-i+x)^2}-\frac{1}{4 (i+x)^4}+\frac{i}{4 (i+x)^3}+\frac{3}{16 (i+x)^2}-\frac{1}{4 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\frac{1}{16 (i-\sinh (x))}+\frac{1}{12 (i+\sinh (x))^3}-\frac{i}{8 (i+\sinh (x))^2}-\frac{3}{16 (i+\sinh (x))}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{4} \tan ^{-1}(\sinh (x))+\frac{1}{16 (i-\sinh (x))}+\frac{1}{12 (i+\sinh (x))^3}-\frac{i}{8 (i+\sinh (x))^2}-\frac{3}{16 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0441689, size = 68, normalized size = 1.13 \[ -\frac{\text{sech}^2(x) \left (6 i \sinh ^2(x)+3 \sinh ^4(x) \tan ^{-1}(\sinh (x))+\sinh ^3(x) \left (3+6 i \tan ^{-1}(\sinh (x))\right )+\sinh (x) \left (-1+6 i \tan ^{-1}(\sinh (x))\right )-3 \tan ^{-1}(\sinh (x))+4 i\right )}{12 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sech[x]^2*(4*I - 3*ArcTan[Sinh[x]] + (-1 + (6*I)*ArcTan[Sinh[x]])*Sinh[x] + (6*I)*Sinh[x]^2 + (3 + (6*I)*Arc
Tan[Sinh[x]])*Sinh[x]^3 + 3*ArcTan[Sinh[x]]*Sinh[x]^4))/(12*(I + Sinh[x])^2)

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Maple [B]  time = 0.05, size = 116, normalized size = 1.9 \begin{align*}{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{\frac{i}{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) +{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{{\frac{7\,i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}-{{\frac{2\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-6}}-{\frac{i}{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) -{{\frac{23\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-5}-{\frac{11}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+{\frac{11}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(I+sinh(x))^2,x)

[Out]

1/8*I/(tanh(1/2*x)-I)^2+1/4*I*ln(tanh(1/2*x)-I)+1/8/(tanh(1/2*x)-I)+7/2*I/(tanh(1/2*x)+I)^4-2/3*I/(tanh(1/2*x)
+I)^6-1/4*I*ln(tanh(1/2*x)+I)-23/8*I/(tanh(1/2*x)+I)^2+2/(tanh(1/2*x)+I)^5-11/3/(tanh(1/2*x)+I)^3+11/8/(tanh(1
/2*x)+I)

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Maxima [B]  time = 1.30324, size = 162, normalized size = 2.7 \begin{align*} -\frac{8 \,{\left (3 \, e^{\left (-x\right )} + 12 i \, e^{\left (-2 \, x\right )} - 13 \, e^{\left (-3 \, x\right )} + 8 i \, e^{\left (-4 \, x\right )} + 13 \, e^{\left (-5 \, x\right )} + 12 i \, e^{\left (-6 \, x\right )} - 3 \, e^{\left (-7 \, x\right )}\right )}}{192 i \, e^{\left (-x\right )} - 192 \, e^{\left (-2 \, x\right )} + 192 i \, e^{\left (-3 \, x\right )} - 480 \, e^{\left (-4 \, x\right )} - 192 i \, e^{\left (-5 \, x\right )} - 192 \, e^{\left (-6 \, x\right )} - 192 i \, e^{\left (-7 \, x\right )} + 48 \, e^{\left (-8 \, x\right )} + 48} - \frac{1}{4} i \, \log \left (i \, e^{\left (-x\right )} + 1\right ) + \frac{1}{4} i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*(3*e^(-x) + 12*I*e^(-2*x) - 13*e^(-3*x) + 8*I*e^(-4*x) + 13*e^(-5*x) + 12*I*e^(-6*x) - 3*e^(-7*x))/(192*I*e
^(-x) - 192*e^(-2*x) + 192*I*e^(-3*x) - 480*e^(-4*x) - 192*I*e^(-5*x) - 192*e^(-6*x) - 192*I*e^(-7*x) + 48*e^(
-8*x) + 48) - 1/4*I*log(I*e^(-x) + 1) + 1/4*I*log(I*e^(-x) - 1)

