3.893 \(\int (-1+\sec ^2(2 x))^3 \sin (2 x) \, dx\)

Optimal. Leaf size=37 \[ \frac{1}{2} \cos (2 x)+\frac{1}{10} \sec ^5(2 x)-\frac{1}{2} \sec ^3(2 x)+\frac{3}{2} \sec (2 x) \]

[Out]

Cos[2*x]/2 + (3*Sec[2*x])/2 - Sec[2*x]^3/2 + Sec[2*x]^5/10

________________________________________________________________________________________

Rubi [A]  time = 0.0391507, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4120, 2590, 270} \[ \frac{1}{2} \cos (2 x)+\frac{1}{10} \sec ^5(2 x)-\frac{1}{2} \sec ^3(2 x)+\frac{3}{2} \sec (2 x) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Sec[2*x]^2)^3*Sin[2*x],x]

[Out]

Cos[2*x]/2 + (3*Sec[2*x])/2 - Sec[2*x]^3/2 + Sec[2*x]^5/10

Rule 4120

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[b^p, Int[ActivateTrig[u*tan[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \left (-1+\sec ^2(2 x)\right )^3 \sin (2 x) \, dx &=\int \sin (2 x) \tan ^6(2 x) \, dx\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (2 x)\right )\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^6}-\frac{3}{x^4}+\frac{3}{x^2}\right ) \, dx,x,\cos (2 x)\right )\right )\\ &=\frac{1}{2} \cos (2 x)+\frac{3}{2} \sec (2 x)-\frac{1}{2} \sec ^3(2 x)+\frac{1}{10} \sec ^5(2 x)\\ \end{align*}

Mathematica [A]  time = 0.0292605, size = 37, normalized size = 1. \[ \frac{1}{2} \cos (2 x)+\frac{1}{10} \sec ^5(2 x)-\frac{1}{2} \sec ^3(2 x)+\frac{3}{2} \sec (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Sec[2*x]^2)^3*Sin[2*x],x]

[Out]

Cos[2*x]/2 + (3*Sec[2*x])/2 - Sec[2*x]^3/2 + Sec[2*x]^5/10

________________________________________________________________________________________

Maple [A]  time = 0.031, size = 32, normalized size = 0.9 \begin{align*}{\frac{1}{10\, \left ( \cos \left ( 2\,x \right ) \right ) ^{5}}}-{\frac{1}{2\, \left ( \cos \left ( 2\,x \right ) \right ) ^{3}}}+{\frac{3}{2\,\cos \left ( 2\,x \right ) }}+{\frac{\cos \left ( 2\,x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+sec(2*x)^2)^3*sin(2*x),x)

[Out]

1/10/cos(2*x)^5-1/2/cos(2*x)^3+3/2/cos(2*x)+1/2*cos(2*x)

________________________________________________________________________________________

Maxima [A]  time = 0.947299, size = 42, normalized size = 1.14 \begin{align*} \frac{3}{2 \, \cos \left (2 \, x\right )} - \frac{1}{2 \, \cos \left (2 \, x\right )^{3}} + \frac{1}{10 \, \cos \left (2 \, x\right )^{5}} + \frac{1}{2} \, \cos \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+sec(2*x)^2)^3*sin(2*x),x, algorithm="maxima")

[Out]

3/2/cos(2*x) - 1/2/cos(2*x)^3 + 1/10/cos(2*x)^5 + 1/2*cos(2*x)

________________________________________________________________________________________

Fricas [A]  time = 2.35321, size = 90, normalized size = 2.43 \begin{align*} \frac{5 \, \cos \left (2 \, x\right )^{6} + 15 \, \cos \left (2 \, x\right )^{4} - 5 \, \cos \left (2 \, x\right )^{2} + 1}{10 \, \cos \left (2 \, x\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+sec(2*x)^2)^3*sin(2*x),x, algorithm="fricas")

[Out]

1/10*(5*cos(2*x)^6 + 15*cos(2*x)^4 - 5*cos(2*x)^2 + 1)/cos(2*x)^5

________________________________________________________________________________________

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+sec(2*x)**2)**3*sin(2*x),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.07006, size = 45, normalized size = 1.22 \begin{align*} \frac{15 \, \cos \left (2 \, x\right )^{4} - 5 \, \cos \left (2 \, x\right )^{2} + 1}{10 \, \cos \left (2 \, x\right )^{5}} + \frac{1}{2} \, \cos \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+sec(2*x)^2)^3*sin(2*x),x, algorithm="giac")

[Out]

1/10*(15*cos(2*x)^4 - 5*cos(2*x)^2 + 1)/cos(2*x)^5 + 1/2*cos(2*x)