3.82 \(\int \frac {1}{1+\tan ^2(x)} \, dx\)

Optimal. Leaf size=14 \[ \frac {x}{2}+\frac {1}{2} \sin (x) \cos (x) \]

[Out]

1/2*x+1/2*cos(x)*sin(x)

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3657, 2635, 8} \[ \frac {x}{2}+\frac {1}{2} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[x]^2)^(-1),x]

[Out]

x/2 + (Cos[x]*Sin[x])/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3657

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \[ \frac {x}{2}+\frac {1}{4} \sin (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[x]^2)^(-1),x]

[Out]

x/2 + Sin[2*x]/4

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fricas [A]  time = 0.44, size = 20, normalized size = 1.43 \[ \frac {x \tan \relax (x)^{2} + x + \tan \relax (x)}{2 \, {\left (\tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x*tan(x)^2 + x + tan(x))/(tan(x)^2 + 1)

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giac [A]  time = 0.92, size = 16, normalized size = 1.14 \[ \frac {1}{2} \, x + \frac {\tan \relax (x)}{2 \, {\left (\tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x + 1/2*tan(x)/(tan(x)^2 + 1)

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maple [A]  time = 0.02, size = 17, normalized size = 1.21 \[ \frac {x}{2}+\frac {\tan \relax (x )}{2 \left (\tan ^{2}\relax (x )\right )+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x)^2+1),x)

[Out]

1/2/(tan(x)^2+1)*tan(x)+1/2*x

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maxima [A]  time = 1.36, size = 16, normalized size = 1.14 \[ \frac {1}{2} \, x + \frac {\tan \relax (x)}{2 \, {\left (\tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x + 1/2*tan(x)/(tan(x)^2 + 1)

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mupad [B]  time = 0.18, size = 10, normalized size = 0.71 \[ \frac {x}{2}+\frac {\sin \left (2\,x\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x)^2 + 1),x)

[Out]

x/2 + sin(2*x)/4

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sympy [B]  time = 0.39, size = 36, normalized size = 2.57 \[ \frac {x \tan ^{2}{\relax (x )}}{2 \tan ^{2}{\relax (x )} + 2} + \frac {x}{2 \tan ^{2}{\relax (x )} + 2} + \frac {\tan {\relax (x )}}{2 \tan ^{2}{\relax (x )} + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tan(x)**2),x)

[Out]

x*tan(x)**2/(2*tan(x)**2 + 2) + x/(2*tan(x)**2 + 2) + tan(x)/(2*tan(x)**2 + 2)

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