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

Optimal. Leaf size=128 \[ \frac {2 i x \text {Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {2 i x \text {Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {2 \text {Li}_3\left (-e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}+\frac {2 \text {Li}_3\left (e^{i x}\right ) \sec (x)}{\sqrt {a \sec ^2(x)}}-\frac {2 x^2 \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt {a \sec ^2(x)}} \]

[Out]

-2*x^2*arctanh(exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)+2*I*x*polylog(2,-exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)-2*I*x*
polylog(2,exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)-2*polylog(3,-exp(I*x))*sec(x)/(a*sec(x)^2)^(1/2)+2*polylog(3,exp
(I*x))*sec(x)/(a*sec(x)^2)^(1/2)

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Rubi [A]  time = 0.59, antiderivative size = 128, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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^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*}

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Mathematica [A]  time = 0.07, size = 99, normalized size = 0.77 \[ \frac {\sec (x) \left (2 i x \text {Li}_2\left (-e^{i x}\right )-2 i x \text {Li}_2\left (e^{i x}\right )-2 \text {Li}_3\left (-e^{i x}\right )+2 \text {Li}_3\left (e^{i x}\right )+x^2 \log \left (1-e^{i x}\right )-x^2 \log \left (1+e^{i x}\right )\right )}{\sqrt {a \sec ^2(x)}} \]

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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fricas [C]  time = 1.62, size = 227, normalized size = 1.77 \[ \frac {2 \, \sqrt {\frac {a}{\cos \relax (x)^{2}}} \cos \relax (x) {\rm polylog}\left (3, \cos \relax (x) + i \, \sin \relax (x)\right ) + 2 \, \sqrt {\frac {a}{\cos \relax (x)^{2}}} \cos \relax (x) {\rm polylog}\left (3, \cos \relax (x) - i \, \sin \relax (x)\right ) - 2 \, \sqrt {\frac {a}{\cos \relax (x)^{2}}} \cos \relax (x) {\rm polylog}\left (3, -\cos \relax (x) + i \, \sin \relax (x)\right ) - 2 \, \sqrt {\frac {a}{\cos \relax (x)^{2}}} \cos \relax (x) {\rm polylog}\left (3, -\cos \relax (x) - i \, \sin \relax (x)\right ) - {\left (x^{2} \cos \relax (x) \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x^{2} \cos \relax (x) \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - x^{2} \cos \relax (x) \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - x^{2} \cos \relax (x) \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + 2 i \, x \cos \relax (x) {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - 2 i \, x \cos \relax (x) {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) + 2 i \, x \cos \relax (x) {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) - 2 i \, x \cos \relax (x) {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right )\right )} \sqrt {\frac {a}{\cos \relax (x)^{2}}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^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)

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maple [A]  time = 0.20, size = 132, normalized size = 1.03 \[ -\frac {2 \left (\frac {{\mathrm e}^{i x} x^{2} \ln \left (1+{\mathrm e}^{i x}\right )}{2}-i {\mathrm e}^{i x} x \polylog \left (2, -{\mathrm e}^{i x}\right )+{\mathrm e}^{i x} \polylog \left (3, -{\mathrm e}^{i x}\right )-\frac {{\mathrm e}^{i x} x^{2} \ln \left (1-{\mathrm e}^{i x}\right )}{2}+i {\mathrm e}^{i x} x \polylog \left (2, {\mathrm e}^{i x}\right )-{\mathrm e}^{i x} \polylog \left (3, {\mathrm e}^{i x}\right )\right )}{\sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*exp(2*I*x)/(exp(2*I*x)+1)^2)^(1/2)/(exp(2*I*x)+1)*(1/2*exp(I*x)*x^2*ln(1+exp(I*x))-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 = 0.68, size = 107, normalized size = 0.84 \[ -\frac {2 i \, x^{2} \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) + 2 i \, x^{2} \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) + x^{2} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) - x^{2} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - 4 i \, x {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + 4 i \, x {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + 4 \, {\rm Li}_{3}(-e^{\left (i \, x\right )}) - 4 \, {\rm Li}_{3}(e^{\left (i \, x\right )})}{2 \, \sqrt {a}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/(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 \[ \int \frac {x^{2} \csc {\relax (x )} \sec {\relax (x )}}{\sqrt {a \sec ^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**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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