### 3.1041 $$\int x \cosh (2 x) \text{sech}^3(x) \, dx$$

Optimal. Leaf size=53 $-\frac{3}{2} i \text{PolyLog}\left (2,-i e^x\right )+\frac{3}{2} i \text{PolyLog}\left (2,i e^x\right )+3 x \tan ^{-1}\left (e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \tanh (x) \text{sech}(x)$

[Out]

3*x*ArcTan[E^x] - ((3*I)/2)*PolyLog[2, (-I)*E^x] + ((3*I)/2)*PolyLog[2, I*E^x] - Sech[x]/2 - (x*Sech[x]*Tanh[x
])/2

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Rubi [A]  time = 0.131812, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {5473, 4180, 2279, 2391, 5455, 4185} $-\frac{3}{2} i \text{PolyLog}\left (2,-i e^x\right )+\frac{3}{2} i \text{PolyLog}\left (2,i e^x\right )+3 x \tan ^{-1}\left (e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \tanh (x) \text{sech}(x)$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Cosh[2*x]*Sech[x]^3,x]

[Out]

3*x*ArcTan[E^x] - ((3*I)/2)*PolyLog[2, (-I)*E^x] + ((3*I)/2)*PolyLog[2, I*E^x] - Sech[x]/2 - (x*Sech[x]*Tanh[x
])/2

Rule 5473

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sinh, Cosh}, F] && MemberQ[{Sech, Csch}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b
/d, 1]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5455

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]*Tanh[(a_.) + (b_.)*(x_)]^(p_), x_Symbol] :> Int[(c + d
*x)^m*Sech[a + b*x]*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sech[a + b*x]^3*Tanh[a + b*x]^(p - 2), x] /; F
reeQ[{a, b, c, d, m}, x] && IGtQ[p/2, 0]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

\begin{align*} \int x \cosh (2 x) \text{sech}^3(x) \, dx &=\int \left (x \text{sech}(x)+x \text{sech}(x) \tanh ^2(x)\right ) \, dx\\ &=\int x \text{sech}(x) \, dx+\int x \text{sech}(x) \tanh ^2(x) \, dx\\ &=2 x \tan ^{-1}\left (e^x\right )-i \int \log \left (1-i e^x\right ) \, dx+i \int \log \left (1+i e^x\right ) \, dx+\int x \text{sech}(x) \, dx-\int x \text{sech}^3(x) \, dx\\ &=4 x \tan ^{-1}\left (e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \text{sech}(x) \tanh (x)-i \int \log \left (1-i e^x\right ) \, dx+i \int \log \left (1+i e^x\right ) \, dx-i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^x\right )+i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^x\right )-\frac{1}{2} \int x \text{sech}(x) \, dx\\ &=3 x \tan ^{-1}\left (e^x\right )-i \text{Li}_2\left (-i e^x\right )+i \text{Li}_2\left (i e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \text{sech}(x) \tanh (x)+\frac{1}{2} i \int \log \left (1-i e^x\right ) \, dx-\frac{1}{2} i \int \log \left (1+i e^x\right ) \, dx-i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^x\right )+i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^x\right )\\ &=3 x \tan ^{-1}\left (e^x\right )-2 i \text{Li}_2\left (-i e^x\right )+2 i \text{Li}_2\left (i e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \text{sech}(x) \tanh (x)+\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^x\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^x\right )\\ &=3 x \tan ^{-1}\left (e^x\right )-\frac{3}{2} i \text{Li}_2\left (-i e^x\right )+\frac{3}{2} i \text{Li}_2\left (i e^x\right )-\frac{\text{sech}(x)}{2}-\frac{1}{2} x \text{sech}(x) \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.0640587, size = 78, normalized size = 1.47 $-\frac{1}{2} i \left (3 \text{PolyLog}\left (2,-i e^{-x}\right )-3 \text{PolyLog}\left (2,i e^{-x}\right )+3 x \log \left (1-i e^{-x}\right )-3 x \log \left (1+i e^{-x}\right )-i \text{sech}(x)-i x \tanh (x) \text{sech}(x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Cosh[2*x]*Sech[x]^3,x]

[Out]

