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

Optimal. Leaf size=220 \[ i x \text {Li}_2\left (-e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}-i x \text {Li}_2\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}-\frac {1}{2} \text {Li}_3\left (-e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} \text {Li}_3\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x^2 \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x^2 \sin ^2(x) \sqrt {a \sec ^4(x)}-2 x^2 \cos ^2(x) \tanh ^{-1}\left (e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\cos ^2(x) \sqrt {a \sec ^4(x)} \log (\cos (x))-x \sin (x) \cos (x) \sqrt {a \sec ^4(x)} \]

[Out]

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

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Rubi [A]  time = 0.54, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {6720, 2620, 14, 4420, 2551, 4419, 4183, 2531, 2282, 6589, 3720, 3475, 30} \[ i x \cos ^2(x) \text {PolyLog}\left (2,-e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-i x \cos ^2(x) \text {PolyLog}\left (2,e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} \cos ^2(x) \text {PolyLog}\left (3,-e^{2 i x}\right ) \sqrt {a \sec ^4(x)}+\frac {1}{2} \cos ^2(x) \text {PolyLog}\left (3,e^{2 i x}\right ) \sqrt {a \sec ^4(x)}+\frac {1}{2} x^2 \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x^2 \sin ^2(x) \sqrt {a \sec ^4(x)}-2 x^2 \cos ^2(x) \tanh ^{-1}\left (e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\cos ^2(x) \sqrt {a \sec ^4(x)} \log (\cos (x))-x \sin (x) \cos (x) \sqrt {a \sec ^4(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && 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 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4420

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Modul
e[{u = IntHide[Csc[a + b*x]^n*Sec[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)*u
, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]

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

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Mathematica [A]  time = 0.63, size = 138, normalized size = 0.63 \[ \frac {1}{24} \cos ^2(x) \sqrt {a \sec ^4(x)} \left (24 i x \text {Li}_2\left (e^{-2 i x}\right )+24 i x \text {Li}_2\left (-e^{2 i x}\right )+12 \text {Li}_3\left (e^{-2 i x}\right )-12 \text {Li}_3\left (-e^{2 i x}\right )+16 i x^3+24 x^2 \log \left (1-e^{-2 i x}\right )-24 x^2 \log \left (1+e^{2 i x}\right )+12 x^2 \sec ^2(x)-24 x \tan (x)-24 \log (\cos (x))-i \pi ^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]^2*Sqrt[a*Sec[x]^4]*((-I)*Pi^3 + (16*I)*x^3 + 24*x^2*Log[1 - E^((-2*I)*x)] - 24*x^2*Log[1 + E^((2*I)*x)
] - 24*Log[Cos[x]] + (24*I)*x*PolyLog[2, E^((-2*I)*x)] + (24*I)*x*PolyLog[2, -E^((2*I)*x)] + 12*PolyLog[3, E^(
(-2*I)*x)] - 12*PolyLog[3, -E^((2*I)*x)] + 12*x^2*Sec[x]^2 - 24*x*Tan[x]))/24

