Optimal. Leaf size=93 \[ \frac {2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.12, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3016, 2754, 12, 2659, 205} \[ \frac {2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac {2 a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 2659
Rule 2754
Rule 3016
Rubi steps
\begin {align*} \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\int \frac {-a+b \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx\\ &=-\frac {2 a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\int \frac {a^2+b^2}{a+b \cos (c+d x)} \, dx}{-a^2+b^2}\\ &=-\frac {2 a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (a^2+b^2\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac {2 a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (2 \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {2 \left (a^2+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {2 a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 91, normalized size = 0.98 \[ \frac {2 \left (\frac {\left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac {a b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}\right )}{d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 346, normalized size = 3.72 \[ \left [\frac {{\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 4 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, \frac {{\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.62, size = 143, normalized size = 1.54 \[ -\frac {2 \, {\left (\frac {2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} - b^{2}\right )}} - \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} + b^{2}\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}}\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 177, normalized size = 1.90 \[ -\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a^{2}}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{2}}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.62, size = 106, normalized size = 1.14 \[ \frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )\,\left (a^2+b^2\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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