\(\int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx\) [815]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=-\frac {i x^3 \sec ^2(x)}{3 \sqrt {a \sec ^4(x)}}+\frac {x^2 \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt {a \sec ^4(x)}}-\frac {i x \operatorname {PolyLog}\left (2,e^{2 i x}\right ) \sec ^2(x)}{\sqrt {a \sec ^4(x)}}+\frac {\operatorname {PolyLog}\left (3,e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt {a \sec ^4(x)}} \] Output:

-1/3*I*x^3*sec(x)^2/(a*sec(x)^4)^(1/2)+x^2*ln(1-exp(2*I*x))*sec(x)^2/(a*se 
c(x)^4)^(1/2)-I*x*polylog(2,exp(2*I*x))*sec(x)^2/(a*sec(x)^4)^(1/2)+1/2*po 
lylog(3,exp(2*I*x))*sec(x)^2/(a*sec(x)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\frac {\left (-i \pi ^3+8 i x^3+24 x^2 \log \left (1-e^{-2 i x}\right )+24 i x \operatorname {PolyLog}\left (2,e^{-2 i x}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i x}\right )\right ) \sec ^2(x)}{24 \sqrt {a \sec ^4(x)}} \] Input:

Integrate[(x^2*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^4],x]
 

Output:

(((-I)*Pi^3 + (8*I)*x^3 + 24*x^2*Log[1 - E^((-2*I)*x)] + (24*I)*x*PolyLog[ 
2, E^((-2*I)*x)] + 12*PolyLog[3, E^((-2*I)*x)])*Sec[x]^2)/(24*Sqrt[a*Sec[x 
]^4])
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.79, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {7271, 3042, 25, 4200, 25, 2620, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \frac {\sec ^2(x) \int x^2 \cot (x)dx}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sec ^2(x) \int -x^2 \tan \left (x+\frac {\pi }{2}\right )dx}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sec ^2(x) \int x^2 \tan \left (x+\frac {\pi }{2}\right )dx}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 4200

\(\displaystyle -\frac {\sec ^2(x) \left (\frac {i x^3}{3}-2 i \int -\frac {e^{2 i x} x^2}{1-e^{2 i x}}dx\right )}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sec ^2(x) \left (2 i \int \frac {e^{2 i x} x^2}{1-e^{2 i x}}dx+\frac {i x^3}{3}\right )}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {\sec ^2(x) \left (2 i \left (\frac {1}{2} i x^2 \log \left (1-e^{2 i x}\right )-i \int x \log \left (1-e^{2 i x}\right )dx\right )+\frac {i x^3}{3}\right )}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {\sec ^2(x) \left (2 i \left (\frac {1}{2} i x^2 \log \left (1-e^{2 i x}\right )-i \left (\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} i \int \operatorname {PolyLog}\left (2,e^{2 i x}\right )dx\right )\right )+\frac {i x^3}{3}\right )}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {\sec ^2(x) \left (2 i \left (\frac {1}{2} i x^2 \log \left (1-e^{2 i x}\right )-i \left (\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \int e^{-2 i x} \operatorname {PolyLog}\left (2,e^{2 i x}\right )de^{2 i x}\right )\right )+\frac {i x^3}{3}\right )}{\sqrt {a \sec ^4(x)}}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {\sec ^2(x) \left (2 i \left (\frac {1}{2} i x^2 \log \left (1-e^{2 i x}\right )-i \left (\frac {1}{2} i x \operatorname {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,e^{2 i x}\right )\right )\right )+\frac {i x^3}{3}\right )}{\sqrt {a \sec ^4(x)}}\)

Input:

Int[(x^2*Csc[x]*Sec[x])/Sqrt[a*Sec[x]^4],x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4200

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol 
] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.40

method	result	size
risch	\(\frac {i {\mathrm e}^{2 i x} x^{3}}{3 \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left (1+{\mathrm e}^{2 i x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 i x}\right )^{2}}-\frac {2 \,{\mathrm e}^{2 i x} \left (\frac {i x^{3}}{3}-\frac {x^{2} \ln \left (1-{\mathrm e}^{i x}\right )}{2}+i x \operatorname {polylog}\left (2, {\mathrm e}^{i x}\right )-\operatorname {polylog}\left (3, {\mathrm e}^{i x}\right )-\frac {x^{2} \ln \left ({\mathrm e}^{i x}+1\right )}{2}+i x \operatorname {polylog}\left (2, -{\mathrm e}^{i x}\right )-\operatorname {polylog}\left (3, -{\mathrm e}^{i x}\right )\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left (1+{\mathrm e}^{2 i x}\right )^{4}}}\, \left (1+{\mathrm e}^{2 i x}\right )^{2}}\)	\(153\)

Input:

int(x^2*csc(x)*sec(x)/(a*sec(x)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (83) = 166\).

Time = 0.09 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.28 \[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\frac {2 \, \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \cos \left (x\right )^{2} {\rm polylog}\left (3, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 2 \, \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \cos \left (x\right )^{2} {\rm polylog}\left (3, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 2 \, \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \cos \left (x\right )^{2} {\rm polylog}\left (3, -\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 2 \, \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \cos \left (x\right )^{2} {\rm polylog}\left (3, -\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + {\left (x^{2} \cos \left (x\right )^{2} \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) + x^{2} \cos \left (x\right )^{2} \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + x^{2} \cos \left (x\right )^{2} \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) + x^{2} \cos \left (x\right )^{2} \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - 2 i \, x \cos \left (x\right )^{2} {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 2 i \, x \cos \left (x\right )^{2} {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 2 i \, x \cos \left (x\right )^{2} {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - 2 i \, x \cos \left (x\right )^{2} {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right )\right )} \sqrt {\frac {a}{\cos \left (x\right )^{4}}}}{2 \, a} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\int \frac {x^{2} \csc {\left (x \right )} \sec {\left (x \right )}}{\sqrt {a \sec ^{4}{\left (x \right )}}}\, dx \] Input:

integrate(x**2*csc(x)*sec(x)/(a*sec(x)**4)**(1/2),x)
 

Output:

Integral(x**2*csc(x)*sec(x)/sqrt(a*sec(x)**4), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.04 \[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\frac {-2 i \, x^{3} + 6 i \, x^{2} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) - 6 i \, x^{2} \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) + 3 \, x^{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + 3 \, x^{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - 12 i \, x {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) - 12 i \, x {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + 12 \, {\rm Li}_{3}(-e^{\left (i \, x\right )}) + 12 \, {\rm Li}_{3}(e^{\left (i \, x\right )})}{6 \, \sqrt {a}} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\int { \frac {x^{2} \csc \left (x\right ) \sec \left (x\right )}{\sqrt {a \sec \left (x\right )^{4}}} \,d x } \] Input:

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

Output:

integrate(x^2*csc(x)*sec(x)/sqrt(a*sec(x)^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\int \frac {x^2}{\cos \left (x\right )\,\sin \left (x\right )\,\sqrt {\frac {a}{{\cos \left (x\right )}^4}}} \,d x \] Input:

int(x^2/(cos(x)*sin(x)*(a/cos(x)^4)^(1/2)),x)
 

Output:

int(x^2/(cos(x)*sin(x)*(a/cos(x)^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x^2 \csc (x) \sec (x)}{\sqrt {a \sec ^4(x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\csc \left (x \right ) x^{2}}{\sec \left (x \right )}d x \right )}{a} \] Input:

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

Output:

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