Optimal. Leaf size=235 \[ -\frac {a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {2 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {a x \left (8 a^4-4 a^2 b^2-b^4\right )}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
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Rubi [A] time = 0.91, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac {\left (-5 a^2 b^2+15 a^4-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {2 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {\left (5 a^2-b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {a \left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}+\frac {a x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^6}-\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}+\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2735
Rule 2889
Rule 3023
Rule 3049
Rule 3050
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac {\sin ^4(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin ^3(c+d x) \left (-4 a+b \sin (c+d x)+5 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b}\\ &=-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin ^2(c+d x) \left (15 a^2-a b \sin (c+d x)-4 \left (5 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2}\\ &=\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {\sin (c+d x) \left (-8 a \left (5 a^2-b^2\right )+b \left (5 a^2+8 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3}\\ &=-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {15 a^2 \left (4 a^2-b^2\right )-a b \left (20 a^2-b^2\right ) \sin (c+d x)-8 \left (15 a^4-5 a^2 b^2-2 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4}\\ &=\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\int \frac {15 a^2 b \left (4 a^2-b^2\right )+15 a \left (8 a^4-4 a^2 b^2-b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\left (a^4 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}-\frac {\left (2 a^4 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}+\frac {\left (4 a^4 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d}\\ &=\frac {a \left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^6}-\frac {2 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {\left (15 a^4-5 a^2 b^2-2 b^4\right ) \cos (c+d x)}{15 b^5 d}-\frac {a \left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}+\frac {\left (5 a^2-b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{15 b^3 d}-\frac {a \cos (c+d x) \sin ^3(c+d x)}{4 b^2 d}+\frac {\cos (c+d x) \sin ^4(c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A] time = 1.77, size = 177, normalized size = 0.75 \[ \frac {-10 \left (4 a^2 b^3+b^5\right ) \cos (3 (c+d x))-960 a^4 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+15 a \left (-8 a^2 b^2 \sin (2 (c+d x))+4 \left (8 a^4-4 a^2 b^2-b^4\right ) (c+d x)+b^4 \sin (4 (c+d x))\right )-60 b \left (-8 a^4+2 a^2 b^2+b^4\right ) \cos (c+d x)+6 b^5 \cos (5 (c+d x))}{480 b^6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 427, normalized size = 1.82 \[ \left [\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 60 \, \sqrt {-a^{2} + b^{2}} a^{4} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, \frac {24 \, b^{5} \cos \left (d x + c\right )^{5} + 120 \, a^{4} b \cos \left (d x + c\right ) + 120 \, \sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 40 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} d x + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.17, size = 467, normalized size = 1.99 \[ \frac {\frac {15 \, {\left (8 \, a^{5} - 4 \, a^{3} b^{2} - a b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{6} - a^{4} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{6}} + \frac {2 \, {\left (60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 15 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 720 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 160 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, a^{4} - 40 \, a^{2} b^{2} - 16 \, b^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} b^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.32, size = 871, normalized size = 3.71 \[ \text {result too large to display} \]
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 = 13.39, size = 376, normalized size = 1.60 \[ \frac {a^4\,\cos \left (c+d\,x\right )}{b^5\,d}-\frac {\frac {\cos \left (c+d\,x\right )}{8}+\frac {\cos \left (3\,c+3\,d\,x\right )}{48}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80}}{b\,d}-\frac {\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32}}{b^2\,d}-\frac {\frac {a^2\,\cos \left (c+d\,x\right )}{4}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{12}}{b^3\,d}-\frac {a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^4\,d}+\frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^6\,d}-\frac {2\,a^4\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}}{b^6\,d} \]
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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