3.7.36 \(\int (\text {sech}(x)-i \tanh (x))^3 \, dx\) [636]

Optimal. Leaf size=28 \[ i \log (i-\sinh (x))+\frac {2 i}{1+i \sinh (x)} \]

[Out]

I*ln(I-sinh(x))+2*I/(1+I*sinh(x))

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Rubi [A]
time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 = {4476, 2746, 45} \begin {gather*} \frac {2 i}{1+i \sinh (x)}+i \log (-\sinh (x)+i) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sech[x] - I*Tanh[x])^3,x]

[Out]

I*Log[I - Sinh[x]] + (2*I)/(1 + I*Sinh[x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 4476

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]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 39, normalized size = 1.39 \begin {gather*} -\text {ArcTan}(\sinh (x))+i \log (\cosh (x))+\frac {3}{2} i \text {sech}^2(x)+2 \text {sech}(x) \tanh (x)-\frac {1}{2} i \tanh ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] - I*Tanh[x])^3,x]

[Out]

-ArcTan[Sinh[x]] + I*Log[Cosh[x]] + ((3*I)/2)*Sech[x]^2 + 2*Sech[x]*Tanh[x] - (I/2)*Tanh[x]^2

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Maple [A]
time = 1.55, size = 26, normalized size = 0.93

method	result	size
risch	\(-i x +\frac {4 \,{\mathrm e}^{x}}{\left ({\mathrm e}^{x}-i\right )^{2}}+2 i \ln \left ({\mathrm e}^{x}-i\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sech(x)-I*tanh(x))^3,x,method=_RETURNVERBOSE)

[Out]

-I*x+4*exp(x)/(exp(x)-I)^2+2*I*ln(exp(x)-I)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (20) = 40\).
time = 0.46, size = 73, normalized size = 2.61 \begin {gather*} -\frac {3}{2} i \, \tanh \left (x\right )^{2} + i \, x + \frac {4 \, {\left (e^{\left (-x\right )} - e^{\left (-3 \, x\right )}\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + \frac {2 i \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + 2 \, \arctan \left (e^{\left (-x\right )}\right ) + i \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)-I*tanh(x))^3,x, algorithm="maxima")

[Out]

-3/2*I*tanh(x)^2 + I*x + 4*(e^(-x) - e^(-3*x))/(2*e^(-2*x) + e^(-4*x) + 1) + 2*I*e^(-2*x)/(2*e^(-2*x) + e^(-4*
x) + 1) + 2*arctan(e^(-x)) + I*log(e^(-2*x) + 1)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (20) = 40\).
time = 0.39, size = 50, normalized size = 1.79 \begin {gather*} \frac {-i \, x e^{\left (2 \, x\right )} - 2 \, {\left (x - 2\right )} e^{x} - 2 \, {\left (-i \, e^{\left (2 \, x\right )} - 2 \, e^{x} + i\right )} \log \left (e^{x} - i\right ) + i \, x}{e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)-I*tanh(x))^3,x, algorithm="fricas")

[Out]

(-I*x*e^(2*x) - 2*(x - 2)*e^x - 2*(-I*e^(2*x) - 2*e^x + I)*log(e^x - I) + I*x)/(e^(2*x) - 2*I*e^x - 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- i \tanh {\left (x \right )} + \operatorname {sech}{\left (x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)-I*tanh(x))**3,x)

[Out]

Integral((-I*tanh(x) + sech(x))**3, x)

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Giac [A]
time = 0.40, size = 21, normalized size = 0.75 \begin {gather*} -i \, x + \frac {4 \, e^{x}}{{\left (e^{x} - i\right )}^{2}} + 2 i \, \log \left (e^{x} - i\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)-I*tanh(x))^3,x, algorithm="giac")

[Out]

-I*x + 4*e^x/(e^x - I)^2 + 2*I*log(e^x - I)

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Mupad [B]
time = 0.16, size = 41, normalized size = 1.46 \begin {gather*} -x\,1{}\mathrm {i}+\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,2{}\mathrm {i}-\frac {4{}\mathrm {i}}{1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}}+\frac {4}{{\mathrm {e}}^x-\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(tanh(x)*1i - 1/cosh(x))^3,x)

[Out]

log(exp(x) - 1i)*2i - x*1i - 4i/(exp(x)*2i - exp(2*x) + 1) + 4/(exp(x) - 1i)

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