3.870 \(\int \frac{x^3 \csc (x) \sec (x)}{\sqrt{a \sec ^2(x)}} \, dx\)

Optimal. Leaf size=186 \[ \frac{3 i x^2 \sec (x) \text{PolyLog}\left (2,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \sec (x) \text{PolyLog}\left (2,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{6 x \sec (x) \text{PolyLog}\left (3,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{6 x \sec (x) \text{PolyLog}\left (3,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{6 i \sec (x) \text{PolyLog}\left (4,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{6 i \sec (x) \text{PolyLog}\left (4,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 x^3 \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt{a \sec ^2(x)}} \]

[Out]

(-2*x^3*ArcTanh[E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + ((3*I)*x^2*PolyLog[2, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] -
((3*I)*x^2*PolyLog[2, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (6*x*PolyLog[3, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]
+ (6*x*PolyLog[3, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - ((6*I)*PolyLog[4, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + (
(6*I)*PolyLog[4, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]

________________________________________________________________________________________

Rubi [A]  time = 0.569598, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6720, 4183, 2531, 6609, 2282, 6589} \[ \frac{3 i x^2 \sec (x) \text{PolyLog}\left (2,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \sec (x) \text{PolyLog}\left (2,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{6 x \sec (x) \text{PolyLog}\left (3,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{6 x \sec (x) \text{PolyLog}\left (3,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{6 i \sec (x) \text{PolyLog}\left (4,-e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{6 i \sec (x) \text{PolyLog}\left (4,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}-\frac{2 x^3 \sec (x) \tanh ^{-1}\left (e^{i x}\right )}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^2],x]

[Out]

(-2*x^3*ArcTanh[E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + ((3*I)*x^2*PolyLog[2, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] -
((3*I)*x^2*PolyLog[2, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - (6*x*PolyLog[3, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2]
+ (6*x*PolyLog[3, E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] - ((6*I)*PolyLog[4, -E^(I*x)]*Sec[x])/Sqrt[a*Sec[x]^2] + (
(6*I)*PolyLog[4, 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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^3 \csc (x) \sec (x)}{\sqrt{a \sec ^2(x)}} \, dx &=\frac{\sec (x) \int x^3 \csc (x) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(3 \sec (x)) \int x^2 \log \left (1-e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}+\frac{(3 \sec (x)) \int x^2 \log \left (1+e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{3 i x^2 \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(6 i \sec (x)) \int x \text{Li}_2\left (-e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}+\frac{(6 i \sec (x)) \int x \text{Li}_2\left (e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{3 i x^2 \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{6 x \text{Li}_3\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{6 x \text{Li}_3\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{(6 \sec (x)) \int \text{Li}_3\left (-e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}-\frac{(6 \sec (x)) \int \text{Li}_3\left (e^{i x}\right ) \, dx}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{3 i x^2 \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{6 x \text{Li}_3\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{6 x \text{Li}_3\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{(6 i \sec (x)) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}+\frac{(6 i \sec (x)) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i x}\right )}{\sqrt{a \sec ^2(x)}}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{3 i x^2 \text{Li}_2\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{6 x \text{Li}_3\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{6 x \text{Li}_3\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}-\frac{6 i \text{Li}_4\left (-e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}+\frac{6 i \text{Li}_4\left (e^{i x}\right ) \sec (x)}{\sqrt{a \sec ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0986147, size = 147, normalized size = 0.79 \[ -\frac{i \sec (x) \left (-24 x^2 \text{PolyLog}\left (2,e^{-i x}\right )-24 x^2 \text{PolyLog}\left (2,-e^{i x}\right )+48 i x \text{PolyLog}\left (3,e^{-i x}\right )-48 i x \text{PolyLog}\left (3,-e^{i x}\right )+48 \text{PolyLog}\left (4,e^{-i x}\right )+48 \text{PolyLog}\left (4,-e^{i x}\right )-2 x^4+8 i x^3 \log \left (1-e^{-i x}\right )-8 i x^3 \log \left (1+e^{i x}\right )+\pi ^4\right )}{8 \sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^2],x]

[Out]

((-I/8)*(Pi^4 - 2*x^4 + (8*I)*x^3*Log[1 - E^((-I)*x)] - (8*I)*x^3*Log[1 + E^(I*x)] - 24*x^2*PolyLog[2, E^((-I)
*x)] - 24*x^2*PolyLog[2, -E^(I*x)] + (48*I)*x*PolyLog[3, E^((-I)*x)] - (48*I)*x*PolyLog[3, -E^(I*x)] + 48*Poly
Log[4, E^((-I)*x)] + 48*PolyLog[4, -E^(I*x)])*Sec[x])/Sqrt[a*Sec[x]^2]

________________________________________________________________________________________

Maple [A]  time = 0.068, size = 172, normalized size = 0.9 \begin{align*}{\frac{2\,i}{1+{{\rm e}^{2\,ix}}} \left ({\frac{i}{2}}{{\rm e}^{ix}}{x}^{3}\ln \left ({{\rm e}^{ix}}+1 \right ) +{\frac{3\,{{\rm e}^{ix}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{ix}} \right ) }{2}}+3\,i{{\rm e}^{ix}}x{\it polylog} \left ( 3,-{{\rm e}^{ix}} \right ) -3\,{{\rm e}^{ix}}{\it polylog} \left ( 4,-{{\rm e}^{ix}} \right ) -{\frac{i}{2}}{{\rm e}^{ix}}{x}^{3}\ln \left ( 1-{{\rm e}^{ix}} \right ) -{\frac{3\,{{\rm e}^{ix}}{x}^{2}{\it polylog} \left ( 2,{{\rm e}^{ix}} \right ) }{2}}-3\,i{{\rm e}^{ix}}x{\it polylog} \left ( 3,{{\rm e}^{ix}} \right ) +3\,{{\rm e}^{ix}}{\it polylog} \left ( 4,{{\rm e}^{ix}} \right ) \right ){\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^3*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x)

