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

Optimal. Leaf size=48 \[ -\frac {e^{-2 i a} \left (1+e^{2 i a} c^4 x^4\right )}{2 c^4 x^3 \cos ^{\frac {3}{2}}(a-2 i \log (c x))} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4484, 4482, 261} \[ -\frac {e^{-2 i a} \left (1+e^{2 i a} c^4 x^4\right )}{2 c^4 x^3 \cos ^{\frac {3}{2}}(a-2 i \log (c x))} \]

Antiderivative was successfully verified.

[In]

Int[Cos[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*Cos[a - (2*I)*Log[c*x]]^(3/2))

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]

Rule 4482

Int[Cos[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_), x_Symbol] :> Dist[(Cos[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 4484

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 82, normalized size = 1.71 \[ -\frac {x (\cos (a)-i \sin (a)) \sqrt {\frac {2 \cos (a) \left (c^4 x^4+1\right )+2 i \sin (a) \left (c^4 x^4-1\right )}{c^2 x^2}}}{\cos (a) \left (c^4 x^4+1\right )+i \sin (a) \left (c^4 x^4-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 39, normalized size = 0.81 \[ -\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {c^{4} x^{4} + e^{\left (-2 i \, a\right )}} e^{\left (-\frac {3}{2} i \, a\right )}}{c^{5} x^{4} + c e^{\left (-2 i \, a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(1/2)*sqrt(c^4*x^4 + e^(-2*I*a))*e^(-3/2*I*a)/(c^5*x^4 + c*e^(-2*I*a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (a - 2 i \, \log \left (c x\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (a -2 i \ln \left (c x \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 187, normalized size = 3.90 \[ -\frac {{\left ({\left (\sqrt {2} \cos \left (\frac {3}{2} \, a\right ) + i \, \sqrt {2} \sin \left (\frac {3}{2} \, a\right )\right )} c^{4} x^{4} + \sqrt {2} \cos \left (\frac {1}{2} \, a\right ) - i \, \sqrt {2} \sin \left (\frac {1}{2} \, a\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (c^{4} x^{4} \sin \left (2 \, a\right ), c^{4} x^{4} \cos \left (2 \, a\right ) + 1\right )\right ) + {\left ({\left (-i \, \sqrt {2} \cos \left (\frac {3}{2} \, a\right ) + \sqrt {2} \sin \left (\frac {3}{2} \, a\right )\right )} c^{4} x^{4} - i \, \sqrt {2} \cos \left (\frac {1}{2} \, a\right ) - \sqrt {2} \sin \left (\frac {1}{2} \, a\right )\right )} \sin \left (\frac {3}{2} \, \arctan \left (c^{4} x^{4} \sin \left (2 \, a\right ), c^{4} x^{4} \cos \left (2 \, a\right ) + 1\right )\right )}{{\left ({\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} c^{8} x^{8} + 2 \, c^{4} x^{4} \cos \left (2 \, a\right ) + 1\right )}^{\frac {3}{4}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((sqrt(2)*cos(3/2*a) + I*sqrt(2)*sin(3/2*a))*c^4*x^4 + sqrt(2)*cos(1/2*a) - I*sqrt(2)*sin(1/2*a))*cos(3/2*ar
ctan2(c^4*x^4*sin(2*a), c^4*x^4*cos(2*a) + 1)) + ((-I*sqrt(2)*cos(3/2*a) + sqrt(2)*sin(3/2*a))*c^4*x^4 - I*sqr
t(2)*cos(1/2*a) - sqrt(2)*sin(1/2*a))*sin(3/2*arctan2(c^4*x^4*sin(2*a), c^4*x^4*cos(2*a) + 1)))/(((cos(2*a)^2
+ sin(2*a)^2)*c^8*x^8 + 2*c^4*x^4*cos(2*a) + 1)^(3/4)*c)

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mupad [B]  time = 2.79, size = 48, normalized size = 1.00 \[ -\frac {2\,x\,\sqrt {\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}}}{2\,c^2\,x^2}+\frac {c^2\,x^2\,{\mathrm {e}}^{a\,1{}\mathrm {i}}}{2}}}{{\mathrm {e}}^{a\,2{}\mathrm {i}}\,c^4\,x^4+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(a - log(c*x)*2i)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos ^{\frac {3}{2}}{\left (a - 2 i \log {\left (c x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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