3.17 \(\int \frac{\sin (2 x)}{a^2-b^2 \cos ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac{\log \left (a^2-b^2 \cos ^2(x)\right )}{b^2} \]

[Out]

Log[a^2 - b^2*Cos[x]^2]/b^2

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Rubi [A]  time = 0.0423405, antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {12, 260} \[ \frac{\log \left (a^2+b^2 \sin ^2(x)-b^2\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]/(a^2 - b^2*Cos[x]^2),x]

[Out]

Log[a^2 - b^2 + b^2*Sin[x]^2]/b^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.0131116, size = 22, normalized size = 1.22 \[ \frac{\log \left (a^2+b^2 \sin ^2(x)-b^2\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]/(a^2 - b^2*Cos[x]^2),x]

[Out]

Log[a^2 - b^2 + b^2*Sin[x]^2]/b^2

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Maple [A]  time = 0.017, size = 19, normalized size = 1.1 \begin{align*}{\frac{\ln \left ({a}^{2}-{b}^{2} \left ( \cos \left ( x \right ) \right ) ^{2} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)/(a^2-b^2*cos(x)^2),x)

[Out]

ln(a^2-b^2*cos(x)^2)/b^2

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Maxima [A]  time = 0.925176, size = 26, normalized size = 1.44 \begin{align*} \frac{\log \left (b^{2} \cos \left (x\right )^{2} - a^{2}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-b^2*cos(x)^2),x, algorithm="maxima")

[Out]

log(b^2*cos(x)^2 - a^2)/b^2

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Fricas [A]  time = 2.6345, size = 39, normalized size = 2.17 \begin{align*} \frac{\log \left (b^{2} \cos \left (x\right )^{2} - a^{2}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-b^2*cos(x)^2),x, algorithm="fricas")

[Out]

log(b^2*cos(x)^2 - a^2)/b^2

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Sympy [A]  time = 2.94719, size = 32, normalized size = 1.78 \begin{align*} 2 \left (\begin{cases} - \frac{\cos ^{2}{\left (x \right )}}{2 a^{2}} & \text{for}\: b^{2} = 0 \\\frac{\log{\left (a^{2} - b^{2} \cos ^{2}{\left (x \right )} \right )}}{2 b^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a**2-b**2*cos(x)**2),x)

[Out]

2*Piecewise((-cos(x)**2/(2*a**2), Eq(b**2, 0)), (log(a**2 - b**2*cos(x)**2)/(2*b**2), True))

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Giac [B]  time = 1.09442, size = 165, normalized size = 9.17 \begin{align*} \frac{{\left (a + b\right )} \log \left ({\left | a - b - \frac{a{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - \frac{b{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} \right |}\right )}{a b^{2} + b^{3}} + \frac{{\left (a - b\right )} \log \left ({\left | -a - b + \frac{a{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - \frac{b{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} \right |}\right )}{a b^{2} - b^{3}} - \frac{2 \, \log \left (-\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} + 1\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)/(a^2-b^2*cos(x)^2),x, algorithm="giac")

[Out]

(a + b)*log(abs(a - b - a*(cos(x) - 1)/(cos(x) + 1) - b*(cos(x) - 1)/(cos(x) + 1)))/(a*b^2 + b^3) + (a - b)*lo
g(abs(-a - b + a*(cos(x) - 1)/(cos(x) + 1) - b*(cos(x) - 1)/(cos(x) + 1)))/(a*b^2 - b^3) - 2*log(-(cos(x) - 1)
/(cos(x) + 1) + 1)/b^2