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

Optimal. Leaf size=15 \[ \frac{\tanh ^{-1}\left (\frac{b \sin (x)}{a}\right )}{a b} \]

[Out]

ArcTanh[(b*Sin[x])/a]/(a*b)

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Rubi [A]  time = 0.027777, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3190, 208} \[ \frac{\tanh ^{-1}\left (\frac{b \sin (x)}{a}\right )}{a b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(a^2 - b^2*Sin[x]^2),x]

[Out]

ArcTanh[(b*Sin[x])/a]/(a*b)

Rule 3190

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]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\cos (x)}{a^2-b^2 \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{a^2-b^2 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{\tanh ^{-1}\left (\frac{b \sin (x)}{a}\right )}{a b}\\ \end{align*}

Mathematica [A]  time = 0.0092904, size = 15, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{b \sin (x)}{a}\right )}{a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(b*Sin[x])/a]/(a*b)

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Maple [B]  time = 0.011, size = 34, normalized size = 2.3 \begin{align*} -{\frac{\ln \left ( b\sin \left ( x \right ) -a \right ) }{2\,ab}}+{\frac{\ln \left ( b\sin \left ( x \right ) +a \right ) }{2\,ab}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/b*ln(b*sin(x)-a)+1/2/a/b*ln(b*sin(x)+a)

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Maxima [B]  time = 0.929799, size = 45, normalized size = 3. \begin{align*} \frac{\log \left (b \sin \left (x\right ) + a\right )}{2 \, a b} - \frac{\log \left (b \sin \left (x\right ) - a\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*sin(x) + a)/(a*b) - 1/2*log(b*sin(x) - a)/(a*b)

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Fricas [A]  time = 2.1602, size = 70, normalized size = 4.67 \begin{align*} \frac{\log \left (b \sin \left (x\right ) + a\right ) - \log \left (-b \sin \left (x\right ) + a\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(b*sin(x) + a) - log(-b*sin(x) + a))/(a*b)

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Sympy [A]  time = 0.732941, size = 44, normalized size = 2.93 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{\sin{\left (x \right )}} & \text{for}\: a = 0 \wedge b = 0 \\\frac{\sin{\left (x \right )}}{a^{2}} & \text{for}\: b = 0 \\\frac{1}{b^{2} \sin{\left (x \right )}} & \text{for}\: a = 0 \\- \frac{\log{\left (- \frac{a}{b} + \sin{\left (x \right )} \right )}}{2 a b} + \frac{\log{\left (\frac{a}{b} + \sin{\left (x \right )} \right )}}{2 a b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/sin(x), Eq(a, 0) & Eq(b, 0)), (sin(x)/a**2, Eq(b, 0)), (1/(b**2*sin(x)), Eq(a, 0)), (-log(-a/b
+ sin(x))/(2*a*b) + log(a/b + sin(x))/(2*a*b), True))

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Giac [B]  time = 1.04545, size = 47, normalized size = 3.13 \begin{align*} \frac{\log \left ({\left | b \sin \left (x\right ) + a \right |}\right )}{2 \, a b} - \frac{\log \left ({\left | b \sin \left (x\right ) - a \right |}\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(b*sin(x) + a))/(a*b) - 1/2*log(abs(b*sin(x) - a))/(a*b)