3.261 \(\int x \sec ^3(a+2 \log (c x^i)) \, dx\)

Optimal. Leaf size=45 \[ \frac{e^{i a} x^2 \left (c x^i\right )^{2 i}}{\left (1+e^{2 i a} \left (c x^i\right )^{4 i}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0427162, antiderivative size = 45, normalized size of antiderivative = 1., 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 = {4509, 4505, 261} \[ \frac{e^{i a} x^2 \left (c x^i\right )^{2 i}}{\left (1+e^{2 i a} \left (c x^i\right )^{4 i}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sec[a + 2*Log[c*x^I]]^3,x]

[Out]

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

Rule 4509

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, 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 x \sec ^3\left (a+2 \log \left (c x^i\right )\right ) \, dx &=-\left (\left (i \left (c x^i\right )^{2 i} x^2\right ) \operatorname{Subst}\left (\int x^{-1-2 i} \sec ^3(a+2 \log (x)) \, dx,x,c x^i\right )\right )\\ &=-\left (\left (8 i e^{3 i a} \left (c x^i\right )^{2 i} x^2\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^i\right )\right )\\ &=\frac{e^{i a} \left (c x^i\right )^{2 i} x^2}{\left (1+e^{2 i a} \left (c x^i\right )^{4 i}\right )^2}\\ \end{align*}

Mathematica [B]  time = 0.151096, size = 127, normalized size = 2.82 \[ -\frac{\sec ^2\left (a+2 \log \left (c x^i\right )\right ) \left (i \left (1-2 x^4\right ) \sin \left (a+2 \log \left (c x^i\right )-2 i \log (x)\right )+\left (2 x^4+1\right ) \cos \left (a+2 \log \left (c x^i\right )-2 i \log (x)\right )\right ) \left (i \sin \left (2 \left (a+2 \log \left (c x^i\right )-2 i \log (x)\right )\right )+\cos \left (2 \left (a+2 \log \left (c x^i\right )-2 i \log (x)\right )\right )\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.199, size = 215, normalized size = 4.8 \begin{align*}{\frac{{x}^{2}{{\rm e}^{-i \left ( i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{i} \right ) +i\pi \,{\it csgn} \left ( ic{x}^{i} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{i} \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{i} \right ) -a \right ) }}}{ \left ({{\rm e}^{-2\,i \left ( i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{3}-i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic{x}^{i} \right ) \right ) ^{2}{\it csgn} \left ( i{x}^{i} \right ) +i\pi \,{\it csgn} \left ( ic{x}^{i} \right ){\it csgn} \left ( ic \right ){\it csgn} \left ( i{x}^{i} \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ({x}^{i} \right ) -a \right ) }}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.11996, size = 189, normalized size = 4.2 \begin{align*} \frac{{\left ({\left (\cos \left (a\right ) + i \, \sin \left (a\right )\right )} \cos \left (2 \, \log \left (c\right )\right ) -{\left (-i \, \cos \left (a\right ) + \sin \left (a\right )\right )} \sin \left (2 \, \log \left (c\right )\right )\right )} x^{2} e^{\left (6 \, \arctan \left (\sin \left (\log \left (x\right )\right ), \cos \left (\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 (\log \left (x\right )\right ), \cos \left (\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 (\log \left (x\right )\right ), \cos \left (\log \left (x\right )\right )\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.449068, size = 127, normalized size = 2.82 \begin{align*} \frac{x^{2} e^{\left (i \, a + 2 i \, \log \left (c x^{i}\right )\right )}}{e^{\left (4 i \, a + 8 i \, \log \left (c x^{i}\right )\right )} + 2 \, e^{\left (2 i \, a + 4 i \, \log \left (c x^{i}\right )\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sec(a + 2*log(c*x^I))^3, x)