3.170 \(\int \frac{\text{sech}^5(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=80 \[ \frac{i}{8 (-\sinh (x)+i)}-\frac{3 i}{16 (\sinh (x)+i)}-\frac{1}{32 (-\sinh (x)+i)^2}+\frac{3}{32 (\sinh (x)+i)^2}+\frac{i}{24 (\sinh (x)+i)^3}-\frac{5}{16} i \tan ^{-1}(\sinh (x)) \]

[Out]

((-5*I)/16)*ArcTan[Sinh[x]] - 1/(32*(I - Sinh[x])^2) + (I/8)/(I - Sinh[x]) + (I/24)/(I + Sinh[x])^3 + 3/(32*(I
+ Sinh[x])^2) - ((3*I)/16)/(I + Sinh[x])

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Rubi [A]  time = 0.0666413, antiderivative size = 80, 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} \[ \frac{i}{8 (-\sinh (x)+i)}-\frac{3 i}{16 (\sinh (x)+i)}-\frac{1}{32 (-\sinh (x)+i)^2}+\frac{3}{32 (\sinh (x)+i)^2}+\frac{i}{24 (\sinh (x)+i)^3}-\frac{5}{16} i \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^5/(I + Sinh[x]),x]

[Out]

((-5*I)/16)*ArcTan[Sinh[x]] - 1/(32*(I - Sinh[x])^2) + (I/8)/(I - Sinh[x]) + (I/24)/(I + Sinh[x])^3 + 3/(32*(I
+ Sinh[x])^2) - ((3*I)/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}^5(x)}{i+\sinh (x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{(i-x)^3 (i+x)^4} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{1}{16 (-i+x)^3}-\frac{i}{8 (-i+x)^2}+\frac{i}{8 (i+x)^4}+\frac{3}{16 (i+x)^3}-\frac{3 i}{16 (i+x)^2}+\frac{5 i}{16 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{32 (i-\sinh (x))^2}+\frac{i}{8 (i-\sinh (x))}+\frac{i}{24 (i+\sinh (x))^3}+\frac{3}{32 (i+\sinh (x))^2}-\frac{3 i}{16 (i+\sinh (x))}-\frac{5}{16} i \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{5}{16} i \tan ^{-1}(\sinh (x))-\frac{1}{32 (i-\sinh (x))^2}+\frac{i}{8 (i-\sinh (x))}+\frac{i}{24 (i+\sinh (x))^3}+\frac{3}{32 (i+\sinh (x))^2}-\frac{3 i}{16 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0671052, size = 94, normalized size = 1.18 \[ -\frac{i \text{sech}^4(x) \left (15 \sinh ^5(x) \tan ^{-1}(\sinh (x))+15 \sinh ^4(x) \left (1+i \tan ^{-1}(\sinh (x))\right )+15 \sinh ^3(x) \left (2 \tan ^{-1}(\sinh (x))+i\right )+5 \sinh ^2(x) \left (5+6 i \tan ^{-1}(\sinh (x))\right )+5 \sinh (x) \left (3 \tan ^{-1}(\sinh (x))+5 i\right )+15 i \tan ^{-1}(\sinh (x))+8\right )}{48 (\sinh (x)+i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^5/(I + Sinh[x]),x]

[Out]

((-I/48)*Sech[x]^4*(8 + (15*I)*ArcTan[Sinh[x]] + 5*(5*I + 3*ArcTan[Sinh[x]])*Sinh[x] + 5*(5 + (6*I)*ArcTan[Sin
h[x]])*Sinh[x]^2 + 15*(I + 2*ArcTan[Sinh[x]])*Sinh[x]^3 + 15*(1 + I*ArcTan[Sinh[x]])*Sinh[x]^4 + 15*ArcTan[Sin
h[x]]*Sinh[x]^5))/(I + Sinh[x])

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Maple [B]  time = 0.042, size = 137, normalized size = 1.7 \begin{align*}{{\frac{3\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}-{\frac{5}{16}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }+{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}+{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-{{\frac{25\,i}{12}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-6}}-{\frac{15}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{\frac{15}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{5}{16}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^5/(I+sinh(x)),x)

[Out]

3/8*I/(tanh(1/2*x)-I)-1/4*I/(tanh(1/2*x)-I)^3+1/8/(tanh(1/2*x)-I)^4-1/2/(tanh(1/2*x)-I)^2-5/16*ln(tanh(1/2*x)-
I)+I/(tanh(1/2*x)+I)^5+I/(tanh(1/2*x)+I)-25/12*I/(tanh(1/2*x)+I)^3+1/3/(tanh(1/2*x)+I)^6-15/8/(tanh(1/2*x)+I)^
4+15/8/(tanh(1/2*x)+I)^2+5/16*ln(tanh(1/2*x)+I)

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Maxima [B]  time = 1.18321, size = 189, normalized size = 2.36 \begin{align*} \frac{32 \,{\left (15 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} + 40 i \, e^{\left (-3 \, x\right )} - 110 \, e^{\left (-4 \, x\right )} + 18 i \, e^{\left (-5 \, x\right )} + 110 \, e^{\left (-6 \, x\right )} + 40 i \, e^{\left (-7 \, x\right )} + 30 \, e^{\left (-8 \, x\right )} + 15 i \, e^{\left (-9 \, x\right )}\right )}}{-1536 i \, e^{\left (-x\right )} - 2304 \, e^{\left (-2 \, x\right )} - 6144 i \, e^{\left (-3 \, x\right )} - 1536 \, e^{\left (-4 \, x\right )} - 9216 i \, e^{\left (-5 \, x\right )} + 1536 \, e^{\left (-6 \, x\right )} - 6144 i \, e^{\left (-7 \, x\right )} + 2304 \, e^{\left (-8 \, x\right )} - 1536 i \, e^{\left (-9 \, x\right )} + 768 \, e^{\left (-10 \, x\right )} - 768} - \frac{5}{16} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{5}{16} \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5/(I+sinh(x)),x, algorithm="maxima")

