3.18 \(\int \frac{1}{(a \cos (x)+b \sin (x))^2} \, dx\)

Optimal. Leaf size=17 \[ \frac{\sin (x)}{a (a \cos (x)+b \sin (x))} \]

[Out]

Sin[x]/(a*(a*Cos[x] + b*Sin[x]))

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Rubi [A]  time = 0.0127537, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3075} \[ \frac{\sin (x)}{a (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[x] + b*Sin[x])^(-2),x]

[Out]

Sin[x]/(a*(a*Cos[x] + b*Sin[x]))

Rule 3075

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-2), x_Symbol] :> Simp[Sin[c + d*x]/(a*d*
(a*Cos[c + d*x] + b*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0205677, size = 17, normalized size = 1. \[ \frac{\sin (x)}{a (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[x]/(a*(a*Cos[x] + b*Sin[x]))

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Maple [A]  time = 0.08, size = 14, normalized size = 0.8 \begin{align*} -{\frac{1}{b \left ( a+b\tan \left ( x \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(x)+b*sin(x))^2,x)

[Out]

-1/b/(a+b*tan(x))

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Maxima [A]  time = 1.12893, size = 19, normalized size = 1.12 \begin{align*} -\frac{1}{b^{2} \tan \left (x\right ) + a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b^2*tan(x) + a*b)

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Fricas [B]  time = 0.46397, size = 95, normalized size = 5.59 \begin{align*} -\frac{b \cos \left (x\right ) - a \sin \left (x\right )}{{\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) +{\left (a^{2} b + b^{3}\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*cos(x) - a*sin(x))/((a^3 + a*b^2)*cos(x) + (a^2*b + b^3)*sin(x))

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Sympy [A]  time = 99.0802, size = 75, normalized size = 4.41 \begin{align*} \begin{cases} - \frac{\tan ^{2}{\left (\frac{x}{2} \right )}}{a b \tan ^{2}{\left (\frac{x}{2} \right )} - a b - 2 b^{2} \tan{\left (\frac{x}{2} \right )}} + \frac{1}{a b \tan ^{2}{\left (\frac{x}{2} \right )} - a b - 2 b^{2} \tan{\left (\frac{x}{2} \right )}} & \text{for}\: b \neq 0 \\- \frac{2 \tan{\left (\frac{x}{2} \right )}}{a^{2} \left (\tan ^{2}{\left (\frac{x}{2} \right )} - 1\right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-tan(x/2)**2/(a*b*tan(x/2)**2 - a*b - 2*b**2*tan(x/2)) + 1/(a*b*tan(x/2)**2 - a*b - 2*b**2*tan(x/2)
), Ne(b, 0)), (-2*tan(x/2)/(a**2*(tan(x/2)**2 - 1)), True))

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Giac [A]  time = 1.1023, size = 18, normalized size = 1.06 \begin{align*} -\frac{1}{{\left (b \tan \left (x\right ) + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((b*tan(x) + a)*b)