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

Optimal. Leaf size=81 \[ -\frac {i \text {Li}_2\left (e^{2 i x}\right ) \sec ^2(x)}{2 \sqrt {a \sec ^4(x)}}-\frac {i x^2 \sec ^2(x)}{2 \sqrt {a \sec ^4(x)}}+\frac {x \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt {a \sec ^4(x)}} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6720, 3717, 2190, 2279, 2391} \[ -\frac {i \sec ^2(x) \text {PolyLog}\left (2,e^{2 i x}\right )}{2 \sqrt {a \sec ^4(x)}}-\frac {i x^2 \sec ^2(x)}{2 \sqrt {a \sec ^4(x)}}+\frac {x \log \left (1-e^{2 i x}\right ) \sec ^2(x)}{\sqrt {a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.62 \[ -\frac {i \sec ^2(x) \left (\text {Li}_2\left (e^{2 i x}\right )+x \left (x+2 i \log \left (1-e^{2 i x}\right )\right )\right )}{2 \sqrt {a \sec ^4(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.10, size = 138, normalized size = 1.70 \[ \frac {{\left (x \cos \relax (x)^{2} \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \cos \relax (x)^{2} \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) + i \, \cos \relax (x)^{2} {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, \cos \relax (x)^{2} {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right )\right )} \sqrt {\frac {a}{\cos \relax (x)^{4}}}}{2 \, a} \]

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(-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*dilo
g(cos(x) - I*sin(x)) + I*cos(x)^2*dilog(-cos(x) + I*sin(x)) - I*cos(x)^2*dilog(-cos(x) - I*sin(x)))*sqrt(a/cos
(x)^4)/a

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

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(x*csc(x)*sec(x)/sqrt(a*sec(x)^4), x)

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maple [B]  time = 0.20, size = 147, normalized size = 1.81 \[ \frac {i {\mathrm e}^{2 i x} x^{2}}{2 \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i x}+1\right )^{2}}-\frac {2 i \left (\frac {{\mathrm e}^{2 i x} x^{2}}{2}+\frac {i {\mathrm e}^{2 i x} x \ln \left (1+{\mathrm e}^{i x}\right )}{2}+\frac {{\mathrm e}^{2 i x} \polylog \left (2, -{\mathrm e}^{i x}\right )}{2}+\frac {i {\mathrm e}^{2 i x} x \ln \left (1-{\mathrm e}^{i x}\right )}{2}+\frac {{\mathrm e}^{2 i x} \polylog \left (2, {\mathrm e}^{i x}\right )}{2}\right )}{\sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i x}+1\right )^{2}} \]

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]

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

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maxima [A]  time = 1.92, size = 83, normalized size = 1.02 \[ \frac {-i \, x^{2} + 2 i \, x \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) - 2 i \, x \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) + x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - 2 i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) - 2 i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right )}{2 \, \sqrt {a}} \]

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]

1/2*(-I*x^2 + 2*I*x*arctan2(sin(x), cos(x) + 1) - 2*I*x*arctan2(sin(x), -cos(x) + 1) + x*log(cos(x)^2 + sin(x)
^2 + 2*cos(x) + 1) + x*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 2*I*dilog(-e^(I*x)) - 2*I*dilog(e^(I*x)))/sqr
t(a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\cos \relax (x)\,\sin \relax (x)\,\sqrt {\frac {a}{{\cos \relax (x)}^4}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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*csc(x)*sec(x)/sqrt(a*sec(x)**4), x)

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