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

Optimal. Leaf size=142 \[ \frac {1}{2} i \text {Li}_2\left (-e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}-\frac {1}{2} i \text {Li}_2\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \sin ^2(x) \sqrt {a \sec ^4(x)}-2 x \cos ^2(x) \tanh ^{-1}\left (e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} \sin (x) \cos (x) \sqrt {a \sec ^4(x)} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {6720, 2620, 14, 4420, 2548, 4419, 4183, 2279, 2391, 3473, 8} \[ \frac {1}{2} i \cos ^2(x) \text {PolyLog}\left (2,-e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} i \cos ^2(x) \text {PolyLog}\left (2,e^{2 i x}\right ) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \sin ^2(x) \sqrt {a \sec ^4(x)}-2 x \cos ^2(x) \tanh ^{-1}\left (e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} \sin (x) \cos (x) \sqrt {a \sec ^4(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 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 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 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 \csc (x) \sec (x) \sqrt {a \sec ^4(x)} \, dx &=\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int x \csc (x) \sec ^3(x) \, dx\\ &=x \cos ^2(x) \log (\tan (x)) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)-\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \left (\log (\tan (x))+\frac {\tan ^2(x)}{2}\right ) \, dx\\ &=x \cos ^2(x) \log (\tan (x)) \sqrt {a \sec ^4(x)}+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)-\frac {1}{2} \left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \tan ^2(x) \, dx-\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \log (\tan (x)) \, dx\\ &=-\frac {1}{2} \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)+\frac {1}{2} \left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int 1 \, dx+\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int x \csc (x) \sec (x) \, dx\\ &=\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}-\frac {1}{2} \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)+\left (2 \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int x \csc (2 x) \, dx\\ &=\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}-2 x \tanh ^{-1}\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}-\frac {1}{2} \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)-\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \log \left (1-e^{2 i x}\right ) \, dx+\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \log \left (1+e^{2 i x}\right ) \, dx\\ &=\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}-2 x \tanh ^{-1}\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}-\frac {1}{2} \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)+\frac {1}{2} \left (i \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i x}\right )-\frac {1}{2} \left (i \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\frac {1}{2} x \cos ^2(x) \sqrt {a \sec ^4(x)}-2 x \tanh ^{-1}\left (e^{2 i x}\right ) \cos ^2(x) \sqrt {a \sec ^4(x)}+\frac {1}{2} i \cos ^2(x) \text {Li}_2\left (-e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} i \cos ^2(x) \text {Li}_2\left (e^{2 i x}\right ) \sqrt {a \sec ^4(x)}-\frac {1}{2} \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {1}{2} x \sqrt {a \sec ^4(x)} \sin ^2(x)\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 85, normalized size = 0.60 \[ \frac {1}{2} \cos ^2(x) \sqrt {a \sec ^4(x)} \left (i \text {Li}_2\left (-e^{2 i x}\right )-i \text {Li}_2\left (e^{2 i x}\right )+2 x \log \left (1-e^{2 i x}\right )-2 x \log \left (1+e^{2 i x}\right )-\tan (x)+x \sec ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.76, size = 270, normalized size = 1.90 \[ \frac {1}{2} \, {\left (x \cos \relax (x)^{2} \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - x \cos \relax (x)^{2} \log \left (i \, \cos \relax (x) + \sin \relax (x) + 1\right ) - x \cos \relax (x)^{2} \log \left (i \, \cos \relax (x) - \sin \relax (x) + 1\right ) - x \cos \relax (x)^{2} \log \left (-i \, \cos \relax (x) + \sin \relax (x) + 1\right ) - x \cos \relax (x)^{2} \log \left (-i \, \cos \relax (x) - \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (i \, \cos \relax (x) + \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (i \, \cos \relax (x) - \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (-i \, \cos \relax (x) + \sin \relax (x)\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (-i \, \cos \relax (x) - \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) - \cos \relax (x) \sin \relax (x) + x\right )} \sqrt {\frac {a}{\cos \relax (x)^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x*cos(x)^2*log(cos(x) + I*sin(x) + 1) + x*cos(x)^2*log(cos(x) - I*sin(x) + 1) - x*cos(x)^2*log(I*cos(x) +
sin(x) + 1) - x*cos(x)^2*log(I*cos(x) - sin(x) + 1) - x*cos(x)^2*log(-I*cos(x) + sin(x) + 1) - x*cos(x)^2*log
(-I*cos(x) - sin(x) + 1) + x*cos(x)^2*log(-cos(x) + I*sin(x) + 1) + x*cos(x)^2*log(-cos(x) - I*sin(x) + 1) - I
*cos(x)^2*dilog(cos(x) + I*sin(x)) + I*cos(x)^2*dilog(cos(x) - I*sin(x)) - I*cos(x)^2*dilog(I*cos(x) + sin(x))
