3.305 \(\int \csc ^3(a+2 \log (c x^{-\frac {i}{2}})) \, dx\)

Optimal. Leaf size=51 \[ \frac {2 i e^{3 i a} x \left (c x^{-\frac {i}{2}}\right )^{6 i}}{\left (1-e^{2 i a} \left (c x^{-\frac {i}{2}}\right )^{4 i}\right )^2} \]

[Out]

2*I*exp(3*I*a)*(c/(x^(1/2*I)))^(6*I)*x/(1-exp(2*I*a)*(c/(x^(1/2*I)))^(4*I))^2

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4504, 4506, 264} \[ \frac {2 i e^{3 i a} x \left (c x^{-\frac {i}{2}}\right )^{6 i}}{\left (1-e^{2 i a} \left (c x^{-\frac {i}{2}}\right )^{4 i}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + 2*Log[c/x^(I/2)]]^3,x]

[Out]

((2*I)*E^((3*I)*a)*(c/x^(I/2))^(6*I)*x)/(1 - E^((2*I)*a)*(c/x^(I/2))^(4*I))^2

Rule 264

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

Rule 4504

Int[Csc[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Csc[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n,
1])

Rule 4506

Int[Csc[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(-2*I)^p*E^(I*a*d*p), Int[(
(e*x)^m*x^(I*b*d*p))/(1 - E^(2*I*a*d)*x^(2*I*b*d))^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \csc ^3\left (a+2 \log \left (c x^{-\frac {i}{2}}\right )\right ) \, dx &=\left (2 i \left (c x^{-\frac {i}{2}}\right )^{-2 i} x\right ) \operatorname {Subst}\left (\int x^{-1+2 i} \csc ^3(a+2 \log (x)) \, dx,x,c x^{-\frac {i}{2}}\right )\\ &=-\left (\left (16 e^{3 i a} \left (c x^{-\frac {i}{2}}\right )^{-2 i} x\right ) \operatorname {Subst}\left (\int \frac {x^{-1+8 i}}{\left (1-e^{2 i a} x^{4 i}\right )^3} \, dx,x,c x^{-\frac {i}{2}}\right )\right )\\ &=\frac {2 i e^{3 i a} \left (c x^{-\frac {i}{2}}\right )^{6 i} x}{\left (1-e^{2 i a} \left (c x^{-\frac {i}{2}}\right )^{4 i}\right )^2}\\ \end {align*}

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Mathematica [B]  time = 0.17, size = 137, normalized size = 2.69 \[ -\frac {\csc ^2\left (a+2 \log \left (c x^{-\frac {i}{2}}\right )\right ) \left (i \left (2 x^2+1\right ) \sin \left (a+2 \log \left (c x^{-\frac {i}{2}}\right )+i \log (x)\right )+\left (2 x^2-1\right ) \cos \left (a+2 \log \left (c x^{-\frac {i}{2}}\right )+i \log (x)\right )\right ) \left (\sin \left (2 \left (a+2 \log \left (c x^{-\frac {i}{2}}\right )+i \log (x)\right )\right )+i \cos \left (2 \left (a+2 \log \left (c x^{-\frac {i}{2}}\right )+i \log (x)\right )\right )\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + 2*Log[c/x^(I/2)]]^3,x]

[Out]

-1/2*(Csc[a + 2*Log[c/x^(I/2)]]^2*((-1 + 2*x^2)*Cos[a + 2*Log[c/x^(I/2)] + I*Log[x]] + I*(1 + 2*x^2)*Sin[a + 2
*Log[c/x^(I/2)] + I*Log[x]])*(I*Cos[2*(a + 2*Log[c/x^(I/2)] + I*Log[x])] + Sin[2*(a + 2*Log[c/x^(I/2)] + I*Log
[x])]))/x^2

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fricas [B]  time = 1.46, size = 56, normalized size = 1.10 \[ \frac {4 i \, x^{2} e^{\left (2 i \, a + 4 i \, \log \relax (c)\right )} - 2 i}{x^{4} e^{\left (5 i \, a + 10 i \, \log \relax (c)\right )} - 2 \, x^{2} e^{\left (3 i \, a + 6 i \, \log \relax (c)\right )} + e^{\left (i \, a + 2 i \, \log \relax (c)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+2*log(c/(x^(1/2*I))))^3,x, algorithm="fricas")

[Out]

(4*I*x^2*e^(2*I*a + 4*I*log(c)) - 2*I)/(x^4*e^(5*I*a + 10*I*log(c)) - 2*x^2*e^(3*I*a + 6*I*log(c)) + e^(I*a +
2*I*log(c)))

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giac [B]  time = 2.61, size = 83, normalized size = 1.63 \[ \frac {4 i \, c^{4 i} x^{2} e^{\left (2 i \, a\right )}}{c^{10 i} x^{4} e^{\left (5 i \, a\right )} - 2 \, c^{6 i} x^{2} e^{\left (3 i \, a\right )} + c^{2 i} e^{\left (i \, a\right )}} - \frac {2 i}{c^{10 i} x^{4} e^{\left (5 i \, a\right )} - 2 \, c^{6 i} x^{2} e^{\left (3 i \, a\right )} + c^{2 i} e^{\left (i \, a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+2*log(c/(x^(1/2*I))))^3,x, algorithm="giac")

[Out]

