3.208 \(\int \frac{\tanh ^4(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=31 \[ -\frac{1}{5} i \tanh ^5(x)-\frac{1}{5} \text{sech}^5(x)+\frac{2 \text{sech}^3(x)}{3}-\text{sech}(x) \]

[Out]

-Sech[x] + (2*Sech[x]^3)/3 - Sech[x]^5/5 - (I/5)*Tanh[x]^5

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Rubi [A]  time = 0.0789464, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2606, 194} \[ -\frac{1}{5} i \tanh ^5(x)-\frac{1}{5} \text{sech}^5(x)+\frac{2 \text{sech}^3(x)}{3}-\text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(I + Sinh[x]),x]

[Out]

-Sech[x] + (2*Sech[x]^3)/3 - Sech[x]^5/5 - (I/5)*Tanh[x]^5

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{i+\sinh (x)} \, dx &=-\left (i \int \text{sech}^2(x) \tanh ^4(x) \, dx\right )+\int \text{sech}(x) \tanh ^5(x) \, dx\\ &=-\operatorname{Subst}\left (\int x^4 \, dx,x,i \tanh (x)\right )-\operatorname{Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\text{sech}(x)\right )\\ &=-\frac{1}{5} i \tanh ^5(x)-\operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\text{sech}(x)\right )\\ &=-\text{sech}(x)+\frac{2 \text{sech}^3(x)}{3}-\frac{\text{sech}^5(x)}{5}-\frac{1}{5} i \tanh ^5(x)\\ \end{align*}

Mathematica [B]  time = 0.135313, size = 96, normalized size = 3.1 \[ -\frac{64 i \sinh (x)+178 i \sinh (2 x)-192 i \sinh (3 x)+89 i \sinh (4 x)-534 \cosh (x)+288 \cosh (2 x)-178 \cosh (3 x)+24 \cosh (4 x)+200}{960 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^5 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(I + Sinh[x]),x]

[Out]

-(200 - 534*Cosh[x] + 288*Cosh[2*x] - 178*Cosh[3*x] + 24*Cosh[4*x] + (64*I)*Sinh[x] + (178*I)*Sinh[2*x] - (192
*I)*Sinh[3*x] + (89*I)*Sinh[4*x])/(960*(Cosh[x/2] - I*Sinh[x/2])^5*(Cosh[x/2] + I*Sinh[x/2])^3)

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Maple [B]  time = 0.07, size = 93, normalized size = 3. \begin{align*}{{\frac{3\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{{\frac{i}{6}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{{\frac{i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{{\frac{2\,i}{5}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{{\frac{3\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(I+sinh(x)),x)

[Out]

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

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Maxima [B]  time = 1.12585, size = 558, normalized size = 18. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x)),x, algorithm="maxima")

[Out]

18*e^(-x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8
*x) - 15) + 42*I*e^(-2*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(
-7*x) + 15*e^(-8*x) - 15) - 26*e^(-3*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6
*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) + 50*I*e^(-4*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(
-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) - 10*e^(-5*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3
*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) + 30*I*e^(-6*x)/(-30*I*e^(-x) - 30*e^(-2
*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) - 30*e^(-7*x)/(-30*I*e^(
-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) + 6*I/(-30
*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15)

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Fricas [B]  time = 2.07948, size = 271, normalized size = 8.74 \begin{align*} -\frac{30 \, e^{\left (7 \, x\right )} + 30 i \, e^{\left (6 \, x\right )} + 10 \, e^{\left (5 \, x\right )} + 50 i \, e^{\left (4 \, x\right )} + 26 \, e^{\left (3 \, x\right )} + 42 i \, e^{\left (2 \, x\right )} - 18 \, e^{x} + 6 i}{15 \, e^{\left (8 \, x\right )} + 30 i \, e^{\left (7 \, x\right )} + 30 \, e^{\left (6 \, x\right )} + 90 i \, e^{\left (5 \, x\right )} + 90 i \, e^{\left (3 \, x\right )} - 30 \, e^{\left (2 \, x\right )} + 30 i \, e^{x} - 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x)),x, algorithm="fricas")

[Out]

-(30*e^(7*x) + 30*I*e^(6*x) + 10*e^(5*x) + 50*I*e^(4*x) + 26*e^(3*x) + 42*I*e^(2*x) - 18*e^x + 6*I)/(15*e^(8*x
) + 30*I*e^(7*x) + 30*e^(6*x) + 90*I*e^(5*x) + 90*I*e^(3*x) - 30*e^(2*x) + 30*I*e^x - 15)

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Sympy [B]  time = 1.30056, size = 116, normalized size = 3.74 \begin{align*} \frac{- 2 e^{7 x} - 2 i e^{6 x} - \frac{2 e^{5 x}}{3} - \frac{10 i e^{4 x}}{3} - \frac{26 e^{3 x}}{15} - \frac{14 i e^{2 x}}{5} + \frac{6 e^{x}}{5} - \frac{2 i}{5}}{e^{8 x} + 2 i e^{7 x} + 2 e^{6 x} + 6 i e^{5 x} + 6 i e^{3 x} - 2 e^{2 x} + 2 i e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**4/(I+sinh(x)),x)

[Out]

(-2*exp(7*x) - 2*I*exp(6*x) - 2*exp(5*x)/3 - 10*I*exp(4*x)/3 - 26*exp(3*x)/15 - 14*I*exp(2*x)/5 + 6*exp(x)/5 -
2*I/5)/(exp(8*x) + 2*I*exp(7*x) + 2*exp(6*x) + 6*I*exp(5*x) + 6*I*exp(3*x) - 2*exp(2*x) + 2*I*exp(x) - 1)

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Giac [B]  time = 1.15486, size = 72, normalized size = 2.32 \begin{align*} -\frac{15 \, e^{\left (2 \, x\right )} - 24 i \, e^{x} - 13}{24 \,{\left (e^{x} - i\right )}^{3}} - \frac{165 \, e^{\left (4 \, x\right )} + 480 i \, e^{\left (3 \, x\right )} - 650 \, e^{\left (2 \, x\right )} - 400 i \, e^{x} + 113}{120 \,{\left (e^{x} + i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^4/(I+sinh(x)),x, algorithm="giac")

[Out]

-1/24*(15*e^(2*x) - 24*I*e^x - 13)/(e^x - I)^3 - 1/120*(165*e^(4*x) + 480*I*e^(3*x) - 650*e^(2*x) - 400*I*e^x
+ 113)/(e^x + I)^5