3.624 \(\int (\text {sech}(x)+i \tanh (x))^5 \, dx\)

Optimal. Leaf size=40 \[ \frac {4 i}{1-i \sinh (x)}-\frac {2 i}{(1-i \sinh (x))^2}+i \log (\sinh (x)+i) \]

[Out]

I*ln(I+sinh(x))-2*I/(1-I*sinh(x))^2+4*I/(1-I*sinh(x))

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Rubi [A]  time = 0.06, antiderivative size = 40, 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 = {4391, 2667, 43} \[ \frac {4 i}{1-i \sinh (x)}-\frac {2 i}{(1-i \sinh (x))^2}+i \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x] + I*Tanh[x])^5,x]

[Out]

I*Log[I + Sinh[x]] - (2*I)/(1 - I*Sinh[x])^2 + (4*I)/(1 - I*Sinh[x])

Rule 43

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 2667

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 62, normalized size = 1.55 \[ -\frac {11}{4} i \tanh ^4(x)-\frac {1}{2} i \tanh ^2(x)-\frac {5}{4} i \text {sech}^4(x)+\tan ^{-1}(\sinh (x))+i \log (\cosh (x))-\tanh (x) \text {sech}^3(x)-5 \tanh ^3(x) \text {sech}(x)+\tanh (x) \text {sech}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] + I*Tanh[x])^5,x]

[Out]

ArcTan[Sinh[x]] + I*Log[Cosh[x]] - ((5*I)/4)*Sech[x]^4 + Sech[x]*Tanh[x] - Sech[x]^3*Tanh[x] - (I/2)*Tanh[x]^2
- 5*Sech[x]*Tanh[x]^3 - ((11*I)/4)*Tanh[x]^4

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fricas [B]  time = 0.42, size = 92, normalized size = 2.30 \[ \frac {-i \, x e^{\left (4 \, x\right )} + 4 \, {\left (x - 2\right )} e^{\left (3 \, x\right )} + {\left (6 i \, x - 8 i\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x - 2\right )} e^{x} + {\left (2 i \, e^{\left (4 \, x\right )} - 8 \, e^{\left (3 \, x\right )} - 12 i \, e^{\left (2 \, x\right )} + 8 \, e^{x} + 2 i\right )} \log \left (e^{x} + i\right ) - i \, x}{e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^5,x, algorithm="fricas")

[Out]

(-I*x*e^(4*x) + 4*(x - 2)*e^(3*x) + (6*I*x - 8*I)*e^(2*x) - 4*(x - 2)*e^x + (2*I*e^(4*x) - 8*e^(3*x) - 12*I*e^
(2*x) + 8*e^x + 2*I)*log(e^x + I) - I*x)/(e^(4*x) + 4*I*e^(3*x) - 6*e^(2*x) - 4*I*e^x + 1)

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giac [A]  time = 0.15, size = 34, normalized size = 0.85 \[ -i \, x - \frac {8 \, {\left (e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x}\right )}}{{\left (e^{x} + i\right )}^{4}} + 2 i \, \log \left (e^{x} + i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^5,x, algorithm="giac")

[Out]

-I*x - 8*(e^(3*x) + I*e^(2*x) - e^x)/(e^x + I)^4 + 2*I*log(e^x + I)

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maple [B]  time = 0.43, size = 78, normalized size = 1.95 \[ \frac {8 \left (\frac {\mathrm {sech}\relax (x )^{3}}{4}+\frac {3 \,\mathrm {sech}\relax (x )}{8}\right ) \tanh \relax (x )}{3}+2 \arctan \left ({\mathrm e}^{x}\right )+\frac {5 i}{4 \cosh \relax (x )^{4}}-\frac {5 \sinh \relax (x )}{3 \cosh \relax (x )^{4}}+\frac {5 i \left (\sinh ^{2}\relax (x )\right )}{\cosh \relax (x )^{4}}-\frac {5 \left (\sinh ^{3}\relax (x )\right )}{\cosh \relax (x )^{4}}+i \ln \left (\cosh \relax (x )\right )-\frac {i \left (\tanh ^{2}\relax (x )\right )}{2}-\frac {i \left (\tanh ^{4}\relax (x )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sech(x)+I*tanh(x))^5,x)

[Out]

8/3*(1/4*sech(x)^3+3/8*sech(x))*tanh(x)+2*arctan(exp(x))+5/4*I/cosh(x)^4-5/3*sinh(x)/cosh(x)^4+5*I*sinh(x)^2/c
osh(x)^4-5*sinh(x)^3/cosh(x)^4+I*ln(cosh(x))-1/2*I*tanh(x)^2-1/4*I*tanh(x)^4

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maxima [B]  time = 0.50, size = 235, normalized size = 5.88 \[ -\frac {5}{2} i \, \tanh \relax (x)^{4} + i \, x - \frac {5 \, {\left (5 \, e^{\left (-x\right )} - 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} - 5 \, e^{\left (-7 \, x\right )}\right )}}{4 \, {\left (4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} + \frac {3 \, e^{\left (-x\right )} + 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} - 3 \, e^{\left (-7 \, x\right )}}{4 \, {\left (4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} - \frac {5 \, {\left (e^{\left (-x\right )} - 7 \, e^{\left (-3 \, x\right )} + 7 \, e^{\left (-5 \, x\right )} - e^{\left (-7 \, x\right )}\right )}}{2 \, {\left (4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} + \frac {4 i \, {\left (e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1} - \frac {20 i}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{4}} - 2 \, \arctan \left (e^{\left (-x\right )}\right ) + i \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))^5,x, algorithm="maxima")

[Out]

-5/2*I*tanh(x)^4 + I*x - 5/4*(5*e^(-x) - 3*e^(-3*x) + 3*e^(-5*x) - 5*e^(-7*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^
(-6*x) + e^(-8*x) + 1) + 1/4*(3*e^(-x) + 11*e^(-3*x) - 11*e^(-5*x) - 3*e^(-7*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*
e^(-6*x) + e^(-8*x) + 1) - 5/2*(e^(-x) - 7*e^(-3*x) + 7*e^(-5*x) - e^(-7*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^(-
6*x) + e^(-8*x) + 1) + 4*I*(e^(-2*x) + e^(-4*x) + e^(-6*x))/(4*e^(-2*x) + 6*e^(-4*x) + 4*e^(-6*x) + e^(-8*x) +
1) - 20*I/(e^(-x) + e^x)^4 - 2*arctan(e^(-x)) + I*log(e^(-2*x) + 1)

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mupad [B]  time = 0.28, size = 90, normalized size = 2.25 \[ -x\,1{}\mathrm {i}+\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,2{}\mathrm {i}+\frac {16{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}}-\frac {8{}\mathrm {i}}{{\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1+{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,4{}\mathrm {i}}-\frac {8}{{\mathrm {e}}^x+1{}\mathrm {i}}+\frac {16}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)*1i + 1/cosh(x))^5,x)

[Out]

log(exp(x) + 1i)*2i - x*1i + 16i/(exp(2*x) + exp(x)*2i - 1) - 8i/(exp(3*x)*4i - 6*exp(2*x) + exp(4*x) - exp(x)
*4i + 1) - 8/(exp(x) + 1i) + 16/(exp(2*x)*3i + exp(3*x) - 3*exp(x) - 1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i \tanh {\relax (x )} + \operatorname {sech}{\relax (x )}\right )^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sech(x)+I*tanh(x))**5,x)

[Out]

Integral((I*tanh(x) + sech(x))**5, x)

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