(-I/2)*(3*x*Log[1 - I/E^x] - 3*x*Log[1 + I/E^x] + 3*PolyLog[2, (-I)/E^x] - 3*PolyLog[2, I/E^x] - I*Sech[x] - I
*x*Sech[x]*Tanh[x])

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Maple [A]  time = 0.037, size = 75, normalized size = 1.4 \begin{align*} -{\frac{{{\rm e}^{x}} \left ( x{{\rm e}^{2\,x}}+{{\rm e}^{2\,x}}-x+1 \right ) }{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}-{\frac{3\,i}{2}}x\ln \left ( 1+i{{\rm e}^{x}} \right ) +{\frac{3\,i}{2}}x\ln \left ( 1-i{{\rm e}^{x}} \right ) -{\frac{3\,i}{2}}{\it dilog} \left ( 1+i{{\rm e}^{x}} \right ) +{\frac{3\,i}{2}}{\it dilog} \left ( 1-i{{\rm e}^{x}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(2*x)*sech(x)^3,x)

[Out]

-exp(x)*(x*exp(2*x)+exp(2*x)-x+1)/(exp(2*x)+1)^2-3/2*I*x*ln(1+I*exp(x))+3/2*I*x*ln(1-I*exp(x))-3/2*I*dilog(1+I
*exp(x))+3/2*I*dilog(1-I*exp(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (x + 1\right )} e^{\left (3 \, x\right )} -{\left (x - 1\right )} e^{x}}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1} + 12 \, \int \frac{x e^{x}}{4 \,{\left (e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(2*x)*sech(x)^3,x, algorithm="maxima")

[Out]

-((x + 1)*e^(3*x) - (x - 1)*e^x)/(e^(4*x) + 2*e^(2*x) + 1) + 12*integrate(1/4*x*e^x/(e^(2*x) + 1), x)

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Fricas [B]  time = 1.93736, size = 1442, normalized size = 27.21 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(2*x)*sech(x)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(x + 1)*cosh(x)^3 + 6*(x + 1)*cosh(x)*sinh(x)^2 + 2*(x + 1)*sinh(x)^3 - 2*(x - 1)*cosh(x) - (3*I*cosh(
x)^4 + 12*I*cosh(x)*sinh(x)^3 + 3*I*sinh(x)^4 + (18*I*cosh(x)^2 + 6*I)*sinh(x)^2 + 6*I*cosh(x)^2 + (12*I*cosh(
x)^3 + 12*I*cosh(x))*sinh(x) + 3*I)*dilog(I*cosh(x) + I*sinh(x)) - (-3*I*cosh(x)^4 - 12*I*cosh(x)*sinh(x)^3 -
3*I*sinh(x)^4 + (-18*I*cosh(x)^2 - 6*I)*sinh(x)^2 - 6*I*cosh(x)^2 + (-12*I*cosh(x)^3 - 12*I*cosh(x))*sinh(x) -
3*I)*dilog(-I*cosh(x) - I*sinh(x)) - (-3*I*x*cosh(x)^4 - 12*I*x*cosh(x)*sinh(x)^3 - 3*I*x*sinh(x)^4 - 6*I*x*c
osh(x)^2 + (-18*I*x*cosh(x)^2 - 6*I*x)*sinh(x)^2 + (-12*I*x*cosh(x)^3 - 12*I*x*cosh(x))*sinh(x) - 3*I*x)*log(I
*cosh(x) + I*sinh(x) + 1) - (3*I*x*cosh(x)^4 + 12*I*x*cosh(x)*sinh(x)^3 + 3*I*x*sinh(x)^4 + 6*I*x*cosh(x)^2 +
(18*I*x*cosh(x)^2 + 6*I*x)*sinh(x)^2 + (12*I*x*cosh(x)^3 + 12*I*x*cosh(x))*sinh(x) + 3*I*x)*log(-I*cosh(x) - I
*sinh(x) + 1) + 2*(3*(x + 1)*cosh(x)^2 - x + 1)*sinh(x))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*c
osh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cosh{\left (2 x \right )} \operatorname{sech}^{3}{\left (x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(2*x)*sech(x)**3,x)

[Out]

Integral(x*cosh(2*x)*sech(x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cosh \left (2 \, x\right ) \operatorname{sech}\left (x\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(2*x)*sech(x)^3,x, algorithm="giac")

[Out]

integrate(x*cosh(2*x)*sech(x)^3, x)