3.221 \(\int \frac{\tanh (x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=36 \[ -\frac{i}{4 (\sinh (x)+i)}-\frac{1}{4 (\sinh (x)+i)^2}-\frac{1}{4} i \tan ^{-1}(\sinh (x)) \]

[Out]

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

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Rubi [A]  time = 0.0365092, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2707, 77, 203} \[ -\frac{i}{4 (\sinh (x)+i)}-\frac{1}{4 (\sinh (x)+i)^2}-\frac{1}{4} i \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(I + Sinh[x])^2,x]

[Out]

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

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.0303892, size = 29, normalized size = 0.81 \[ -\frac{i \left (\sinh (x)+(\sinh (x)+i)^2 \tan ^{-1}(\sinh (x))\right )}{4 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(I + Sinh[x])^2,x]

[Out]

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

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Maple [B]  time = 0.051, size = 66, normalized size = 1.8 \begin{align*} -{\frac{1}{4}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}-{\frac{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{1}{4}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(I+sinh(x))^2,x)

[Out]

-1/4*ln(tanh(1/2*x)-I)+2*I/(tanh(1/2*x)+I)^3-1/2*I/(tanh(1/2*x)+I)+1/(tanh(1/2*x)+I)^4-3/2/(tanh(1/2*x)+I)^2+1
/4*ln(tanh(1/2*x)+I)

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Maxima [B]  time = 1.07141, size = 82, normalized size = 2.28 \begin{align*} \frac{-i \, e^{\left (-x\right )} + i \, e^{\left (-3 \, x\right )}}{8 i \, e^{\left (-x\right )} - 12 \, e^{\left (-2 \, x\right )} - 8 i \, e^{\left (-3 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + 2} - \frac{1}{4} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{1}{4} \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06011, size = 284, normalized size = 7.89 \begin{align*} \frac{{\left (e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1\right )} \log \left (e^{x} + i\right ) -{\left (e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1\right )} \log \left (e^{x} - i\right ) - 2 i \, e^{\left (3 \, x\right )} + 2 i \, e^{x}}{4 \, e^{\left (4 \, x\right )} + 16 i \, e^{\left (3 \, x\right )} - 24 \, e^{\left (2 \, x\right )} - 16 i \, e^{x} + 4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((e^(4*x) + 4*I*e^(3*x) - 6*e^(2*x) - 4*I*e^x + 1)*log(e^x + I) - (e^(4*x) + 4*I*e^(3*x) - 6*e^(2*x) - 4*I*e^x
+ 1)*log(e^x - I) - 2*I*e^(3*x) + 2*I*e^x)/(4*e^(4*x) + 16*I*e^(3*x) - 24*e^(2*x) - 16*I*e^x + 4)

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Sympy [B]  time = 0.486307, size = 60, normalized size = 1.67 \begin{align*} \frac{- \frac{i e^{3 x}}{2} + \frac{i e^{x}}{2}}{e^{4 x} + 4 i e^{3 x} - 6 e^{2 x} - 4 i e^{x} + 1} - \frac{\log{\left (e^{x} - i \right )}}{4} + \frac{\log{\left (e^{x} + i \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(I+sinh(x))**2,x)

[Out]

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

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Giac [B]  time = 1.13077, size = 89, normalized size = 2.47 \begin{align*} -\frac{3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 20 i \, e^{\left (-x\right )} + 20 i \, e^{x} - 12}{16 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{2}} + \frac{1}{8} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{1}{8} \, \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(tanh(x)/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/16*(3*(e^(-x) - e^x)^2 - 20*I*e^(-x) + 20*I*e^x - 12)/(e^(-x) - e^x - 2*I)^2 + 1/8*log(-e^(-x) + e^x + 2*I)
- 1/8*log(-e^(-x) + e^x - 2*I)