3.103 \(\int \frac{\tanh ^4(x)}{i+\text{csch}(x)} \, dx\)

Optimal. Leaf size=52 \[ -i x+\frac{1}{5} \tanh ^5(x) (-\text{csch}(x)+i)+\frac{1}{15} \tanh ^3(x) (-4 \text{csch}(x)+5 i)+\frac{1}{15} \tanh (x) (-8 \text{csch}(x)+15 i) \]

[Out]

(-I)*x + ((15*I - 8*Csch[x])*Tanh[x])/15 + ((5*I - 4*Csch[x])*Tanh[x]^3)/15 + ((I - Csch[x])*Tanh[x]^5)/5

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Rubi [A]  time = 0.0934978, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3888, 3882, 8} \[ -i x+\frac{1}{5} \tanh ^5(x) (-\text{csch}(x)+i)+\frac{1}{15} \tanh ^3(x) (-4 \text{csch}(x)+5 i)+\frac{1}{15} \tanh (x) (-8 \text{csch}(x)+15 i) \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(I + Csch[x]),x]

[Out]

(-I)*x + ((15*I - 8*Csch[x])*Tanh[x])/15 + ((5*I - 4*Csch[x])*Tanh[x]^3)/15 + ((I - Csch[x])*Tanh[x]^5)/5

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{i+\text{csch}(x)} \, dx &=\int (-i+\text{csch}(x)) \tanh ^6(x) \, dx\\ &=\frac{1}{5} (i-\text{csch}(x)) \tanh ^5(x)-\frac{1}{5} \int (5 i-4 \text{csch}(x)) \tanh ^4(x) \, dx\\ &=\frac{1}{15} (5 i-4 \text{csch}(x)) \tanh ^3(x)+\frac{1}{5} (i-\text{csch}(x)) \tanh ^5(x)+\frac{1}{15} \int (-15 i+8 \text{csch}(x)) \tanh ^2(x) \, dx\\ &=\frac{1}{15} (15 i-8 \text{csch}(x)) \tanh (x)+\frac{1}{15} (5 i-4 \text{csch}(x)) \tanh ^3(x)+\frac{1}{5} (i-\text{csch}(x)) \tanh ^5(x)-\frac{1}{15} \int 15 i \, dx\\ &=-i x+\frac{1}{15} (15 i-8 \text{csch}(x)) \tanh (x)+\frac{1}{15} (5 i-4 \text{csch}(x)) \tanh ^3(x)+\frac{1}{5} (i-\text{csch}(x)) \tanh ^5(x)\\ \end{align*}

Mathematica [B]  time = 0.125434, size = 126, normalized size = 2.42 \[ \frac{64 i \sinh (x)+240 x \sinh (2 x)+178 i \sinh (2 x)+128 i \sinh (3 x)+120 x \sinh (4 x)+89 i \sinh (4 x)+6 (89-120 i x) \cosh (x)-128 \cosh (2 x)-240 i x \cosh (3 x)+178 \cosh (3 x)-184 \cosh (4 x)-200}{960 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^3 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(I + Csch[x]),x]

[Out]

(-200 + 6*(89 - (120*I)*x)*Cosh[x] - 128*Cosh[2*x] + 178*Cosh[3*x] - (240*I)*x*Cosh[3*x] - 184*Cosh[4*x] + (64
*I)*Sinh[x] + (178*I)*Sinh[2*x] + 240*x*Sinh[2*x] + (128*I)*Sinh[3*x] + (89*I)*Sinh[4*x] + 120*x*Sinh[4*x])/(9
60*(Cosh[x/2] - I*Sinh[x/2])^3*(Cosh[x/2] + I*Sinh[x/2])^5)

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Maple [B]  time = 0.056, size = 99, normalized size = 1.9 \begin{align*} -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +{{\frac{11\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{{\frac{2\,i}{5}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-5}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}+i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +{{\frac{5\,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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02734, size = 130, normalized size = 2.5 \begin{align*} -i \, x - \frac{62 \, e^{\left (-x\right )} + 62 i \, e^{\left (-2 \, x\right )} + 146 \, e^{\left (-3 \, x\right )} + 50 i \, e^{\left (-4 \, x\right )} + 130 \, e^{\left (-5 \, x\right )} - 30 i \, e^{\left (-6 \, x\right )} + 30 \, e^{\left (-7 \, x\right )} + 46 i}{30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} + 90 i \, e^{\left (-3 \, x\right )} + 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} + 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x - (62*e^(-x) + 62*I*e^(-2*x) + 146*e^(-3*x) + 50*I*e^(-4*x) + 130*e^(-5*x) - 30*I*e^(-6*x) + 30*e^(-7*x)
+ 46*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 = 1.67382, size = 394, normalized size = 7.58 \begin{align*} \frac{-15 i \, x e^{\left (8 \, x\right )} - 30 \,{\left (x + 1\right )} e^{\left (7 \, x\right )} +{\left (-30 i \, x - 30 i\right )} e^{\left (6 \, x\right )} - 10 \,{\left (9 \, x + 13\right )} e^{\left (5 \, x\right )} - 2 \,{\left (45 \, x + 73\right )} e^{\left (3 \, x\right )} +{\left (30 i \, x + 62 i\right )} e^{\left (2 \, x\right )} - 2 \,{\left (15 \, x + 31\right )} e^{x} + 15 i \, x + 50 i \, e^{\left (4 \, x\right )} + 46 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+csch(x)),x, algorithm="fricas")

[Out]

(-15*I*x*e^(8*x) - 30*(x + 1)*e^(7*x) + (-30*I*x - 30*I)*e^(6*x) - 10*(9*x + 13)*e^(5*x) - 2*(45*x + 73)*e^(3*
x) + (30*I*x + 62*I)*e^(2*x) - 2*(15*x + 31)*e^x + 15*I*x + 50*I*e^(4*x) + 46*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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{4}{\left (x \right )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(x)**4/(csch(x) + I), x)

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Giac [A]  time = 1.21189, size = 84, normalized size = 1.62 \begin{align*} -\frac{21 i \, e^{\left (2 \, x\right )} - 36 \, e^{x} - 19 i}{24 \,{\left (i \, e^{x} - 1\right )}^{3}} - \frac{115 \, e^{\left (4 \, x\right )} - 380 i \, e^{\left (3 \, x\right )} - 530 \, e^{\left (2 \, x\right )} + 340 i \, e^{x} + 91}{40 \,{\left (e^{x} - i\right )}^{5}} - i \, \log \left (i \, e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(21*I*e^(2*x) - 36*e^x - 19*I)/(I*e^x - 1)^3 - 1/40*(115*e^(4*x) - 380*I*e^(3*x) - 530*e^(2*x) + 340*I*e
^x + 91)/(e^x - I)^5 - I*log(I*e^x)