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Fricas [B]  time = 1.85669, size = 625, normalized size = 10.42 \begin{align*} \frac{{\left (-3 i \, e^{\left (8 \, x\right )} + 12 \, e^{\left (7 \, x\right )} + 12 i \, e^{\left (6 \, x\right )} + 12 \, e^{\left (5 \, x\right )} + 30 i \, e^{\left (4 \, x\right )} - 12 \, e^{\left (3 \, x\right )} + 12 i \, e^{\left (2 \, x\right )} - 12 \, e^{x} - 3 i\right )} \log \left (e^{x} + i\right ) +{\left (3 i \, e^{\left (8 \, x\right )} - 12 \, e^{\left (7 \, x\right )} - 12 i \, e^{\left (6 \, x\right )} - 12 \, e^{\left (5 \, x\right )} - 30 i \, e^{\left (4 \, x\right )} + 12 \, e^{\left (3 \, x\right )} - 12 i \, e^{\left (2 \, x\right )} + 12 \, e^{x} + 3 i\right )} \log \left (e^{x} - i\right ) - 6 \, e^{\left (7 \, x\right )} - 24 i \, e^{\left (6 \, x\right )} + 26 \, e^{\left (5 \, x\right )} - 16 i \, e^{\left (4 \, x\right )} - 26 \, e^{\left (3 \, x\right )} - 24 i \, e^{\left (2 \, x\right )} + 6 \, e^{x}}{12 \, e^{\left (8 \, x\right )} + 48 i \, e^{\left (7 \, x\right )} - 48 \, e^{\left (6 \, x\right )} + 48 i \, e^{\left (5 \, x\right )} - 120 \, e^{\left (4 \, x\right )} - 48 i \, e^{\left (3 \, x\right )} - 48 \, e^{\left (2 \, x\right )} - 48 i \, e^{x} + 12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-3*I*e^(8*x) + 12*e^(7*x) + 12*I*e^(6*x) + 12*e^(5*x) + 30*I*e^(4*x) - 12*e^(3*x) + 12*I*e^(2*x) - 12*e^x -
3*I)*log(e^x + I) + (3*I*e^(8*x) - 12*e^(7*x) - 12*I*e^(6*x) - 12*e^(5*x) - 30*I*e^(4*x) + 12*e^(3*x) - 12*I*e
^(2*x) + 12*e^x + 3*I)*log(e^x - I) - 6*e^(7*x) - 24*I*e^(6*x) + 26*e^(5*x) - 16*I*e^(4*x) - 26*e^(3*x) - 24*I
*e^(2*x) + 6*e^x)/(12*e^(8*x) + 48*I*e^(7*x) - 48*e^(6*x) + 48*I*e^(5*x) - 120*e^(4*x) - 48*I*e^(3*x) - 48*e^(
2*x) - 48*I*e^x + 12)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.32622, size = 142, normalized size = 2.37 \begin{align*} \frac{-i \, e^{\left (-x\right )} + i \, e^{x} + 3}{8 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}} + \frac{11 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 84 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 228 i \, e^{\left (-x\right )} + 228 i \, e^{x} - 240}{48 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{3}} - \frac{1}{8} i \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) + \frac{1}{8} i \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-I*e^(-x) + I*e^x + 3)/(e^(-x) - e^x + 2*I) + 1/48*(11*I*(e^(-x) - e^x)^3 + 84*(e^(-x) - e^x)^2 - 228*I*e
^(-x) + 228*I*e^x - 240)/(e^(-x) - e^x - 2*I)^3 - 1/8*I*log(-e^(-x) + e^x + 2*I) + 1/8*I*log(-e^(-x) + e^x - 2
*I)