4*I*c^(4*I)*x^2*e^(2*I*a)/(c^(10*I)*x^4*e^(5*I*a) - 2*c^(6*I)*x^2*e^(3*I*a) + c^(2*I)*e^(I*a)) - 2*I/(c^(10*I)
*x^4*e^(5*I*a) - 2*c^(6*I)*x^2*e^(3*I*a) + c^(2*I)*e^(I*a))

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maple [C]  time = 0.18, size = 239, normalized size = 4.69 \[ \frac {2 i x \left (x^{\frac {i}{2}}\right )^{-6 i} c^{6 i} {\mathrm e}^{3 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{3}-3 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{2} \mathrm {csgn}\left (i c \right )-3 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{2} \mathrm {csgn}\left (i x^{-\frac {i}{2}}\right )+3 \pi \,\mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{-\frac {i}{2}}\right )+3 i a}}{\left (\left (x^{\frac {i}{2}}\right )^{-4 i} c^{4 i} {\mathrm e}^{2 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{3}} {\mathrm e}^{-2 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{2} \mathrm {csgn}\left (i c \right )} {\mathrm e}^{-2 \pi \mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right )^{2} \mathrm {csgn}\left (i x^{-\frac {i}{2}}\right )} {\mathrm e}^{2 \pi \,\mathrm {csgn}\left (i c \,x^{-\frac {i}{2}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{-\frac {i}{2}}\right )} {\mathrm e}^{2 i a}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(a+2*ln(c/(x^(1/2*I))))^3,x)

[Out]

2*I*x*((x^(1/2*I))^(-2*I))^3*(c^(2*I))^3*exp(3*Pi*csgn(I*c/(x^(1/2*I)))^3-3*Pi*csgn(I*c/(x^(1/2*I)))^2*csgn(I*
c)-3*Pi*csgn(I*c/(x^(1/2*I)))^2*csgn(I/(x^(1/2*I)))+3*Pi*csgn(I*c/(x^(1/2*I)))*csgn(I*c)*csgn(I/(x^(1/2*I)))+3
*I*a)/(((x^(1/2*I))^(-2*I))^2*(c^(2*I))^2*exp(2*Pi*csgn(I*c/(x^(1/2*I)))^3)*exp(-2*Pi*csgn(I*c/(x^(1/2*I)))^2*
csgn(I*c))*exp(-2*Pi*csgn(I*c/(x^(1/2*I)))^2*csgn(I/(x^(1/2*I))))*exp(2*Pi*csgn(I*c/(x^(1/2*I)))*csgn(I*c)*csg
n(I/(x^(1/2*I))))*exp(2*I*a)-1)^2

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maxima [B]  time = 0.40, size = 166, normalized size = 3.25 \[ \frac {{\left (2 \, {\left (i \, \cos \left (3 \, a\right ) - \sin \left (3 \, a\right )\right )} \cos \left (6 \, \log \relax (c)\right ) - {\left (2 \, \cos \left (3 \, a\right ) + 2 i \, \sin \left (3 \, a\right )\right )} \sin \left (6 \, \log \relax (c)\right )\right )} x e^{\left (6 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \relax (x)\right ), \cos \left (\frac {1}{2} \, \log \relax (x)\right )\right )\right )}}{{\left ({\left (\cos \left (4 \, a\right ) + i \, \sin \left (4 \, a\right )\right )} \cos \left (8 \, \log \relax (c)\right ) - {\left (-i \, \cos \left (4 \, a\right ) + \sin \left (4 \, a\right )\right )} \sin \left (8 \, \log \relax (c)\right )\right )} e^{\left (8 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \relax (x)\right ), \cos \left (\frac {1}{2} \, \log \relax (x)\right )\right )\right )} - {\left ({\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )} \cos \left (4 \, \log \relax (c)\right ) - 2 \, {\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \sin \left (4 \, \log \relax (c)\right )\right )} e^{\left (4 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \relax (x)\right ), \cos \left (\frac {1}{2} \, \log \relax (x)\right )\right )\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+2*log(c/(x^(1/2*I))))^3,x, algorithm="maxima")

[Out]

(2*(I*cos(3*a) - sin(3*a))*cos(6*log(c)) - (2*cos(3*a) + 2*I*sin(3*a))*sin(6*log(c)))*x*e^(6*arctan2(sin(1/2*l
og(x)), cos(1/2*log(x))))/(((cos(4*a) + I*sin(4*a))*cos(8*log(c)) - (-I*cos(4*a) + sin(4*a))*sin(8*log(c)))*e^
(8*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) - ((2*cos(2*a) + 2*I*sin(2*a))*cos(4*log(c)) - 2*(-I*cos(2*a) +
sin(2*a))*sin(4*log(c)))*e^(4*arctan2(sin(1/2*log(x)), cos(1/2*log(x)))) + 1)

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mupad [B]  time = 6.32, size = 38, normalized size = 0.75 \[ \frac {x\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (\frac {c}{x^{\frac {1}{2}{}\mathrm {i}}}\right )}^{6{}\mathrm {i}}\,2{}\mathrm {i}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (\frac {c}{x^{\frac {1}{2}{}\mathrm {i}}}\right )}^{4{}\mathrm {i}}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(a + 2*log(c/x^(1i/2)))^3,x)

[Out]

(x*exp(a*3i)*(c/x^(1i/2))^6i*2i)/(exp(a*2i)*(c/x^(1i/2))^4i - 1)^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc ^{3}{\left (a + 2 \log {\left (c x^{- \frac {i}{2}} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(a+2*ln(c/(x**(1/2*I))))**3,x)

[Out]

Integral(csc(a + 2*log(c*x**(-I/2)))**3, x)

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