### 3.625 $$\int (\text{sech}(x)+i \tanh (x))^4 \, dx$$

Optimal. Leaf size=38 $x-\frac{2 i \cosh ^3(x)}{3 (1-i \sinh (x))^3}+\frac{2 i \cosh (x)}{1-i \sinh (x)}$

[Out]

x - (((2*I)/3)*Cosh[x]^3)/(1 - I*Sinh[x])^3 + ((2*I)*Cosh[x])/(1 - I*Sinh[x])

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Rubi [A]  time = 0.112979, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.364, Rules used = {4391, 2670, 2680, 8} $x-\frac{2 i \cosh ^3(x)}{3 (1-i \sinh (x))^3}+\frac{2 i \cosh (x)}{1-i \sinh (x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x] + I*Tanh[x])^4,x]

[Out]

x - (((2*I)/3)*Cosh[x]^3)/(1 - I*Sinh[x])^3 + ((2*I)*Cosh[x])/(1 - I*Sinh[x])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2670

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.126771, size = 74, normalized size = 1.95 $\frac{3 (3 x-8 i) \cosh \left (\frac{x}{2}\right )+(-3 x+16 i) \cosh \left (\frac{3 x}{2}\right )-6 i \sinh \left (\frac{x}{2}\right ) (2 x+x \cosh (x)-4 i)}{6 \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x] + I*Tanh[x])^4,x]

[Out]

(3*(-8*I + 3*x)*Cosh[x/2] + (16*I - 3*x)*Cosh[(3*x)/2] - (6*I)*(-4*I + 2*x + x*Cosh[x])*Sinh[x/2])/(6*(Cosh[x/
2] - I*Sinh[x/2])^3)

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Maple [B]  time = 0.054, size = 89, normalized size = 2.3 \begin{align*} -2\, \left ( 2/3+1/3\, \left ({\rm sech} \left (x\right ) \right ) ^{2} \right ) \tanh \left ( x \right ) +4\,i \left ({\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( x \right ) \right ) ^{3}}}+{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{3\,\cosh \left ( x \right ) }}-{\frac{\cosh \left ( x \right ) }{3}} \right ) +3\,{\frac{\sinh \left ( x \right ) }{ \left ( \cosh \left ( x \right ) \right ) ^{3}}}-4\,i \left ( -{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( x \right ) \right ) ^{3}}}+{\frac{2\, \left ( \sinh \left ( x \right ) \right ) ^{2}}{3\,\cosh \left ( x \right ) }}-{\frac{2\,\cosh \left ( x \right ) }{3}} \right ) +x-\tanh \left ( x \right ) -{\frac{ \left ( \tanh \left ( x \right ) \right ) ^{3}}{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((sech(x)+I*tanh(x))^4,x)

[Out]

-2*(2/3+1/3*sech(x)^2)*tanh(x)+4*I*(1/3*sinh(x)^2/cosh(x)^3+1/3*sinh(x)^2/cosh(x)-1/3*cosh(x))+3*sinh(x)/cosh(
x)^3-4*I*(-1/3*sinh(x)^2/cosh(x)^3+2/3*sinh(x)^2/cosh(x)-2/3*cosh(x))+x-tanh(x)-1/3*tanh(x)^3

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Maxima [B]  time = 1.056, size = 244, normalized size = 6.42 \begin{align*} -2 \, \tanh \left (x\right )^{3} + x - \frac{4 \,{\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + 2\right )}}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac{8 i \, e^{\left (-x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac{4 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac{16 i \, e^{\left (-3 \, x\right )}}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac{8 i \, e^{\left (-5 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1} + \frac{4}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} - \frac{32 i}{3 \,{\left (e^{\left (-x\right )} + e^{x}\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^4,x, algorithm="maxima")

[Out]

-2*tanh(x)^3 + x - 4/3*(3*e^(-2*x) + 3*e^(-4*x) + 2)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 8*I*e^(-x)/(3*
e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 4*e^(-2*x)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 16/3*I*e^(-3*x)/
(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 8*I*e^(-5*x)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 4/3/(3*e^(-
2*x) + 3*e^(-4*x) + e^(-6*x) + 1) - 32/3*I/(e^(-x) + e^x)^3

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Fricas [A]  time = 2.32624, size = 153, normalized size = 4.03 \begin{align*} \frac{3 \, x e^{\left (3 \, x\right )} +{\left (9 i \, x + 24 i\right )} e^{\left (2 \, x\right )} - 3 \,{\left (3 \, x + 8\right )} e^{x} - 3 i \, x - 16 i}{3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^4,x, algorithm="fricas")

[Out]

(3*x*e^(3*x) + (9*I*x + 24*I)*e^(2*x) - 3*(3*x + 8)*e^x - 3*I*x - 16*I)/(3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (i \tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )}\right )^{4}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))**4,x)

[Out]

Integral((I*tanh(x) + sech(x))**4, x)

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Giac [A]  time = 1.14611, size = 30, normalized size = 0.79 \begin{align*} x - \frac{-24 i \, e^{\left (2 \, x\right )} + 24 \, e^{x} + 16 i}{3 \,{\left (e^{x} + i\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^4,x, algorithm="giac")

[Out]

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