3.845 \(\int \frac {x \text {csch}(x) \text {sech}(x)}{\sqrt {a \text {sech}^4(x)}} \, dx\)

Optimal. Leaf size=73 \[ \frac {\text {Li}_2\left (e^{2 x}\right ) \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}+\frac {x \log \left (1-e^{2 x}\right ) \text {sech}^2(x)}{\sqrt {a \text {sech}^4(x)}} \]

[Out]

-1/2*x^2*sech(x)^2/(a*sech(x)^4)^(1/2)+x*ln(1-exp(2*x))*sech(x)^2/(a*sech(x)^4)^(1/2)+1/2*polylog(2,exp(2*x))*
sech(x)^2/(a*sech(x)^4)^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6720, 3716, 2190, 2279, 2391} \[ \frac {\text {sech}^2(x) \text {PolyLog}\left (2,e^{2 x}\right )}{2 \sqrt {a \text {sech}^4(x)}}-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}+\frac {x \log \left (1-e^{2 x}\right ) \text {sech}^2(x)}{\sqrt {a \text {sech}^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^4],x]

[Out]

-(x^2*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4]) + (x*Log[1 - E^(2*x)]*Sech[x]^2)/Sqrt[a*Sech[x]^4] + (PolyLog[2, E^(2*x
)]*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4])

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && 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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {x \text {csch}(x) \text {sech}(x)}{\sqrt {a \text {sech}^4(x)}} \, dx &=\frac {\text {sech}^2(x) \int x \coth (x) \, dx}{\sqrt {a \text {sech}^4(x)}}\\ &=-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}-\frac {\left (2 \text {sech}^2(x)\right ) \int \frac {e^{2 x} x}{1-e^{2 x}} \, dx}{\sqrt {a \text {sech}^4(x)}}\\ &=-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}+\frac {x \log \left (1-e^{2 x}\right ) \text {sech}^2(x)}{\sqrt {a \text {sech}^4(x)}}-\frac {\text {sech}^2(x) \int \log \left (1-e^{2 x}\right ) \, dx}{\sqrt {a \text {sech}^4(x)}}\\ &=-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}+\frac {x \log \left (1-e^{2 x}\right ) \text {sech}^2(x)}{\sqrt {a \text {sech}^4(x)}}-\frac {\text {sech}^2(x) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 x}\right )}{2 \sqrt {a \text {sech}^4(x)}}\\ &=-\frac {x^2 \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}+\frac {x \log \left (1-e^{2 x}\right ) \text {sech}^2(x)}{\sqrt {a \text {sech}^4(x)}}+\frac {\text {Li}_2\left (e^{2 x}\right ) \text {sech}^2(x)}{2 \sqrt {a \text {sech}^4(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 44, normalized size = 0.60 \[ \frac {\text {sech}^2(x) \left (x \left (x+2 \log \left (1-e^{-2 x}\right )\right )-\text {Li}_2\left (e^{-2 x}\right )\right )}{2 \sqrt {a \text {sech}^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Csch[x]*Sech[x])/Sqrt[a*Sech[x]^4],x]

[Out]

((x*(x + 2*Log[1 - E^(-2*x)]) - PolyLog[2, E^(-2*x)])*Sech[x]^2)/(2*Sqrt[a*Sech[x]^4])

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fricas [B]  time = 0.45, size = 152, normalized size = 2.08 \[ -\frac {{\left (x^{2} e^{\left (4 \, x\right )} + 2 \, x^{2} e^{\left (2 \, x\right )} + x^{2} - 2 \, {\left (e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1\right )} {\rm Li}_2\left (\cosh \relax (x) + \sinh \relax (x)\right ) - 2 \, {\left (e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1\right )} {\rm Li}_2\left (-\cosh \relax (x) - \sinh \relax (x)\right ) - 2 \, {\left (x e^{\left (4 \, x\right )} + 2 \, x e^{\left (2 \, x\right )} + x\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - 2 \, {\left (x e^{\left (4 \, x\right )} + 2 \, x e^{\left (2 \, x\right )} + x\right )} \log \left (-\cosh \relax (x) - \sinh \relax (x) + 1\right )\right )} \sqrt {\frac {a}{e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x\right )} + 6 \, e^{\left (4 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 1}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(x^2*e^(4*x) + 2*x^2*e^(2*x) + x^2 - 2*(e^(4*x) + 2*e^(2*x) + 1)*dilog(cosh(x) + sinh(x)) - 2*(e^(4*x) +
2*e^(2*x) + 1)*dilog(-cosh(x) - sinh(x)) - 2*(x*e^(4*x) + 2*x*e^(2*x) + x)*log(cosh(x) + sinh(x) + 1) - 2*(x*e
^(4*x) + 2*x*e^(2*x) + x)*log(-cosh(x) - sinh(x) + 1))*sqrt(a/(e^(8*x) + 4*e^(6*x) + 6*e^(4*x) + 4*e^(2*x) + 1
))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {csch}\relax (x) \operatorname {sech}\relax (x)}{\sqrt {a \operatorname {sech}\relax (x)^{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(x*csch(x)*sech(x)/sqrt(a*sech(x)^4), x)

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maple [B]  time = 0.37, size = 175, normalized size = 2.40 \[ -\frac {{\mathrm e}^{2 x} x^{2}}{2 \sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2}}+\frac {{\mathrm e}^{2 x} x \ln \left ({\mathrm e}^{x}+1\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2}}+\frac {{\mathrm e}^{2 x} \polylog \left (2, -{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2}}+\frac {{\mathrm e}^{2 x} x \ln \left (1-{\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2}}+\frac {{\mathrm e}^{2 x} \polylog \left (2, {\mathrm e}^{x}\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 x}}{\left (1+{\mathrm e}^{2 x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 x}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x)

[Out]

-1/2/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)/(1+exp(2*x))^2*exp(2*x)*x^2+1/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)/(1+exp(
2*x))^2*exp(2*x)*x*ln(exp(x)+1)+1/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)/(1+exp(2*x))^2*exp(2*x)*polylog(2,-exp(x))
+1/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2)/(1+exp(2*x))^2*exp(2*x)*x*ln(1-exp(x))+1/(a*exp(4*x)/(1+exp(2*x))^4)^(1/2
)/(1+exp(2*x))^2*exp(2*x)*polylog(2,exp(x))

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maxima [A]  time = 0.46, size = 43, normalized size = 0.59 \[ -\frac {x^{2}}{2 \, \sqrt {a}} + \frac {x \log \left (e^{x} + 1\right ) + {\rm Li}_2\left (-e^{x}\right )}{\sqrt {a}} + \frac {x \log \left (-e^{x} + 1\right ) + {\rm Li}_2\left (e^{x}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)^4)^(1/2),x, algorithm="maxima")

[Out]

-1/2*x^2/sqrt(a) + (x*log(e^x + 1) + dilog(-e^x))/sqrt(a) + (x*log(-e^x + 1) + dilog(e^x))/sqrt(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)\,\sqrt {\frac {a}{{\mathrm {cosh}\relax (x)}^4}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(cosh(x)*sinh(x)*(a/cosh(x)^4)^(1/2)),x)

[Out]

int(x/(cosh(x)*sinh(x)*(a/cosh(x)^4)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {csch}{\relax (x )} \operatorname {sech}{\relax (x )}}{\sqrt {a \operatorname {sech}^{4}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csch(x)*sech(x)/(a*sech(x)**4)**(1/2),x)

[Out]

Integral(x*csch(x)*sech(x)/sqrt(a*sech(x)**4), x)

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