3.873 \(\int \frac{x^3 \csc (x) \sec (x)}{\sqrt{a \sec ^4(x)}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{3 i x^2 \sec ^2(x) \text{PolyLog}\left (2,e^{2 i x}\right )}{2 \sqrt{a \sec ^4(x)}}+\frac{3 x \sec ^2(x) \text{PolyLog}\left (3,e^{2 i x}\right )}{2 \sqrt{a \sec ^4(x)}}+\frac{3 i \sec ^2(x) \text{PolyLog}\left (4,e^{2 i x}\right )}{4 \sqrt{a \sec ^4(x)}}-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}} \]

[Out]

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

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Rubi [A]  time = 0.610126, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {6720, 3717, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 i x^2 \sec ^2(x) \text{PolyLog}\left (2,e^{2 i x}\right )}{2 \sqrt{a \sec ^4(x)}}+\frac{3 x \sec ^2(x) \text{PolyLog}\left (3,e^{2 i x}\right )}{2 \sqrt{a \sec ^4(x)}}+\frac{3 i \sec ^2(x) \text{PolyLog}\left (4,e^{2 i x}\right )}{4 \sqrt{a \sec ^4(x)}}-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^4],x]

[Out]

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

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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 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 ^4(x)}} \, dx &=\frac{\sec ^2(x) \int x^3 \cot (x) \, dx}{\sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}-\frac{\left (2 i \sec ^2(x)\right ) \int \frac{e^{2 i x} x^3}{1-e^{2 i x}} \, dx}{\sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}}-\frac{\left (3 \sec ^2(x)\right ) \int x^2 \log \left (1-e^{2 i x}\right ) \, dx}{\sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\left (3 i \sec ^2(x)\right ) \int x \text{Li}_2\left (e^{2 i x}\right ) \, dx}{\sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{3 x \text{Li}_3\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}-\frac{\left (3 \sec ^2(x)\right ) \int \text{Li}_3\left (e^{2 i x}\right ) \, dx}{2 \sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{3 x \text{Li}_3\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{\left (3 i \sec ^2(x)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i x}\right )}{4 \sqrt{a \sec ^4(x)}}\\ &=-\frac{i x^4 \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}+\frac{x^3 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt{a \sec ^4(x)}}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{3 x \text{Li}_3\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt{a \sec ^4(x)}}+\frac{3 i \text{Li}_4\left (e^{2 i x}\right ) \sec ^2(x)}{4 \sqrt{a \sec ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0693036, size = 87, normalized size = 0.61 \[ -\frac{i \sec ^2(x) \left (-96 x^2 \text{PolyLog}\left (2,e^{-2 i x}\right )+96 i x \text{PolyLog}\left (3,e^{-2 i x}\right )+48 \text{PolyLog}\left (4,e^{-2 i x}\right )-16 x^4+64 i x^3 \log \left (1-e^{-2 i x}\right )+\pi ^4\right )}{64 \sqrt{a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^4],x]

[Out]

((-I/64)*(Pi^4 - 16*x^4 + (64*I)*x^3*Log[1 - E^((-2*I)*x)] - 96*x^2*PolyLog[2, E^((-2*I)*x)] + (96*I)*x*PolyLo
g[3, E^((-2*I)*x)] + 48*PolyLog[4, E^((-2*I)*x)])*Sec[x]^2)/Sqrt[a*Sec[x]^4]

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Maple [A]  time = 0.069, size = 221, normalized size = 1.6 \begin{align*}{\frac{{\frac{i}{4}}{{\rm e}^{2\,ix}}{x}^{4}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}{\frac{1}{\sqrt{{\frac{a{{\rm e}^{4\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{4}}}}}}}+{\frac{2\,i}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}} \left ( -{\frac{{{\rm e}^{2\,ix}}{x}^{4}}{4}}-{\frac{i}{2}}{{\rm e}^{2\,ix}}{x}^{3}\ln \left ({{\rm e}^{ix}}+1 \right ) -{\frac{3\,{{\rm e}^{2\,ix}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{ix}} \right ) }{2}}-3\,i{{\rm e}^{2\,ix}}x{\it polylog} \left ( 3,-{{\rm e}^{ix}} \right ) +3\,{{\rm e}^{2\,ix}}{\it polylog} \left ( 4,-{{\rm e}^{ix}} \right ) -{\frac{i}{2}}{{\rm e}^{2\,ix}}{x}^{3}\ln \left ( 1-{{\rm e}^{ix}} \right ) -{\frac{3\,{{\rm e}^{2\,ix}}{x}^{2}{\it polylog} \left ( 2,{{\rm e}^{ix}} \right ) }{2}}-3\,i{{\rm e}^{2\,ix}}x{\it polylog} \left ( 3,{{\rm e}^{ix}} \right ) +3\,{{\rm e}^{2\,ix}}{\it polylog} \left ( 4,{{\rm e}^{ix}} \right ) \right ){\frac{1}{\sqrt{{\frac{a{{\rm e}^{4\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{4}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*csc(x)*sec(x)/(a*sec(x)^4)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.55314, size = 185, normalized size = 1.29 \begin{align*} \frac{-i \, x^{4} + 4 i \, x^{3} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) - 4 i \, x^{3} \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) + 2 \, x^{3} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + 2 \, x^{3} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - 12 i \, x^{2}{\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) - 12 i \, x^{2}{\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + 24 \, x{\rm Li}_{3}(-e^{\left (i \, x\right )}) + 24 \, x{\rm Li}_{3}(e^{\left (i \, x\right )}) + 24 i \,{\rm Li}_{4}(-e^{\left (i \, x\right )}) + 24 i \,{\rm Li}_{4}(e^{\left (i \, x\right )})}{4 \, \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)^4)^(1/2),x, algorithm="maxima")

[Out]

1/4*(-I*x^4 + 4*I*x^3*arctan2(sin(x), cos(x) + 1) - 4*I*x^3*arctan2(sin(x), -cos(x) + 1) + 2*x^3*log(cos(x)^2
+ sin(x)^2 + 2*cos(x) + 1) + 2*x^3*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 12*I*x^2*dilog(-e^(I*x)) - 12*I*x
^2*dilog(e^(I*x)) + 24*x*polylog(3, -e^(I*x)) + 24*x*polylog(3, e^(I*x)) + 24*I*polylog(4, -e^(I*x)) + 24*I*po
lylog(4, e^(I*x)))/sqrt(a)

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Fricas [C]  time = 2.69629, size = 1180, normalized size = 8.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*csc(x)*sec(x)/(a*sec(x)^4)^(1/2),x, algorithm="fricas")

[Out]

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

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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)**4)**(1/2),x)

[Out]

Timed out

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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 )^{4}}}\,{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)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3*csc(x)*sec(x)/sqrt(a*sec(x)^4), x)