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

Optimal. Leaf size=128 \[ \frac{2 i x \sec (x) \text{PolyLog}\left (2,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 i x \sec (x) \text{PolyLog}\left (2,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 \sec (x) \text{PolyLog}\left (3,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{2 \sec (x) \text{PolyLog}\left (3,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 x^2 \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt{a \sec ^2(x)}} \]

[Out]

(-2*x^2*ArcTanh[E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + ((2*I)*x*PolyLog[2, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (
(2*I)*x*PolyLog[2, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (2*PolyLog[3, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + (2*P
olyLog[3, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]

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Rubi [A]  time = 0.591856, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6720, 4183, 2531, 2282, 6589} \[ \frac{2 i x \sec (x) \text{PolyLog}\left (2,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 i x \sec (x) \text{PolyLog}\left (2,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 \sec (x) \text{PolyLog}\left (3,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{2 \sec (x) \text{PolyLog}\left (3,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 x^2 \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*ArcTanh[E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + ((2*I)*x*PolyLog[2, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (
(2*I)*x*PolyLog[2, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (2*PolyLog[3, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + (2*P
olyLog[3, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]

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])

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{x^2 \csc (x) \sec (x)}{\sqrt{a \sec ^2(x)}} \, dx &=\frac{\sec (x) \int x^2 \csc (x) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^2 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(2 \sec (x)) \int x \log \left (1-e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}+\frac{(2 \sec (x)) \int x \log \left (1+e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^2 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{2 i x \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{2 i x \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(2 i \sec (x)) \int \text{Li}_2\left (-e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}+\frac{(2 i \sec (x)) \int \text{Li}_2\left (e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^2 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{2 i x \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{2 i x \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(2 \sec (x)) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{(2 \sec (x)) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^2 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{2 i x \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{2 i x \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{2 \text{Li}_3\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{2 \text{Li}_3\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0687644, size = 99, normalized size = 0.77 \[ \frac{\sec (x) \left (2 i x \text{PolyLog}\left (2,-e^{i x}\right )-2 i x \text{PolyLog}\left (2,e^{i x}\right )-2 \text{PolyLog}\left (3,-e^{i x}\right )+2 \text{PolyLog}\left (3,e^{i x}\right )+x^2 \log \left (1-e^{i x}\right )-x^2 \log \left (1+e^{i x}\right )\right )}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*Log[1 - E^(I*x)] - x^2*Log[1 + E^(I*x)] + (2*I)*x*PolyLog[2, -E^(I*x)] - (2*I)*x*PolyLog[2, E^(I*x)] - 2
*PolyLog[3, -E^(I*x)] + 2*PolyLog[3, E^(I*x)])*Sec[x])/Sqrt[a*Sec[x]^2]

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Maple [A]  time = 0.066, size = 132, normalized size = 1. \begin{align*} -2\,{\frac{1/2\,{{\rm e}^{ix}}{x}^{2}\ln \left ({{\rm e}^{ix}}+1 \right ) -i{{\rm e}^{ix}}x{\it polylog} \left ( 2,-{{\rm e}^{ix}} \right ) +{{\rm e}^{ix}}{\it polylog} \left ( 3,-{{\rm e}^{ix}} \right ) -1/2\,{{\rm e}^{ix}}{x}^{2}\ln \left ( 1-{{\rm e}^{ix}} \right ) +i{{\rm e}^{ix}}x{\it polylog} \left ( 2,{{\rm e}^{ix}} \right ) -{{\rm e}^{ix}}{\it polylog} \left ( 3,{{\rm e}^{ix}} \right ) }{1+{{\rm e}^{2\,ix}}}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{2\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*exp(2*I*x)/(1+exp(2*I*x))^2)^(1/2)/(1+exp(2*I*x))*(1/2*exp(I*x)*x^2*ln(exp(I*x)+1)-I*exp(I*x)*x*polylog(
2,-exp(I*x))+exp(I*x)*polylog(3,-exp(I*x))-1/2*exp(I*x)*x^2*ln(1-exp(I*x))+I*exp(I*x)*x*polylog(2,exp(I*x))-ex
p(I*x)*polylog(3,exp(I*x)))

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Maxima [A]  time = 1.52118, size = 144, normalized size = 1.12 \begin{align*} -\frac{2 i \, x^{2} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) + 2 i \, x^{2} \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) + x^{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - x^{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - 4 i \, x{\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + 4 i \, x{\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + 4 \,{\rm Li}_{3}(-e^{\left (i \, x\right )}) - 4 \,{\rm Li}_{3}(e^{\left (i \, x\right )})}{2 \, \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*I*x^2*arctan2(sin(x), cos(x) + 1) + 2*I*x^2*arctan2(sin(x), -cos(x) + 1) + x^2*log(cos(x)^2 + sin(x)^2
+ 2*cos(x) + 1) - x^2*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 4*I*x*dilog(-e^(I*x)) + 4*I*x*dilog(e^(I*x))
+ 4*polylog(3, -e^(I*x)) - 4*polylog(3, e^(I*x)))/sqrt(a)

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Fricas [C]  time = 2.4141, size = 788, normalized size = 6.16 \begin{align*} \frac{2 \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 2 \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) - 2 \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 2 \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) -{\left (x^{2} \cos \left (x\right ) \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) + x^{2} \cos \left (x\right ) \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - x^{2} \cos \left (x\right ) \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - x^{2} \cos \left (x\right ) \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + 2 i \, x \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 2 i \, x \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 2 i \, x \cos \left (x\right ){\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 2 i \, x \cos \left (x\right ){\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right )\right )} \sqrt{\frac{a}{\cos \left (x\right )^{2}}}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(a/cos(x)^2)*cos(x)*polylog(3, cos(x) + I*sin(x)) + 2*sqrt(a/cos(x)^2)*cos(x)*polylog(3, cos(x) - I
*sin(x)) - 2*sqrt(a/cos(x)^2)*cos(x)*polylog(3, -cos(x) + I*sin(x)) - 2*sqrt(a/cos(x)^2)*cos(x)*polylog(3, -co
s(x) - I*sin(x)) - (x^2*cos(x)*log(cos(x) + I*sin(x) + 1) + x^2*cos(x)*log(cos(x) - I*sin(x) + 1) - x^2*cos(x)
*log(-cos(x) + I*sin(x) + 1) - x^2*cos(x)*log(-cos(x) - I*sin(x) + 1) + 2*I*x*cos(x)*dilog(cos(x) + I*sin(x))
- 2*I*x*cos(x)*dilog(cos(x) - I*sin(x)) + 2*I*x*cos(x)*dilog(-cos(x) + I*sin(x)) - 2*I*x*cos(x)*dilog(-cos(x)
- I*sin(x)))*sqrt(a/cos(x)^2))/a

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \csc{\left (x \right )} \sec{\left (x \right )}}{\sqrt{a \sec ^{2}{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \csc \left (x\right ) \sec \left (x\right )}{\sqrt{a \sec \left (x\right )^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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