Optimal. Leaf size=193 \[ -\frac {2 a^5 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d \sqrt {a-b} \sqrt {a+b}}-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}+\frac {x \left (8 a^4+4 a^2 b^2+3 b^4\right )}{8 b^5}-\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]
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Rubi [A] time = 0.54, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2793, 3049, 3023, 2735, 2659, 205} \[ -\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac {2 a^5 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}+\frac {x \left (4 a^2 b^2+8 a^4+3 b^4\right )}{8 b^5}-\frac {a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 2659
Rule 2735
Rule 2793
Rule 3023
Rule 3049
Rubi steps
\begin {align*} \int \frac {\cos ^5(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a+3 b \cos (c+d x)-4 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b}\\ &=-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos (c+d x) \left (-8 a^2+a b \cos (c+d x)+3 \left (4 a^2+3 b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2}\\ &=\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {3 a \left (4 a^2+3 b^2\right )-b \left (4 a^2-9 b^2\right ) \cos (c+d x)-8 a \left (3 a^2+2 b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3}\\ &=-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {3 a b \left (4 a^2+3 b^2\right )+3 \left (8 a^4+4 a^2 b^2+3 b^4\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4}\\ &=\frac {\left (8 a^4+4 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac {a^5 \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^5}\\ &=\frac {\left (8 a^4+4 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac {\left (2 a^5\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 )}{b^5 d}\\ &=\frac {\left (8 a^4+4 a^2 b^2+3 b^4\right ) x}{8 b^5}-\frac {2 a^5 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^5 \sqrt {a+b} d}-\frac {a \left (3 a^2+2 b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d}\\ \end {align*}
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Mathematica [A] time = 0.64, size = 153, normalized size = 0.79 \[ \frac {-24 a b \left (4 a^2+3 b^2\right ) \sin (c+d x)+24 b^2 \left (a^2+b^2\right ) \sin (2 (c+d x))+\frac {192 a^5 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+12 \left (8 a^4+4 a^2 b^2+3 b^4\right ) (c+d x)-8 a b^3 \sin (3 (c+d x))+3 b^4 \sin (4 (c+d x))}{96 b^5 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.72, size = 479, normalized size = 2.48 \[ \left [-\frac {12 \, \sqrt {-a^{2} + b^{2}} a^{5} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 3 \, {\left (8 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} d x + {\left (24 \, a^{5} b - 8 \, a^{3} b^{3} - 16 \, a b^{5} - 6 \, {\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}, -\frac {24 \, \sqrt {a^{2} - b^{2}} a^{5} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \, {\left (8 \, a^{6} - 4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} d x + {\left (24 \, a^{5} b - 8 \, a^{3} b^{3} - 16 \, a b^{5} - 6 \, {\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, a^{4} b^{2} - a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.62, size = 393, normalized size = 2.04 \[ \frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{5}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {3 \, {\left (8 \, a^{4} + 4 \, a^{2} b^{2} + 3 \, b^{4}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {2 \, {\left (24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 672, normalized size = 3.48 \[ -\frac {2 a^{5} \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 )}{d \,b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {5 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {6 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{d \,b^{5}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b} \]
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 = 1.72, size = 474, normalized size = 2.46 \[ \frac {\sin \left (2\,c+2\,d\,x\right )}{4\,b\,d}+\frac {\sin \left (4\,c+4\,d\,x\right )}{32\,b\,d}+\frac {3\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{4\,b\,d}-\frac {a\,\sin \left (3\,c+3\,d\,x\right )}{12\,b^2\,d}-\frac {a^3\,\sin \left (c+d\,x\right )}{b^4\,d}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^3\,d}+\frac {a^2\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{b^3\,d}+\frac {2\,a^4\,\mathrm {atan}\left (\frac {40\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^6+15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^8+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^{10}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (40\,a^4\,b^5+15\,a^2\,b^7+9\,b^9\right )}\right )}{b^5\,d}-\frac {3\,a\,\sin \left (c+d\,x\right )}{4\,b^2\,d}-\frac {a^5\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^5\,d\,\sqrt {b^2-a^2}} \]
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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