3.112 \(\int \cos ^2(x) (1-\tan ^2(x)) \, dx\)

Optimal. Leaf size=5 \[ \sin (x) \cos (x) \]

[Out]

cos(x)*sin(x)

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Rubi [A]  time = 0.02, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3675, 383} \[ \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2*(1 - Tan[x]^2),x]

[Out]

Cos[x]*Sin[x]

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*x*(a + b*x^n)^(p + 1))/a, x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a*d - b*c*(n*(p + 1) + 1), 0]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 8, normalized size = 1.60 \[ \frac {1}{2} \sin (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2*(1 - Tan[x]^2),x]

[Out]

Sin[2*x]/2

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fricas [A]  time = 0.43, size = 5, normalized size = 1.00 \[ \cos \relax (x) \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tan(x)^2)/sec(x)^2,x, algorithm="fricas")

[Out]

cos(x)*sin(x)

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giac [A]  time = 1.03, size = 9, normalized size = 1.80 \[ \frac {1}{\frac {1}{\tan \relax (x)} + \tan \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tan(x)^2)/sec(x)^2,x, algorithm="giac")

[Out]

1/(1/tan(x) + tan(x))

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maple [A]  time = 0.03, size = 6, normalized size = 1.20 \[ \cos \relax (x ) \sin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-tan(x)^2)/sec(x)^2,x)

[Out]

cos(x)*sin(x)

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maxima [B]  time = 0.42, size = 11, normalized size = 2.20 \[ \frac {\tan \relax (x)}{\tan \relax (x)^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tan(x)^2)/sec(x)^2,x, algorithm="maxima")

[Out]

tan(x)/(tan(x)^2 + 1)

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mupad [B]  time = 0.17, size = 6, normalized size = 1.20 \[ \frac {\sin \left (2\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-cos(x)^2*(tan(x)^2 - 1),x)

[Out]

sin(2*x)/2

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sympy [A]  time = 0.44, size = 7, normalized size = 1.40 \[ \frac {\tan {\relax (x )}}{\sec ^{2}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tan(x)**2)/sec(x)**2,x)

[Out]

tan(x)/sec(x)**2

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