[Out]

32*(15*I*e^(-x) - 30*e^(-2*x) + 40*I*e^(-3*x) - 110*e^(-4*x) + 18*I*e^(-5*x) + 110*e^(-6*x) + 40*I*e^(-7*x) +
30*e^(-8*x) + 15*I*e^(-9*x))/(-1536*I*e^(-x) - 2304*e^(-2*x) - 6144*I*e^(-3*x) - 1536*e^(-4*x) - 9216*I*e^(-5*
x) + 1536*e^(-6*x) - 6144*I*e^(-7*x) + 2304*e^(-8*x) - 1536*I*e^(-9*x) + 768*e^(-10*x) - 768) - 5/16*log(e^(-x
) + I) + 5/16*log(e^(-x) - I)

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Fricas [B]  time = 2.14083, size = 801, normalized size = 10.01 \begin{align*} \frac{{\left (15 \, e^{\left (10 \, x\right )} + 30 i \, e^{\left (9 \, x\right )} + 45 \, e^{\left (8 \, x\right )} + 120 i \, e^{\left (7 \, x\right )} + 30 \, e^{\left (6 \, x\right )} + 180 i \, e^{\left (5 \, x\right )} - 30 \, e^{\left (4 \, x\right )} + 120 i \, e^{\left (3 \, x\right )} - 45 \, e^{\left (2 \, x\right )} + 30 i \, e^{x} - 15\right )} \log \left (e^{x} + i\right ) -{\left (15 \, e^{\left (10 \, x\right )} + 30 i \, e^{\left (9 \, x\right )} + 45 \, e^{\left (8 \, x\right )} + 120 i \, e^{\left (7 \, x\right )} + 30 \, e^{\left (6 \, x\right )} + 180 i \, e^{\left (5 \, x\right )} - 30 \, e^{\left (4 \, x\right )} + 120 i \, e^{\left (3 \, x\right )} - 45 \, e^{\left (2 \, x\right )} + 30 i \, e^{x} - 15\right )} \log \left (e^{x} - i\right ) - 30 i \, e^{\left (9 \, x\right )} + 60 \, e^{\left (8 \, x\right )} - 80 i \, e^{\left (7 \, x\right )} + 220 \, e^{\left (6 \, x\right )} - 36 i \, e^{\left (5 \, x\right )} - 220 \, e^{\left (4 \, x\right )} - 80 i \, e^{\left (3 \, x\right )} - 60 \, e^{\left (2 \, x\right )} - 30 i \, e^{x}}{48 \, e^{\left (10 \, x\right )} + 96 i \, e^{\left (9 \, x\right )} + 144 \, e^{\left (8 \, x\right )} + 384 i \, e^{\left (7 \, x\right )} + 96 \, e^{\left (6 \, x\right )} + 576 i \, e^{\left (5 \, x\right )} - 96 \, e^{\left (4 \, x\right )} + 384 i \, e^{\left (3 \, x\right )} - 144 \, e^{\left (2 \, x\right )} + 96 i \, e^{x} - 48} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5/(I+sinh(x)),x, algorithm="fricas")

[Out]

((15*e^(10*x) + 30*I*e^(9*x) + 45*e^(8*x) + 120*I*e^(7*x) + 30*e^(6*x) + 180*I*e^(5*x) - 30*e^(4*x) + 120*I*e^
(3*x) - 45*e^(2*x) + 30*I*e^x - 15)*log(e^x + I) - (15*e^(10*x) + 30*I*e^(9*x) + 45*e^(8*x) + 120*I*e^(7*x) +
30*e^(6*x) + 180*I*e^(5*x) - 30*e^(4*x) + 120*I*e^(3*x) - 45*e^(2*x) + 30*I*e^x - 15)*log(e^x - I) - 30*I*e^(9
*x) + 60*e^(8*x) - 80*I*e^(7*x) + 220*e^(6*x) - 36*I*e^(5*x) - 220*e^(4*x) - 80*I*e^(3*x) - 60*e^(2*x) - 30*I*
e^x)/(48*e^(10*x) + 96*I*e^(9*x) + 144*e^(8*x) + 384*I*e^(7*x) + 96*e^(6*x) + 576*I*e^(5*x) - 96*e^(4*x) + 384
*I*e^(3*x) - 144*e^(2*x) + 96*I*e^x - 48)

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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)**5/(I+sinh(x)),x)

[Out]

Timed out

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Giac [B]  time = 1.26678, size = 159, normalized size = 1.99 \begin{align*} \frac{15 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 76 i \, e^{\left (-x\right )} - 76 i \, e^{x} - 100}{64 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{2}} - \frac{55 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 402 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 1020 \, e^{\left (-x\right )} + 1020 \, e^{x} + 936 i}{192 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{3}} + \frac{5}{32} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{5}{32} \, \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)^5/(I+sinh(x)),x, algorithm="giac")

[Out]

1/64*(15*(e^(-x) - e^x)^2 + 76*I*e^(-x) - 76*I*e^x - 100)/(e^(-x) - e^x + 2*I)^2 - 1/192*(55*(e^(-x) - e^x)^3
- 402*I*(e^(-x) - e^x)^2 - 1020*e^(-x) + 1020*e^x + 936*I)/(e^(-x) - e^x - 2*I)^3 + 5/32*log(-e^(-x) + e^x + 2
*I) - 5/32*log(-e^(-x) + e^x - 2*I)