3.219 \(\int \frac{\tanh ^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=66 \[ -\frac{i}{16 (-\sinh (x)+i)}-\frac{3 i}{16 (\sinh (x)+i)}-\frac{1}{4 (\sinh (x)+i)^2}+\frac{i}{12 (\sinh (x)+i)^3}-\frac{1}{8} i \tan ^{-1}(\sinh (x)) \]

[Out]

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

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Rubi [A]  time = 0.0645879, antiderivative size = 66, 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 = {2707, 88, 203} \[ -\frac{i}{16 (-\sinh (x)+i)}-\frac{3 i}{16 (\sinh (x)+i)}-\frac{1}{4 (\sinh (x)+i)^2}+\frac{i}{12 (\sinh (x)+i)^3}-\frac{1}{8} i \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(I + Sinh[x])^2,x]

[Out]

(-I/8)*ArcTan[Sinh[x]] - (I/16)/(I - Sinh[x]) + (I/12)/(I + Sinh[x])^3 - 1/(4*(I + Sinh[x])^2) - ((3*I)/16)/(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A]  time = 0.0997977, size = 52, normalized size = 0.79 \[ \frac{1}{48} \left (\frac{2 \left (-3 i \sinh ^3(x)-6 \sinh ^2(x)-7 i \sinh (x)+2\right )}{(\sinh (x)-i) (\sinh (x)+i)^3}-6 i \tan ^{-1}(\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*ArcTan[Sinh[x]] + (2*(2 - (7*I)*Sinh[x] - 6*Sinh[x]^2 - (3*I)*Sinh[x]^3))/((-I + Sinh[x])*(I + Sinh[x]
)^3))/48

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Maple [B]  time = 0.078, size = 114, normalized size = 1.7 \begin{align*}{-{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}-{\frac{1}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{{\frac{2\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{{\frac{i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+{\frac{2}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-6}}-2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-4}-{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{1}{8}\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)^3/(I+sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.11928, size = 155, normalized size = 2.35 \begin{align*} \frac{-3 i \, e^{\left (-x\right )} - 12 \, e^{\left (-2 \, x\right )} - 19 i \, e^{\left (-3 \, x\right )} + 40 \, e^{\left (-4 \, x\right )} + 19 i \, e^{\left (-5 \, x\right )} - 12 \, e^{\left (-6 \, x\right )} + 3 i \, e^{\left (-7 \, x\right )}}{48 i \, e^{\left (-x\right )} - 48 \, e^{\left (-2 \, x\right )} + 48 i \, e^{\left (-3 \, x\right )} - 120 \, e^{\left (-4 \, x\right )} - 48 i \, e^{\left (-5 \, x\right )} - 48 \, e^{\left (-6 \, x\right )} - 48 i \, e^{\left (-7 \, x\right )} + 12 \, e^{\left (-8 \, x\right )} + 12} - \frac{1}{8} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{1}{8} \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*I*e^(-x) - 12*e^(-2*x) - 19*I*e^(-3*x) + 40*e^(-4*x) + 19*I*e^(-5*x) - 12*e^(-6*x) + 3*I*e^(-7*x))/(48*I*e
^(-x) - 48*e^(-2*x) + 48*I*e^(-3*x) - 120*e^(-4*x) - 48*I*e^(-5*x) - 48*e^(-6*x) - 48*I*e^(-7*x) + 12*e^(-8*x)
+ 12) - 1/8*log(e^(-x) + I) + 1/8*log(e^(-x) - I)

