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

Optimal. Leaf size=19 \[ -\frac{\log \left (a^2-b^2 \sin ^2(x)\right )}{b^2} \]

[Out]

-(Log[a^2 - b^2*Sin[x]^2]/b^2)

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Rubi [A]  time = 0.066265, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1 \[ -\frac{\log \left (a^2-b^2 \sin ^2(x)\right )}{b^2} \]

Antiderivative was successfully verified.

[In]  Int[Sin[2*x]/(a^2 - b^2*Sin[x]^2),x]

[Out]

-(Log[a^2 - b^2*Sin[x]^2]/b^2)

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Rubi in Sympy [A]  time = 4.00925, size = 17, normalized size = 0.89 \[ - \frac{\log{\left (a^{2} - b^{2} \sin ^{2}{\left (x \right )} \right )}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(sin(2*x)/(a**2-b**2*sin(x)**2),x)

[Out]

-log(a**2 - b**2*sin(x)**2)/b**2

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Mathematica [A]  time = 0.0216296, size = 25, normalized size = 1.32 \[ -\frac{\log \left (2 a^2+b^2 \cos (2 x)-b^2\right )}{b^2} \]

Antiderivative was successfully verified.

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

[Out]

-(Log[2*a^2 - b^2 + b^2*Cos[2*x]]/b^2)

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Maple [A]  time = 0.032, size = 20, normalized size = 1.1 \[ -{\frac{\ln \left ({a}^{2}-{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }{{b}^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(sin(2*x)/(a^2-b^2*sin(x)^2),x)

[Out]

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

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Maxima [A]  time = 1.42873, size = 27, normalized size = 1.42 \[ -\frac{\log \left (b^{2} \sin \left (x\right )^{2} - a^{2}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-log(b^2*sin(x)^2 - a^2)/b^2

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Fricas [A]  time = 0.247466, size = 31, normalized size = 1.63 \[ -\frac{\log \left (b^{2} \cos \left (x\right )^{2} + a^{2} - b^{2}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-log(b^2*cos(x)^2 + a^2 - b^2)/b^2

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Sympy [A]  time = 3.74202, size = 34, normalized size = 1.79 \[ 2 \left (\begin{cases} - \frac{\cos ^{2}{\left (x \right )}}{2 a^{2}} & \text{for}\: b^{2} = 0 \\- \frac{\log{\left (a^{2} - b^{2} \sin ^{2}{\left (x \right )} \right )}}{2 b^{2}} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sin(2*x)/(a**2-b**2*sin(x)**2),x)

[Out]

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

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GIAC/XCAS [A]  time = 0.20716, size = 104, normalized size = 5.47 \[ -\frac{{\rm ln}\left (a^{2} - \frac{2 \, a^{2}{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} + \frac{4 \, b^{2}{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} + \frac{a^{2}{\left (\cos \left (x\right ) - 1\right )}^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{b^{2}} + \frac{2 \,{\rm ln}\left (-\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} + 1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-ln(a^2 - 2*a^2*(cos(x) - 1)/(cos(x) + 1) + 4*b^2*(cos(x) - 1)/(cos(x) + 1) + a^
2*(cos(x) - 1)^2/(cos(x) + 1)^2)/b^2 + 2*ln(-(cos(x) - 1)/(cos(x) + 1) + 1)/b^2