3.262 \(\int \sec ^3(a+2 \log (c x^{\frac{i}{2}})) \, dx\)

Optimal. Leaf size=58 \[ \frac{1}{2} x \sec \left (a+2 \log \left (c x^{\frac{i}{2}}\right )\right )-\frac{1}{2} i x \tan \left (a+2 \log \left (c x^{\frac{i}{2}}\right )\right ) \sec \left (a+2 \log \left (c x^{\frac{i}{2}}\right )\right ) \]

[Out]

(x*Sec[a + 2*Log[c*x^(I/2)]])/2 - (I/2)*x*Sec[a + 2*Log[c*x^(I/2)]]*Tan[a + 2*Log[c*x^(I/2)]]

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Rubi [A]  time = 0.0349121, antiderivative size = 48, normalized size of antiderivative = 0.83, 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 = {4503, 4505, 261} \[ \frac{2 e^{i a} x \left (c x^{\frac{i}{2}}\right )^{2 i}}{\left (1+e^{2 i a} \left (c x^{\frac{i}{2}}\right )^{4 i}\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[Sec[a + 2*Log[c*x^(I/2)]]^3,x]

[Out]

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

Rule 4503

Int[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sec[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 4505

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^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]

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 \sec ^3\left (a+2 \log \left (c x^{\frac{i}{2}}\right )\right ) \, dx &=-\left (\left (2 i \left (c x^{\frac{i}{2}}\right )^{2 i} x\right ) \operatorname{Subst}\left (\int x^{-1-2 i} \sec ^3(a+2 \log (x)) \, dx,x,c x^{\frac{i}{2}}\right )\right )\\ &=-\left (\left (16 i e^{3 i a} \left (c x^{\frac{i}{2}}\right )^{2 i} x\right ) \operatorname{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 )\right )\\ &=\frac{2 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]  time = 0.113546, size = 137, normalized size = 2.36 \[ -\frac{\sec ^2\left (a+2 \log \left (c x^{\frac{i}{2}}\right )\right ) \left (i \left (1-2 x^2\right ) \sin \left (a+2 \log \left (c x^{\frac{i}{2}}\right )-i \log (x)\right )+\left (2 x^2+1\right ) \cos \left (a+2 \log \left (c x^{\frac{i}{2}}\right )-i \log (x)\right )\right ) \left (i \sin \left (2 \left (a+2 \log \left (c x^{\frac{i}{2}}\right )-i \log (x)\right )\right )+\cos \left (2 \left (a+2 \log \left (c x^{\frac{i}{2}}\right )-i \log (x)\right )\right )\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[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]])*(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)

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Maple [C]  time = 0.204, size = 214, normalized size = 3.7 \begin{align*} 2\,{\frac{x{{\rm e}^{-i \left ( i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{i/2} \right ) +i\pi \,{\it csgn} \left ( ic{x}^{i/2} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{i/2} \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{i/2} \right ) -a \right ) }}}{ \left ({{\rm e}^{-2\,i \left ( i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{i/2} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{i/2} \right ) +i\pi \,{\it csgn} \left ( ic{x}^{i/2} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{i/2} \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{i/2} \right ) -a \right ) }}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.24468, size = 208, normalized size = 3.59 \begin{align*} \frac{{\left ({\left (2 \, \cos \left (a\right ) + 2 i \, \sin \left (a\right )\right )} \cos \left (2 \, \log \left (c\right )\right ) + 2 \,{\left (i \, \cos \left (a\right ) - \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 ) +{\left ({\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )} \cos \left (4 \, \log \left (c\right )\right ) - 2 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.449053, size = 151, normalized size = 2.6 \begin{align*} \frac{2 \, x e^{\left (i \, a + 2 i \, \log \left (c x^{\frac{1}{2} i}\right )\right )}}{e^{\left (4 i \, a + 8 i \, \log \left (c x^{\frac{1}{2} i}\right )\right )} + 2 \, e^{\left (2 i \, a + 4 i \, \log \left (c x^{\frac{1}{2} i}\right )\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*e^(I*a + 2*I*log(c*x^(1/2*I)))/(e^(4*I*a + 8*I*log(c*x^(1/2*I))) + 2*e^(2*I*a + 4*I*log(c*x^(1/2*I))) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{3}{\left (a + 2 \log{\left (c x^{\frac{i}{2}} \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 4.75822, size = 100, normalized size = 1.72 \begin{align*} -\frac{2 \, 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 \, 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*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*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)