Optimal. Leaf size=235 \[ \frac {2 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {a \left (4 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}-\frac {a x \left (8 a^4-12 a^2 b^2+3 b^4\right )}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}-\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
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Rubi [A] time = 0.72, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2895, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac {\left (-20 a^2 b^2+15 a^4+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {2 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}-\frac {\left (5 a^2-6 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{15 b^3 d}+\frac {a \left (4 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^4 d}-\frac {a x \left (-12 a^2 b^2+8 a^4+3 b^4\right )}{8 b^6}+\frac {a \sin ^3(c+d x) \cos (c+d x)}{4 b^2 d}-\frac {\sin ^4(c+d x) \cos (c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2735
Rule 2895
Rule 3023
Rule 3049
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\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 (5 \left (3 a^2-4 b^2\right )-a b \sin (c+d x)-4 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 b^2}\\ &=-\frac {\left (5 a^2-6 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^2-6 b^2\right )+b \left (5 a^2-12 b^2\right ) \sin (c+d x)+15 a \left (4 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 b^3}\\ &=\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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^2-5 b^2\right )-a b \left (20 a^2-21 b^2\right ) \sin (c+d x)-8 \left (15 a^4-20 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^4}\\ &=-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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^2-5 b^2\right )+15 a \left (8 a^4-12 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{120 b^5}\\ &=-\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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^2 \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6}\\ &=-\frac {a \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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^2 \left (a^2-b^2\right )^2\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^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}-\frac {\left (15 a^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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^2 \left (a^2-b^2\right )^2\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^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {2 a^2 \left (a^2-b^2\right )^{3/2} \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^4-20 a^2 b^2+3 b^4\right ) \cos (c+d x)}{15 b^5 d}+\frac {a \left (4 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^4 d}-\frac {\left (5 a^2-6 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 = 2.04, size = 186, normalized size = 0.79 \[ \frac {960 a^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+10 \left (4 a^2 b^3-3 b^5\right ) \cos (3 (c+d x))-15 a \left (\left (8 b^4-8 a^2 b^2\right ) \sin (2 (c+d x))+4 \left (8 a^4-12 a^2 b^2+3 b^4\right ) (c+d x)+b^4 \sin (4 (c+d x))\right )-60 b \left (8 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x)-6 b^5 \cos (5 (c+d x))}{480 b^6 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.79, size = 457, normalized size = 1.94 \[ \left [-\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 60 \, {\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 120 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}, -\frac {24 \, b^{5} \cos \left (d x + c\right )^{5} - 40 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} + 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x + 120 \, {\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 120 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) + 15 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, b^{6} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.17, size = 458, normalized size = 1.95 \[ -\frac {\frac {15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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} - 75 \, 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} - 240 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 120 \, b^{4} \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} - 30 \, 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} - 720 \, a^{2} b^{2} \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} - 880 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, 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} + 30 \, 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} - 560 \, a^{2} b^{2} \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 ) + 75 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, a^{4} - 160 \, a^{2} b^{2} + 24 \, 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 = 0.31, size = 941, normalized size = 4.00 \[ \text {result too large to display} \]
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 = 13.37, size = 511, normalized size = 2.17 \[ \frac {5\,a^2\,\cos \left (c+d\,x\right )}{4\,b^3\,d}-\frac {\cos \left (3\,c+3\,d\,x\right )}{16\,b\,d}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80\,b\,d}-\frac {\cos \left (c+d\,x\right )}{8\,b\,d}-\frac {a^4\,\cos \left (c+d\,x\right )}{b^5\,d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^2\,d}-\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{32\,b^2\,d}+\frac {3\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}-\frac {2\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^6\,d}+\frac {a^2\,\cos \left (3\,c+3\,d\,x\right )}{12\,b^3\,d}+\frac {a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,b^4\,d}-\frac {3\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,b^2\,d}+\frac {2\,a^2\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{b^6\,d} \]
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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