\(\int \csc ^3(a+2 \log (c x^{\frac {i}{2}})) \, dx\) [304]

Optimal result

Integrand size = 17, antiderivative size = 58 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=\frac {1}{2} x \csc \left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right )+\frac {1}{2} i x \cot \left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \csc \left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \]

[Out]

1/2*x*csc(a+2*ln(c*x^(1/2*I)))+1/2*I*x*cot(a+2*ln(c*x^(1/2*I)))*csc(a+2*ln(c*x^(1/2*I)))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4600, 4602, 267} \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=-\frac {2 i e^{i a} x \left (c x^{\frac {i}{2}}\right )^{2 i}}{\left (1-e^{2 i a} \left (c x^{\frac {i}{2}}\right )^{4 i}\right )^2} \]

[In]

Int[Csc[a + 2*Log[c*x^(I/2)]]^3,x]

[Out]

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

Rule 267

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

Rule 4600

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 4602

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(137\) vs. \(2(58)=116\).

Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.36 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=\frac {\csc ^2\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \left (i \left (-1+2 x^2\right ) \cos \left (a+2 \log \left (c x^{\frac {i}{2}}\right )-i \log (x)\right )+\left (1+2 x^2\right ) \sin \left (a+2 \log \left (c x^{\frac {i}{2}}\right )-i \log (x)\right )\right ) \left (\cos \left (2 \left (a+2 \log \left (c x^{\frac {i}{2}}\right )-i \log (x)\right )\right )+i \sin \left (2 \left (a+2 \log \left (c x^{\frac {i}{2}}\right )-i \log (x)\right )\right )\right )}{2 x^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 265.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17

method	result	size
parallelrisch	\(-\frac {x \left (i {\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {i}{2}}\right )\right )}^{4}-2 {\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {i}{2}}\right )\right )}^{3}-i-2 \tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {i}{2}}\right )\right )\right )}{8 {\tan \left (\frac {a}{2}+\ln \left (c \,x^{\frac {i}{2}}\right )\right )}^{2}}\)	\(68\)
risch	\(-\frac {2 i x \left (x^{\frac {i}{2}}\right )^{2 i} c^{2 i} {\mathrm e}^{-\operatorname {csgn}\left (i x^{\frac {i}{2}}\right ) \pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{2}+\operatorname {csgn}\left (i x^{\frac {i}{2}}\right ) \pi \,\operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right ) \operatorname {csgn}\left (i c \right )+\pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{3}-\pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{2} \operatorname {csgn}\left (i c \right )+i a}}{\left (c^{4 i} \left (x^{\frac {i}{2}}\right )^{4 i} {\mathrm e}^{-2 \,\operatorname {csgn}\left (i x^{\frac {i}{2}}\right ) \pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{2}} {\mathrm e}^{2 \,\operatorname {csgn}\left (i x^{\frac {i}{2}}\right ) \pi \,\operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right ) \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 \pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{3}} {\mathrm e}^{-2 \pi \operatorname {csgn}\left (i c \,x^{\frac {i}{2}}\right )^{2} \operatorname {csgn}\left (i c \right )} {\mathrm e}^{2 i a}-1\right )^{2}}\)	\(209\)

[In]

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

[Out]

-1/8*x*(I*tan(1/2*a+ln(c*x^(1/2*I)))^4-2*tan(1/2*a+ln(c*x^(1/2*I)))^3-I-2*tan(1/2*a+ln(c*x^(1/2*I))))/tan(1/2*
a+ln(c*x^(1/2*I)))^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=-\frac {2 \, {\left (2 i \, x^{2} e^{\left (3 i \, a + 6 i \, \log \left (c\right )\right )} - i \, e^{\left (5 i \, a + 10 i \, \log \left (c\right )\right )}\right )}}{x^{4} - 2 \, x^{2} e^{\left (2 i \, a + 4 i \, \log \left (c\right )\right )} + e^{\left (4 i \, a + 8 i \, \log \left (c\right )\right )}} \]

[In]

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

[Out]

-2*(2*I*x^2*e^(3*I*a + 6*I*log(c)) - I*e^(5*I*a + 10*I*log(c)))/(x^4 - 2*x^2*e^(2*I*a + 4*I*log(c)) + e^(4*I*a
+ 8*I*log(c)))

Sympy [F]

\[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=\int \csc ^{3}{\left (a + 2 \log {\left (c x^{\frac {i}{2}} \right )} \right )}\, dx \]

[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)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (40) = 80\).

Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.64 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=-\frac {2 \, {\left ({\left (i \, \cos \left (a\right ) - \sin \left (a\right )\right )} \cos \left (2 \, \log \left (c\right )\right ) - {\left (\cos \left (a\right ) + i \, \sin \left (a\right )\right )} \sin \left (2 \, \log \left (c\right )\right )\right )} x e^{\left (6 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \left (x\right )\right ), \cos \left (\frac {1}{2} \, \log \left (x\right )\right )\right )\right )}}{{\left (\cos \left (4 \, a\right ) + i \, \sin \left (4 \, a\right )\right )} \cos \left (8 \, \log \left (c\right )\right ) - 2 \, {\left ({\left (\cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} \cos \left (4 \, \log \left (c\right )\right ) + {\left (i \, \cos \left (2 \, a\right ) - \sin \left (2 \, a\right )\right )} \sin \left (4 \, \log \left (c\right )\right )\right )} e^{\left (4 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \left (x\right )\right ), \cos \left (\frac {1}{2} \, \log \left (x\right )\right )\right )\right )} + {\left (i \, \cos \left (4 \, a\right ) - \sin \left (4 \, a\right )\right )} \sin \left (8 \, \log \left (c\right )\right ) + e^{\left (8 \, \arctan \left (\sin \left (\frac {1}{2} \, \log \left (x\right )\right ), \cos \left (\frac {1}{2} \, \log \left (x\right )\right )\right )\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=\frac {2 i \, c^{10 i} e^{\left (5 i \, a\right )}}{c^{8 i} e^{\left (4 i \, a\right )} - 2 \, c^{4 i} x^{2} e^{\left (2 i \, a\right )} + x^{4}} - \frac {4 i \, c^{6 i} x^{2} e^{\left (3 i \, a\right )}}{c^{8 i} e^{\left (4 i \, a\right )} - 2 \, c^{4 i} x^{2} e^{\left (2 i \, a\right )} + x^{4}} \]

[In]

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

[Out]

2*I*c^(10*I)*e^(5*I*a)/(c^(8*I)*e^(4*I*a) - 2*c^(4*I)*x^2*e^(2*I*a) + x^4) - 4*I*c^(6*I)*x^2*e^(3*I*a)/(c^(8*I
)*e^(4*I*a) - 2*c^(4*I)*x^2*e^(2*I*a) + x^4)

Mupad [B] (verification not implemented)

Time = 28.87 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \csc ^3\left (a+2 \log \left (c x^{\frac {i}{2}}\right )\right ) \, dx=-\frac {x\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^{\frac {1}{2}{}\mathrm {i}}\right )}^{2{}\mathrm {i}}\,2{}\mathrm {i}}{1+{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\left (c\,x^{\frac {1}{2}{}\mathrm {i}}\right )}^{8{}\mathrm {i}}-2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^{\frac {1}{2}{}\mathrm {i}}\right )}^{4{}\mathrm {i}}} \]

[In]

int(1/sin(a + 2*log(c*x^(1i/2)))^3,x)

[Out]

-(x*exp(a*1i)*(c*x^(1i/2))^2i*2i)/(exp(a*4i)*(c*x^(1i/2))^8i - 2*exp(a*2i)*(c*x^(1i/2))^4i + 1)