Optimal. Leaf size=206 \[ \frac {a \left (a^2+4 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.28, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2790, 2754, 12, 2659, 205} \[ \frac {a \left (a^2+4 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\left (-10 a^2 b^2+a^4-6 b^4\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 2659
Rule 2754
Rule 2790
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx &=-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\int \frac {3 a b+\left (a^2-3 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\int \frac {-2 b \left (2 a^2+3 b^2\right )-a \left (a^2-6 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2}\\ &=-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\int \frac {3 a b \left (a^2+4 b^2\right )}{a+b \cos (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3}\\ &=-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+4 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+4 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 )^3 d}\\ &=\frac {a \left (a^2+4 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)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.20, size = 162, normalized size = 0.79 \[ \frac {\frac {6 a \left (a^2+4 b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}+\frac {\sin (c+d x) \left (-13 a^4 b-2 a^2 b^3+b \left (a^4-10 a^2 b^2-6 b^4\right ) \cos ^2(c+d x)+3 a \left (a^4-9 a^2 b^2-2 b^4\right ) \cos (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.92, size = 893, normalized size = 4.33 \[ \left [\frac {3 \, {\left (a^{6} + 4 \, a^{4} b^{2} + {\left (a^{3} b^{3} + 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b + 4 \, a^{3} 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 ) - 2 \, {\left (13 \, a^{6} b - 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - {\left (a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{7} - 10 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}, \frac {3 \, {\left (a^{6} + 4 \, a^{4} b^{2} + {\left (a^{3} b^{3} + 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{5} b + 4 \, a^{3} 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 ) - {\left (13 \, a^{6} b - 11 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - {\left (a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{7} - 10 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right ) + {\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.12, size = 427, normalized size = 2.07 \[ -\frac {\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 28 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 930, normalized size = 4.51 \[ \text {result too large to display} \]
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 = 4.15, size = 381, normalized size = 1.85 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (a^3+4\,a\,b^2\right )}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^3+6\,a^2\,b+2\,a\,b^2+2\,b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b+3\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3-6\,a^2\,b+2\,a\,b^2-2\,b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]
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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