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fricas [C]  time = 2.05, size = 550, normalized size = 2.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, cos(x) + I*sin(x)) + sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, cos(x) - I*sin(
x)) - sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, I*cos(x) + sin(x)) - sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, I*cos(x)
- sin(x)) - sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, -I*cos(x) + sin(x)) - sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, -I
*cos(x) - sin(x)) + sqrt(a/cos(x)^4)*cos(x)^2*polylog(3, -cos(x) + I*sin(x)) + sqrt(a/cos(x)^4)*cos(x)^2*polyl
og(3, -cos(x) - I*sin(x)) + 1/2*(x^2*cos(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(I*cos(x) + sin(x) + 1) - x^2*cos(x)^2*log(I*cos(x) - sin(x) + 1) - x^2*cos(x)^2*log(-I
*cos(x) + sin(x) + 1) - x^2*cos(x)^2*log(-I*cos(x) - 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*dilog(I*cos(x) + sin(x)) + 2*I*x*cos(x)^2*dilog(I*cos(x) - sin(x)) + 2*I*x*cos
(x)^2*dilog(-I*cos(x) + sin(x)) - 2*I*x*cos(x)^2*dilog(-I*cos(x) - sin(x)) + 2*I*x*cos(x)^2*dilog(-cos(x) + I*
sin(x)) - 2*I*x*cos(x)^2*dilog(-cos(x) - I*sin(x)) - cos(x)^2*log(cos(x) + I*sin(x) + I) - cos(x)^2*log(cos(x)
- I*sin(x) + I) - cos(x)^2*log(-cos(x) + I*sin(x) + I) - cos(x)^2*log(-cos(x) - I*sin(x) + I) - 2*x*cos(x)*si
n(x) + x^2)*sqrt(a/cos(x)^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sec \relax (x)^{4}} x^{2} \csc \relax (x) \sec \relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.23, size = 254, normalized size = 1.15 \[ 2 \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, x \left (x -i-i {\mathrm e}^{-2 i x}\right )+2 \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i x}+1\right )^{2} \left (-\frac {{\mathrm e}^{-2 i x} \ln \left ({\mathrm e}^{2 i x}+1\right )}{2}-{\mathrm e}^{-2 i x} \Im \relax (x )+{\mathrm e}^{-2 i x} \ln \left ({\mathrm e}^{i \Re \relax (x )}\right )+\frac {{\mathrm e}^{-2 i x} x^{2} \ln \left (1+{\mathrm e}^{i x}\right )}{2}-i {\mathrm e}^{-2 i x} x \polylog \left (2, -{\mathrm e}^{i x}\right )+{\mathrm e}^{-2 i x} \polylog \left (3, -{\mathrm e}^{i x}\right )-\frac {{\mathrm e}^{-2 i x} x^{2} \ln \left ({\mathrm e}^{2 i x}+1\right )}{2}+\frac {i {\mathrm e}^{-2 i x} x \polylog \left (2, -{\mathrm e}^{2 i x}\right )}{2}-\frac {{\mathrm e}^{-2 i x} \polylog \left (3, -{\mathrm e}^{2 i x}\right )}{4}+\frac {{\mathrm e}^{-2 i x} x^{2} \ln \left (1-{\mathrm e}^{i x}\right )}{2}-i {\mathrm e}^{-2 i x} x \polylog \left (2, {\mathrm e}^{i x}\right )+{\mathrm e}^{-2 i x} \polylog \left (3, {\mathrm e}^{i x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.52, size = 653, normalized size = 2.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*x^2 + 2*(x^2 + 1)*cos(4*x) + 4*(x^2 + 1)*cos(2*x) + (2*I*x^2 + 2*I)*sin(4*x) + (4*I*x^2 + 4*I)*sin(2*x) +
2)*arctan2(sin(2*x), cos(2*x) + 1) - (2*x^2*cos(4*x) + 4*x^2*cos(2*x) + 2*I*x^2*sin(4*x) + 4*I*x^2*sin(2*x) +
2*x^2)*arctan2(sin(x), cos(x) + 1) + (2*x^2*cos(4*x) + 4*x^2*cos(2*x) + 2*I*x^2*sin(4*x) + 4*I*x^2*sin(2*x) +
2*x^2)*arctan2(sin(x), -cos(x) + 1) - 4*x*cos(4*x) + (4*I*x^2 - 4*x)*cos(2*x) - (2*x*cos(4*x) + 4*x*cos(2*x)
+ 2*I*x*sin(4*x) + 4*I*x*sin(2*x) + 2*x)*dilog(-e^(2*I*x)) + (4*x*cos(4*x) + 8*x*cos(2*x) + 4*I*x*sin(4*x) + 8
*I*x*sin(2*x) + 4*x)*dilog(-e^(I*x)) + (4*x*cos(4*x) + 8*x*cos(2*x) + 4*I*x*sin(4*x) + 8*I*x*sin(2*x) + 4*x)*d
ilog(e^(I*x)) + (-I*x^2 + (-I*x^2 - I)*cos(4*x) + (-2*I*x^2 - 2*I)*cos(2*x) + (x^2 + 1)*sin(4*x) + 2*(x^2 + 1)
*sin(2*x) - I)*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1) + (I*x^2*cos(4*x) + 2*I*x^2*cos(2*x) - x^2*sin(4*
x) - 2*x^2*sin(2*x) + I*x^2)*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + (I*x^2*cos(4*x) + 2*I*x^2*cos(2*x) - x^
2*sin(4*x) - 2*x^2*sin(2*x) + I*x^2)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + (-I*cos(4*x) - 2*I*cos(2*x) + s
in(4*x) + 2*sin(2*x) - I)*polylog(3, -e^(2*I*x)) + (4*I*cos(4*x) + 8*I*cos(2*x) - 4*sin(4*x) - 8*sin(2*x) + 4*
I)*polylog(3, -e^(I*x)) + (4*I*cos(4*x) + 8*I*cos(2*x) - 4*sin(4*x) - 8*sin(2*x) + 4*I)*polylog(3, e^(I*x)) -
4*I*x*sin(4*x) - 4*(x^2 + I*x)*sin(2*x))*sqrt(a)/(-2*I*cos(4*x) - 4*I*cos(2*x) + 2*sin(4*x) + 4*sin(2*x) - 2*I
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\sqrt {\frac {a}{{\cos \relax (x)}^4}}}{\cos \relax (x)\,\sin \relax (x)} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {a \sec ^{4}{\relax (x )}} \csc {\relax (x )} \sec {\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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