Optimal. Leaf size=15 \[ \frac {\tanh ^{-1}\left (\frac {b \sin (x)}{a}\right )}{a b} \]
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Rubi [A] time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, 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.
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Rule 208
Rule 3190
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*}
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Mathematica [A] time = 0.01, size = 15, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {b \sin (x)}{a}\right )}{a b} \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos (x)}{a^2-b^2 \sin ^2(x)} \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 0.87, size = 26, normalized size = 1.73 \[ \frac {\log \left (b \sin \relax (x) + a\right ) - \log \left (-b \sin \relax (x) + a\right )}{2 \, a b} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.09, size = 35, normalized size = 2.33 \[ \frac {\log \left ({\left | b \sin \relax (x) + a \right |}\right )}{2 \, a b} - \frac {\log \left ({\left | b \sin \relax (x) - a \right |}\right )}{2 \, a b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 33, normalized size = 2.20
method | result | size |
derivativedivides | \(-\frac {\ln \left (-b \sin \relax (x )+a \right )}{2 a b}+\frac {\ln \left (b \sin \relax (x )+a \right )}{2 a b}\) | \(33\) |
default | \(-\frac {\ln \left (-b \sin \relax (x )+a \right )}{2 a b}+\frac {\ln \left (b \sin \relax (x )+a \right )}{2 a b}\) | \(33\) |
norman | \(-\frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 b \tan \left (\frac {x}{2}\right )+a \right )}{2 a b}+\frac {\ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}{2 a b}\) | \(54\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{b}-1\right )}{2 a b}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{b}-1\right )}{2 a b}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 33, normalized size = 2.20 \[ \frac {\log \left (b \sin \relax (x) + a\right )}{2 \, a b} - \frac {\log \left (b \sin \relax (x) - a\right )}{2 \, a b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.18, size = 15, normalized size = 1.00 \[ \frac {\mathrm {atanh}\left (\frac {b\,\sin \relax (x)}{a}\right )}{a\,b} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.73, size = 44, normalized size = 2.93 \[ \begin {cases} \frac {\tilde {\infty }}{\sin {\relax (x )}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {1}{b^{2} \sin {\relax (x )}} & \text {for}\: a = 0 \\\frac {\sin {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \\- \frac {\log {\left (- \frac {a}{b} + \sin {\relax (x )} \right )}}{2 a b} + \frac {\log {\left (\frac {a}{b} + \sin {\relax (x )} \right )}}{2 a b} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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