3.868 \(\int \frac {x \csc (x) \sec (x)}{\sqrt {a \sec ^2(x)}} \, dx\)

Optimal. Leaf size=76 \[ \frac {i \text {Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {i \text {Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {2 x \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt {a \sec ^2(x)}} \]

[Out]

-2*x*arctanh(exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)+I*polylog(2,-exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)-I*polylog(2,
exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6720, 4183, 2279, 2391} \[ \frac {i \sec (x) \text {PolyLog}\left (2,-e^{i x}\right )}{\sqrt {a \sec ^2(x)}}-\frac {i \sec (x) \text {PolyLog}\left (2,e^{i x}\right )}{\sqrt {a \sec ^2(x)}}-\frac {2 x \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^2],x]

[Out]

(-2*x*ArcTanh[E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + (I*PolyLog[2, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (I*PolyLo
g[2, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]

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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && 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 \csc (x) \sec (x)}{\sqrt {a \sec ^2(x)}} \, dx &=\frac {\sec (x) \int x \csc (x) \, dx}{\sqrt {a \sec ^2(x)}}\\ &=-\frac {2 x \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {\sec (x) \int \log \left (1-e^{i x}\right ) \, dx}{\sqrt {a \sec ^2(x)}}+\frac {\sec (x) \int \log \left (1+e^{i x}\right ) \, dx}{\sqrt {a \sec ^2(x)}}\\ &=-\frac {2 x \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}+\frac {(i \sec (x)) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i x}\right )}{\sqrt {a \sec ^2(x)}}-\frac {(i \sec (x)) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i x}\right )}{\sqrt {a \sec ^2(x)}}\\ &=-\frac {2 x \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}+\frac {i \text {Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {i \text {Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 69, normalized size = 0.91 \[ \frac {\sec (x) \left (i \text {Li}_2\left (-e^{i x}\right )-i \text {Li}_2\left (e^{i x}\right )+x \left (\log \left (1-e^{i x}\right )-\log \left (1+e^{i x}\right )\right )\right )}{\sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^2],x]

[Out]

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

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fricas [B]  time = 1.27, size = 124, normalized size = 1.63 \[ -\frac {{\left (x \cos \relax (x) \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \cos \relax (x) \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - x \cos \relax (x) \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - x \cos \relax (x) \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + i \, \cos \relax (x) {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, \cos \relax (x) {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) + i \, \cos \relax (x) {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, \cos \relax (x) {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right )\right )} \sqrt {\frac {a}{\cos \relax (x)^{2}}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(x*cos(x)*log(cos(x) + I*sin(x) + 1) + x*cos(x)*log(cos(x) - I*sin(x) + 1) - x*cos(x)*log(-cos(x) + I*sin
(x) + 1) - x*cos(x)*log(-cos(x) - I*sin(x) + 1) + I*cos(x)*dilog(cos(x) + I*sin(x)) - I*cos(x)*dilog(cos(x) -
I*sin(x)) + I*cos(x)*dilog(-cos(x) + I*sin(x)) - I*cos(x)*dilog(-cos(x) - I*sin(x)))*sqrt(a/cos(x)^2)/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \csc \relax (x) \sec \relax (x)}{\sqrt {a \sec \relax (x)^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x*csc(x)*sec(x)/sqrt(a*sec(x)^2), x)

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maple [A]  time = 0.23, size = 98, normalized size = 1.29 \[ -\frac {2 i \left (-\frac {i {\mathrm e}^{i x} x \ln \left (1+{\mathrm e}^{i x}\right )}{2}-\frac {{\mathrm e}^{i x} \polylog \left (2, -{\mathrm e}^{i x}\right )}{2}+\frac {i {\mathrm e}^{i x} x \ln \left (1-{\mathrm e}^{i x}\right )}{2}+\frac {{\mathrm e}^{i x} \polylog \left (2, {\mathrm e}^{i x}\right )}{2}\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x)

[Out]

-2*I/(a*exp(2*I*x)/(exp(2*I*x)+1)^2)^(1/2)/(exp(2*I*x)+1)*(-1/2*I*exp(I*x)*x*ln(1+exp(I*x))-1/2*exp(I*x)*polyl
og(2,-exp(I*x))+1/2*I*exp(I*x)*x*ln(1-exp(I*x))+1/2*exp(I*x)*polylog(2,exp(I*x)))

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maxima [A]  time = 0.85, size = 79, normalized size = 1.04 \[ -\frac {2 i \, x \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) + 2 i \, x \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) + x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) - x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - 2 i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + 2 i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right )}{2 \, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*(2*I*x*arctan2(sin(x), cos(x) + 1) + 2*I*x*arctan2(sin(x), -cos(x) + 1) + x*log(cos(x)^2 + sin(x)^2 + 2*c
os(x) + 1) - x*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 2*I*dilog(-e^(I*x)) + 2*I*dilog(e^(I*x)))/sqrt(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\cos \relax (x)\,\sin \relax (x)\,\sqrt {\frac {a}{{\cos \relax (x)}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(cos(x)*sin(x)*(a/cos(x)^2)^(1/2)),x)

[Out]

int(x/(cos(x)*sin(x)*(a/cos(x)^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \csc {\relax (x )} \sec {\relax (x )}}{\sqrt {a \sec ^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)*sec(x)/(a*sec(x)**2)**(1/2),x)

[Out]

Integral(x*csc(x)*sec(x)/sqrt(a*sec(x)**2), x)

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