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

Optimal. Leaf size=105 \[ i \cos (x) \text{PolyLog}\left (2,-e^{i x}\right ) \sqrt{a \sec ^2(x)}-i \cos (x) \text{PolyLog}\left (2,e^{i x}\right ) \sqrt{a \sec ^2(x)}+x \sqrt{a \sec ^2(x)}-2 x \cos (x) \tanh ^{-1}\left (e^{i x}\right ) \sqrt{a \sec ^2(x)}-\cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]

[Out]

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

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Rubi [A]  time = 0.34337, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6720, 2622, 321, 207, 4420, 6271, 4183, 2279, 2391, 3770} \[ i \cos (x) \text{PolyLog}\left (2,-e^{i x}\right ) \sqrt{a \sec ^2(x)}-i \cos (x) \text{PolyLog}\left (2,e^{i x}\right ) \sqrt{a \sec ^2(x)}+x \sqrt{a \sec ^2(x)}-2 x \cos (x) \tanh ^{-1}\left (e^{i x}\right ) \sqrt{a \sec ^2(x)}-\cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Sqrt[a*Sec[x]^2] - 2*x*ArcTanh[E^(I*x)]*Cos[x]*Sqrt[a*Sec[x]^2] - ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2] +
I*Cos[x]*PolyLog[2, -E^(I*x)]*Sqrt[a*Sec[x]^2] - I*Cos[x]*PolyLog[2, E^(I*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 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

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 6271

Int[ArcTanh[u_], x_Symbol] :> Simp[x*ArcTanh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 - u^2), x], x] /; I
nverseFunctionFreeQ[u, x]

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 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 3770

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

Rubi steps

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

Mathematica [A]  time = 0.0792283, size = 108, normalized size = 1.03 \[ \sqrt{a \sec ^2(x)} \left (i \cos (x) \left (\text{PolyLog}\left (2,-e^{i x}\right )-\text{PolyLog}\left (2,e^{i x}\right )\right )+x+x \left (\log \left (1-e^{i x}\right )-\log \left (1+e^{i x}\right )\right ) \cos (x)+\cos (x) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\cos (x) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.107, size = 86, normalized size = 0.8 \begin{align*} 2\,\sqrt{{\frac{a{{\rm e}^{2\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}}}x+4\,\sqrt{{\frac{a{{\rm e}^{2\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}}} \left ( i\arctan \left ({{\rm e}^{ix}} \right ) +i/2{\it dilog} \left ({{\rm e}^{ix}}+1 \right ) -1/2\,x\ln \left ({{\rm e}^{ix}}+1 \right ) +i/2{\it dilog} \left ({{\rm e}^{ix}} \right ) \right ) \cos \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a*exp(2*I*x)/(1+exp(2*I*x))^2)^(1/2)*x+4*(a*exp(2*I*x)/(1+exp(2*I*x))^2)^(1/2)*(I*arctan(exp(I*x))+1/2*I*di
log(exp(I*x)+1)-1/2*x*ln(exp(I*x)+1)+1/2*I*dilog(exp(I*x)))*cos(x)

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Maxima [B]  time = 1.57098, size = 405, normalized size = 3.86 \begin{align*} \frac{{\left (2 \,{\left (\cos \left (2 \, x\right ) + i \, \sin \left (2 \, x\right ) + 1\right )} \arctan \left (\cos \left (x\right ), \sin \left (x\right ) + 1\right ) + 2 \,{\left (\cos \left (2 \, x\right ) + i \, \sin \left (2 \, x\right ) + 1\right )} \arctan \left (\cos \left (x\right ), -\sin \left (x\right ) + 1\right ) -{\left (2 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) -{\left (2 \, x \cos \left (2 \, x\right ) + 2 i \, x \sin \left (2 \, x\right ) + 2 \, x\right )} \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) - 4 i \, x \cos \left (x\right ) + 2 \,{\left (\cos \left (2 \, x\right ) + i \, \sin \left (2 \, x\right ) + 1\right )}{\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) - 2 \,{\left (\cos \left (2 \, x\right ) + i \, \sin \left (2 \, x\right ) + 1\right )}{\rm Li}_2\left (e^{\left (i \, x\right )}\right ) -{\left (-i \, x \cos \left (2 \, x\right ) + x \sin \left (2 \, x\right ) - i \, x\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) -{\left (i \, x \cos \left (2 \, x\right ) - x \sin \left (2 \, x\right ) + i \, x\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) -{\left (-i \, \cos \left (2 \, x\right ) + \sin \left (2 \, x\right ) - i\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) -{\left (i \, \cos \left (2 \, x\right ) - \sin \left (2 \, x\right ) + i\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) + 4 \, x \sin \left (x\right )\right )} \sqrt{a}}{-2 i \, \cos \left (2 \, x\right ) + 2 \, \sin \left (2 \, x\right ) - 2 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(cos(2*x) + I*sin(2*x) + 1)*arctan2(cos(x), sin(x) + 1) + 2*(cos(2*x) + I*sin(2*x) + 1)*arctan2(cos(x), -si
n(x) + 1) - (2*x*cos(2*x) + 2*I*x*sin(2*x) + 2*x)*arctan2(sin(x), cos(x) + 1) - (2*x*cos(2*x) + 2*I*x*sin(2*x)
+ 2*x)*arctan2(sin(x), -cos(x) + 1) - 4*I*x*cos(x) + 2*(cos(2*x) + I*sin(2*x) + 1)*dilog(-e^(I*x)) - 2*(cos(2
*x) + I*sin(2*x) + 1)*dilog(e^(I*x)) - (-I*x*cos(2*x) + x*sin(2*x) - I*x)*log(cos(x)^2 + sin(x)^2 + 2*cos(x) +
1) - (I*x*cos(2*x) - x*sin(2*x) + I*x)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - (-I*cos(2*x) + sin(2*x) - I)
*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - (I*cos(2*x) - sin(2*x) + I)*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1)
+ 4*x*sin(x))*sqrt(a)/(-2*I*cos(2*x) + 2*sin(2*x) - 2*I)

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Fricas [A]  time = 2.41178, size = 500, normalized size = 4.76 \begin{align*} -\frac{1}{2} \,{\left (x \cos \left (x\right ) \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) + x \cos \left (x\right ) \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - x \cos \left (x\right ) \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - x \cos \left (x\right ) \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + i \, \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - i \, \cos \left (x\right ){\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + i \, \cos \left (x\right ){\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - i \, \cos \left (x\right ){\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) + \cos \left (x\right ) \log \left (-\frac{\sin \left (x\right ) + 1}{\sin \left (x\right ) - 1}\right ) - 2 \, x\right )} \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{a \sec ^{2}{\left (x \right )}} \csc{\left (x \right )} \sec{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sec \left (x\right )^{2}} x \csc \left (x\right ) \sec \left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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