3.69 \(\int \frac{1}{\sin ^{\frac{3}{2}}(a-2 i \log (c x))} \, dx\)

Optimal. Leaf size=49 \[ \frac{e^{-2 i a} \left (1-e^{2 i a} c^4 x^4\right )}{2 c^4 x^3 \sin ^{\frac{3}{2}}(a-2 i \log (c x))} \]

[Out]

(1 - c^4*E^((2*I)*a)*x^4)/(2*c^4*E^((2*I)*a)*x^3*Sin[a - (2*I)*Log[c*x]]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0389924, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4483, 4481, 261} \[ \frac{e^{-2 i a} \left (1-e^{2 i a} c^4 x^4\right )}{2 c^4 x^3 \sin ^{\frac{3}{2}}(a-2 i \log (c x))} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a - (2*I)*Log[c*x]]^(-3/2),x]

[Out]

(1 - c^4*E^((2*I)*a)*x^4)/(2*c^4*E^((2*I)*a)*x^3*Sin[a - (2*I)*Log[c*x]]^(3/2))

Rule 4483

Int[Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sin[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 4481

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

Rule 261

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]

Rubi steps

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

Mathematica [A]  time = 0.131781, size = 81, normalized size = 1.65 \[ \frac{x (\cos (a)-i \sin (a)) \sqrt{\frac{2 \sin (a) \left (c^4 x^4+1\right )-2 i \cos (a) \left (c^4 x^4-1\right )}{c^2 x^2}}}{\cos (a) \left (c^4 x^4-1\right )+i \sin (a) \left (c^4 x^4+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Cos[a] - I*Sin[a])*Sqrt[((-2*I)*(-1 + c^4*x^4)*Cos[a] + 2*(1 + c^4*x^4)*Sin[a])/(c^2*x^2)])/((-1 + c^4*x^4
)*Cos[a] + I*(1 + c^4*x^4)*Sin[a])

________________________________________________________________________________________

Maple [F]  time = 0.415, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( a-2\,i\ln \left ( cx \right ) \right ) \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(a-2*I*ln(c*x))^(3/2),x)

[Out]

int(1/sin(a-2*I*ln(c*x))^(3/2),x)

________________________________________________________________________________________

Maxima [B]  time = 2.44169, size = 543, normalized size = 11.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((cos(a)^2 + sin(a)^2)*c^4*x^4 + 2*c^2*x^2*cos(a) + 1)^(1/4)*((cos(a)^2 + sin(a)^2)*c^4*x^4 - 2*c^2*x^2*cos(a)
+ 1)^(1/4)*(((c^4*x^4*((I + 1)*cos(3/2*a) + (I - 1)*sin(3/2*a)) - (I + 1)*cos(1/2*a) + (I - 1)*sin(1/2*a))*co
s(3/2*arctan2(c^2*x^2*sin(a), -c^2*x^2*cos(a) + 1)) + (c^4*x^4*((I - 1)*cos(3/2*a) - (I + 1)*sin(3/2*a)) - (I
- 1)*cos(1/2*a) - (I + 1)*sin(1/2*a))*sin(3/2*arctan2(c^2*x^2*sin(a), -c^2*x^2*cos(a) + 1)))*cos(3/2*arctan2(c
^2*x^2*sin(a), c^2*x^2*cos(a) + 1)) + ((c^4*x^4*(-(I - 1)*cos(3/2*a) + (I + 1)*sin(3/2*a)) + (I - 1)*cos(1/2*a
) + (I + 1)*sin(1/2*a))*cos(3/2*arctan2(c^2*x^2*sin(a), -c^2*x^2*cos(a) + 1)) + (c^4*x^4*((I + 1)*cos(3/2*a) +
(I - 1)*sin(3/2*a)) - (I + 1)*cos(1/2*a) + (I - 1)*sin(1/2*a))*sin(3/2*arctan2(c^2*x^2*sin(a), -c^2*x^2*cos(a
) + 1)))*sin(3/2*arctan2(c^2*x^2*sin(a), c^2*x^2*cos(a) + 1)))/(((cos(a)^4 + 2*cos(a)^2*sin(a)^2 + sin(a)^4)*c
^8*x^8 - 2*(cos(a)^2 - sin(a)^2)*c^4*x^4 + 1)*c)

________________________________________________________________________________________

Fricas [A]  time = 0.458693, size = 146, normalized size = 2.98 \begin{align*} -\frac{2 \, \sqrt{\frac{1}{2}} x \sqrt{i \, e^{\left (-2 i \, a - 4 \, \log \left (c x\right )\right )} - i} e^{\left (-\frac{3}{2} i \, a - 3 \, \log \left (c x\right )\right )}}{e^{\left (-2 i \, a - 4 \, \log \left (c x\right )\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(1/2)*x*sqrt(I*e^(-2*I*a - 4*log(c*x)) - I)*e^(-3/2*I*a - 3*log(c*x))/(e^(-2*I*a - 4*log(c*x)) - 1)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(a-2*I*ln(c*x))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sin \left (a - 2 i \, \log \left (c x\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a - 2*I*log(c*x))^(-3/2), x)