3.1.10 \(\int \frac {\sin (2 x)}{1+\cos (2 x)} \, dx\) [10]

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

[Out]

-ln(cos(x))

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(11\) vs. \(2(5)=10\).
time = 0.01, antiderivative size = 11, normalized size of antiderivative = 2.20, 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 = {2746, 31} \begin {gather*} -\frac {1}{2} \log (\cos (2 x)+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/(1 + Cos[2*x]),x]

[Out]

-1/2*Log[1 + Cos[2*x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 5, normalized size = 1.00 \begin {gather*} -\log (\cos (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]/(1 + Cos[2*x]),x]

[Out]

-Log[Cos[x]]

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Maple [A]
time = 0.04, size = 10, normalized size = 2.00

method	result	size
derivativedivides	\(-\frac {\ln \left (\cos \left (2 x \right )+1\right )}{2}\)	\(10\)
default	\(-\frac {\ln \left (\cos \left (2 x \right )+1\right )}{2}\)	\(10\)
norman	\(\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{2}\)	\(10\)
risch	\(i x -\ln \left ({\mathrm e}^{2 i x}+1\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(cos(2*x)+1)

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Maxima [A]
time = 0.31, size = 9, normalized size = 1.80 \begin {gather*} -\frac {1}{2} \, \log \left (\cos \left (2 \, x\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(cos(2*x) + 1)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (5) = 10\).
time = 0.38, size = 11, normalized size = 2.20 \begin {gather*} -\frac {1}{2} \, \log \left (\frac {1}{2} \, \cos \left (2 \, x\right ) + \frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(1/2*cos(2*x) + 1/2)

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Sympy [A]
time = 0.04, size = 10, normalized size = 2.00 \begin {gather*} - \frac {\log {\left (\cos {\left (2 x \right )} + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(cos(2*x) + 1)/2

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Giac [A]
time = 0.42, size = 9, normalized size = 1.80 \begin {gather*} -\frac {1}{2} \, \log \left (\cos \left (2 \, x\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(1+cos(2*x)),x, algorithm="giac")

[Out]

-1/2*log(cos(2*x) + 1)

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Mupad [B]
time = 0.26, size = 7, normalized size = 1.40 \begin {gather*} -\frac {\ln \left ({\cos \left (x\right )}^2\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(cos(2*x) + 1),x)

[Out]

-log(cos(x)^2)/2

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