+ I*cos(x)^2*dilog(I*cos(x) - sin(x)) + I*cos(x)^2*dilog(-I*cos(x) + sin(x)) - I*cos(x)^2*dilog(-I*cos(x) - s
in(x)) + I*cos(x)^2*dilog(-cos(x) + I*sin(x)) - I*cos(x)^2*dilog(-cos(x) - I*sin(x)) - cos(x)*sin(x) + x)*sqrt
(a/cos(x)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 165, normalized size = 1.16 \[ \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left (-i+2 x -i {\mathrm e}^{-2 i x}\right )-4 i \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i x}+1\right )^{2} \left (\frac {i {\mathrm e}^{-2 i x} x \ln \left (1+{\mathrm e}^{i x}\right )}{4}+\frac {{\mathrm e}^{-2 i x} \polylog \left (2, -{\mathrm e}^{i x}\right )}{4}-\frac {i {\mathrm e}^{-2 i x} x \ln \left ({\mathrm e}^{2 i x}+1\right )}{4}-\frac {{\mathrm e}^{-2 i x} \polylog \left (2, -{\mathrm e}^{2 i x}\right )}{8}+\frac {i {\mathrm e}^{-2 i x} x \ln \left (1-{\mathrm e}^{i x}\right )}{4}+\frac {{\mathrm e}^{-2 i x} \polylog \left (2, {\mathrm e}^{i x}\right )}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*exp(4*I*x)/(exp(2*I*x)+1)^4)^(1/2)*(-I+2*x-I*exp(-2*I*x))-4*I*(a*exp(4*I*x)/(exp(2*I*x)+1)^4)^(1/2)*(exp(2*
I*x)+1)^2*(1/4*I*exp(-2*I*x)*x*ln(1+exp(I*x))+1/4*exp(-2*I*x)*polylog(2,-exp(I*x))-1/4*I*exp(-2*I*x)*x*ln(exp(
2*I*x)+1)-1/8*exp(-2*I*x)*polylog(2,-exp(2*I*x))+1/4*I*exp(-2*I*x)*x*ln(1-exp(I*x))+1/4*exp(-2*I*x)*polylog(2,
exp(I*x)))

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maxima [B]  time = 0.50, size = 432, normalized size = 3.04 \[ -\frac {{\left ({\left (2 \, x \cos \left (4 \, x\right ) + 4 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (4 \, x\right ) + 4 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) - {\left (2 \, x \cos \left (4 \, x\right ) + 4 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (4 \, x\right ) + 4 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) + {\left (2 \, x \cos \left (4 \, x\right ) + 4 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (4 \, x\right ) + 4 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) - 2 \, {\left (-2 i \, x - 1\right )} \cos \left (2 \, x\right ) - {\left (\cos \left (4 \, x\right ) + 2 \, \cos \left (2 \, x\right ) + i \, \sin \left (4 \, x\right ) + 2 i \, \sin \left (2 \, x\right ) + 1\right )} {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) + {\left (2 \, \cos \left (4 \, x\right ) + 4 \, \cos \left (2 \, x\right ) + 2 i \, \sin \left (4 \, x\right ) + 4 i \, \sin \left (2 \, x\right ) + 2\right )} {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + {\left (2 \, \cos \left (4 \, x\right ) + 4 \, \cos \left (2 \, x\right ) + 2 i \, \sin \left (4 \, x\right ) + 4 i \, \sin \left (2 \, x\right ) + 2\right )} {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + {\left (-i \, x \cos \left (4 \, x\right ) - 2 i \, x \cos \left (2 \, x\right ) + x \sin \left (4 \, x\right ) + 2 \, x \sin \left (2 \, x\right ) - i \, x\right )} \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) + {\left (i \, x \cos \left (4 \, x\right ) + 2 i \, x \cos \left (2 \, x\right ) - x \sin \left (4 \, x\right ) - 2 \, x \sin \left (2 \, x\right ) + i \, x\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + {\left (i \, x \cos \left (4 \, x\right ) + 2 i \, x \cos \left (2 \, x\right ) - x \sin \left (4 \, x\right ) - 2 \, x \sin \left (2 \, x\right ) + i \, x\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - {\left (4 \, x - 2 i\right )} \sin \left (2 \, x\right ) + 2\right )} \sqrt {a}}{-2 i \, \cos \left (4 \, x\right ) - 4 i \, \cos \left (2 \, x\right ) + 2 \, \sin \left (4 \, x\right ) + 4 \, \sin \left (2 \, x\right ) - 2 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*x*cos(4*x) + 4*x*cos(2*x) + 2*I*x*sin(4*x) + 4*I*x*sin(2*x) + 2*x)*arctan2(sin(2*x), cos(2*x) + 1) - (2*x
*cos(4*x) + 4*x*cos(2*x) + 2*I*x*sin(4*x) + 4*I*x*sin(2*x) + 2*x)*arctan2(sin(x), cos(x) + 1) + (2*x*cos(4*x)
+ 4*x*cos(2*x) + 2*I*x*sin(4*x) + 4*I*x*sin(2*x) + 2*x)*arctan2(sin(x), -cos(x) + 1) - 2*(-2*I*x - 1)*cos(2*x)
- (cos(4*x) + 2*cos(2*x) + I*sin(4*x) + 2*I*sin(2*x) + 1)*dilog(-e^(2*I*x)) + (2*cos(4*x) + 4*cos(2*x) + 2*I*
sin(4*x) + 4*I*sin(2*x) + 2)*dilog(-e^(I*x)) + (2*cos(4*x) + 4*cos(2*x) + 2*I*sin(4*x) + 4*I*sin(2*x) + 2)*dil
og(e^(I*x)) + (-I*x*cos(4*x) - 2*I*x*cos(2*x) + x*sin(4*x) + 2*x*sin(2*x) - I*x)*log(cos(2*x)^2 + sin(2*x)^2 +
2*cos(2*x) + 1) + (I*x*cos(4*x) + 2*I*x*cos(2*x) - x*sin(4*x) - 2*x*sin(2*x) + I*x)*log(cos(x)^2 + sin(x)^2 +
2*cos(x) + 1) + (I*x*cos(4*x) + 2*I*x*cos(2*x) - x*sin(4*x) - 2*x*sin(2*x) + I*x)*log(cos(x)^2 + sin(x)^2 - 2
*cos(x) + 1) - (4*x - 2*I)*sin(2*x) + 2)*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 = -0.01 \[ \int \frac {x\,\sqrt {\frac {a}{{\cos \relax (x)}^4}}}{\cos \relax (x)\,\sin \relax (x)} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*(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 \[ \int x \sqrt {a \sec ^{4}{\relax (x )}} \csc {\relax (x )} \sec {\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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