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Fricas [B]  time = 2.02832, size = 621, normalized size = 9.41 \begin{align*} \frac{{\left (3 \, e^{\left (8 \, x\right )} + 12 i \, e^{\left (7 \, x\right )} - 12 \, e^{\left (6 \, x\right )} + 12 i \, e^{\left (5 \, x\right )} - 30 \, e^{\left (4 \, x\right )} - 12 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 12 i \, e^{x} + 3\right )} \log \left (e^{x} + i\right ) -{\left (3 \, e^{\left (8 \, x\right )} + 12 i \, e^{\left (7 \, x\right )} - 12 \, e^{\left (6 \, x\right )} + 12 i \, e^{\left (5 \, x\right )} - 30 \, e^{\left (4 \, x\right )} - 12 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 12 i \, e^{x} + 3\right )} \log \left (e^{x} - i\right ) - 6 i \, e^{\left (7 \, x\right )} - 24 \, e^{\left (6 \, x\right )} - 38 i \, e^{\left (5 \, x\right )} + 80 \, e^{\left (4 \, x\right )} + 38 i \, e^{\left (3 \, x\right )} - 24 \, e^{\left (2 \, x\right )} + 6 i \, e^{x}}{24 \, e^{\left (8 \, x\right )} + 96 i \, e^{\left (7 \, x\right )} - 96 \, e^{\left (6 \, x\right )} + 96 i \, e^{\left (5 \, x\right )} - 240 \, e^{\left (4 \, x\right )} - 96 i \, e^{\left (3 \, x\right )} - 96 \, e^{\left (2 \, x\right )} - 96 i \, e^{x} + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*e^(8*x) + 12*I*e^(7*x) - 12*e^(6*x) + 12*I*e^(5*x) - 30*e^(4*x) - 12*I*e^(3*x) - 12*e^(2*x) - 12*I*e^x + 3
)*log(e^x + I) - (3*e^(8*x) + 12*I*e^(7*x) - 12*e^(6*x) + 12*I*e^(5*x) - 30*e^(4*x) - 12*I*e^(3*x) - 12*e^(2*x
) - 12*I*e^x + 3)*log(e^x - I) - 6*I*e^(7*x) - 24*e^(6*x) - 38*I*e^(5*x) + 80*e^(4*x) + 38*I*e^(3*x) - 24*e^(2
*x) + 6*I*e^x)/(24*e^(8*x) + 96*I*e^(7*x) - 96*e^(6*x) + 96*I*e^(5*x) - 240*e^(4*x) - 96*I*e^(3*x) - 96*e^(2*x
) - 96*I*e^x + 24)

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Sympy [B]  time = 1.5879, size = 129, normalized size = 1.95 \begin{align*} \frac{- \frac{i e^{7 x}}{4} - e^{6 x} - \frac{19 i e^{5 x}}{12} + \frac{10 e^{4 x}}{3} + \frac{19 i e^{3 x}}{12} - e^{2 x} + \frac{i e^{x}}{4}}{e^{8 x} + 4 i e^{7 x} - 4 e^{6 x} + 4 i e^{5 x} - 10 e^{4 x} - 4 i e^{3 x} - 4 e^{2 x} - 4 i e^{x} + 1} - \frac{\log{\left (e^{x} - i \right )}}{8} + \frac{\log{\left (e^{x} + i \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*exp(7*x)/4 - exp(6*x) - 19*I*exp(5*x)/12 + 10*exp(4*x)/3 + 19*I*exp(3*x)/12 - exp(2*x) + I*exp(x)/4)/(exp(
8*x) + 4*I*exp(7*x) - 4*exp(6*x) + 4*I*exp(5*x) - 10*exp(4*x) - 4*I*exp(3*x) - 4*exp(2*x) - 4*I*exp(x) + 1) -
log(exp(x) - I)/8 + log(exp(x) + I)/8

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Giac [B]  time = 1.18423, size = 138, normalized size = 2.09 \begin{align*} \frac{e^{\left (-x\right )} - e^{x}}{16 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}} - \frac{11 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} - 102 i \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 180 \, e^{\left (-x\right )} + 180 \, e^{x} + 104 i}{96 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{3}} + \frac{1}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{1}{16} \, \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)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

1/16*(e^(-x) - e^x)/(e^(-x) - e^x + 2*I) - 1/96*(11*(e^(-x) - e^x)^3 - 102*I*(e^(-x) - e^x)^2 - 180*e^(-x) + 1
80*e^x + 104*I)/(e^(-x) - e^x - 2*I)^3 + 1/16*log(-e^(-x) + e^x + 2*I) - 1/16*log(-e^(-x) + e^x - 2*I)