[Out]

2*I/(a*exp(2*I*x)/(1+exp(2*I*x))^2)^(1/2)/(1+exp(2*I*x))*(1/2*I*exp(I*x)*x^3*ln(exp(I*x)+1)+3/2*exp(I*x)*x^2*p
olylog(2,-exp(I*x))+3*I*exp(I*x)*x*polylog(3,-exp(I*x))-3*exp(I*x)*polylog(4,-exp(I*x))-1/2*I*exp(I*x)*x^3*ln(
1-exp(I*x))-3/2*exp(I*x)*x^2*polylog(2,exp(I*x))-3*I*exp(I*x)*x*polylog(3,exp(I*x))+3*exp(I*x)*polylog(4,exp(I
*x)))

________________________________________________________________________________________

Maxima [A]  time = 1.56115, size = 177, normalized size = 0.95 \begin{align*} -\frac{2 i \, x^{3} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) + 2 i \, x^{3} \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) + x^{3} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - x^{3} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - 6 i \, x^{2}{\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + 6 i \, x^{2}{\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + 12 \, x{\rm Li}_{3}(-e^{\left (i \, x\right )}) - 12 \, x{\rm Li}_{3}(e^{\left (i \, x\right )}) + 12 i \,{\rm Li}_{4}(-e^{\left (i \, x\right )}) - 12 i \,{\rm Li}_{4}(e^{\left (i \, x\right )})}{2 \, \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*(2*I*x^3*arctan2(sin(x), cos(x) + 1) + 2*I*x^3*arctan2(sin(x), -cos(x) + 1) + x^3*log(cos(x)^2 + sin(x)^2
+ 2*cos(x) + 1) - x^3*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 6*I*x^2*dilog(-e^(I*x)) + 6*I*x^2*dilog(e^(I*
x)) + 12*x*polylog(3, -e^(I*x)) - 12*x*polylog(3, e^(I*x)) + 12*I*polylog(4, -e^(I*x)) - 12*I*polylog(4, e^(I*
x)))/sqrt(a)

________________________________________________________________________________________

Fricas [C]  time = 2.70255, size = 1137, normalized size = 6.11 \begin{align*} \frac{6 \, x \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 ) + 6 \, x \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 ) - 6 \, x \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 ) - 6 \, x \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 ) + 6 i \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (4, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 6 i \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (4, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 6 i \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (4, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 6 i \, \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \cos \left (x\right ){\rm polylog}\left (4, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) -{\left (x^{3} \cos \left (x\right ) \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) + x^{3} \cos \left (x\right ) \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - x^{3} \cos \left (x\right ) \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - x^{3} \cos \left (x\right ) \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + 3 i \, x^{2} \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 3 i \, x^{2} \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 3 i \, x^{2} \cos \left (x\right ){\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 3 i \, x^{2} \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^3*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/2*(6*x*sqrt(a/cos(x)^2)*cos(x)*polylog(3, cos(x) + I*sin(x)) + 6*x*sqrt(a/cos(x)^2)*cos(x)*polylog(3, cos(x)
- I*sin(x)) - 6*x*sqrt(a/cos(x)^2)*cos(x)*polylog(3, -cos(x) + I*sin(x)) - 6*x*sqrt(a/cos(x)^2)*cos(x)*polylo
g(3, -cos(x) - I*sin(x)) + 6*I*sqrt(a/cos(x)^2)*cos(x)*polylog(4, cos(x) + I*sin(x)) - 6*I*sqrt(a/cos(x)^2)*co
s(x)*polylog(4, cos(x) - I*sin(x)) + 6*I*sqrt(a/cos(x)^2)*cos(x)*polylog(4, -cos(x) + I*sin(x)) - 6*I*sqrt(a/c
os(x)^2)*cos(x)*polylog(4, -cos(x) - I*sin(x)) - (x^3*cos(x)*log(cos(x) + I*sin(x) + 1) + x^3*cos(x)*log(cos(x
) - I*sin(x) + 1) - x^3*cos(x)*log(-cos(x) + I*sin(x) + 1) - x^3*cos(x)*log(-cos(x) - I*sin(x) + 1) + 3*I*x^2*
cos(x)*dilog(cos(x) + I*sin(x)) - 3*I*x^2*cos(x)*dilog(cos(x) - I*sin(x)) + 3*I*x^2*cos(x)*dilog(-cos(x) + I*s
in(x)) - 3*I*x^2*cos(x)*dilog(-cos(x) - I*sin(x)))*sqrt(a/cos(x)^2))/a

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*csc(x)*sec(x)/(a*sec(x)**2)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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^3*csc(x)*sec(x)/(a*sec(x)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3*csc(x)*sec(x)/sqrt(a*sec(x)^2), x)