3.3.78 \(\int \frac {\cos (x)}{1+\sin ^2(x)} \, dx\) [278]

Optimal. Leaf size=3 \[ \tan ^{-1}(\sin (x)) \]

[Out]

arctan(sin(x))

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Rubi [A]
time = 0.01, antiderivative size = 3, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3269, 209} \begin {gather*} \tan ^{-1}(\sin (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(1 + Sin[x]^2),x]

[Out]

ArcTan[Sin[x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 3, normalized size = 1.00 \begin {gather*} \tan ^{-1}(\sin (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/(1 + Sin[x]^2),x]

[Out]

ArcTan[Sin[x]]

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Mathics [A]
time = 1.60, size = 3, normalized size = 1.00 \begin {gather*} \text {ArcTan}\left [\text {Sin}\left [x\right ]\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Cos[x]/(1 + Sin[x]^2),x]')

[Out]

ArcTan[Sin[x]]

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Maple [A]
time = 0.04, size = 4, normalized size = 1.33

method	result	size
derivativedivides	\(\arctan \left (\sin \left (x \right )\right )\)	\(4\)
default	\(\arctan \left (\sin \left (x \right )\right )\)	\(4\)
risch	\(\frac {i \ln \left ({\mathrm e}^{2 i x}-2 \,{\mathrm e}^{i x}-1\right )}{2}-\frac {i \ln \left ({\mathrm e}^{2 i x}+2 \,{\mathrm e}^{i x}-1\right )}{2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(1+sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

arctan(sin(x))

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Maxima [A]
time = 0.35, size = 3, normalized size = 1.00 \begin {gather*} \arctan \left (\sin \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sin(x))

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Fricas [A]
time = 0.35, size = 3, normalized size = 1.00 \begin {gather*} \arctan \left (\sin \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sin(x))

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Sympy [A]
time = 0.09, size = 3, normalized size = 1.00 \begin {gather*} \operatorname {atan}{\left (\sin {\left (x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(sin(x))

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Giac [A]
time = 0.00, size = 3, normalized size = 1.00 \begin {gather*} \arctan \left (\sin x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(1+sin(x)^2),x)

[Out]

arctan(sin(x))

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Mupad [B]
time = 0.07, size = 3, normalized size = 1.00 \begin {gather*} \mathrm {atan}\left (\sin \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(sin(x)^2 + 1),x)

[Out]

atan(sin(x))

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