Optimal. Leaf size=164 \[ \frac {2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (5 B-C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 a d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.38, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3045, 2968, 3023, 2751, 2649, 206} \[ \frac {2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (5 B-C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 a d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 2649
Rule 2751
Rule 2968
Rule 3023
Rule 3045
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\cos (c+d x) \left (\frac {1}{2} a (5 A+4 C)+\frac {1}{2} a (5 B-C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\frac {1}{2} a (5 A+4 C) \cos (c+d x)+\frac {1}{2} a (5 B-C) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac {4 \int \frac {\frac {1}{4} a^2 (5 B-C)+\frac {1}{4} a^2 (15 A-10 B+14 C) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac {2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+(-A+B-C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac {(2 (A-B+C)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} (A-B+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 98, normalized size = 0.60 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right ) (30 A+2 (5 B-C) \cos (c+d x)-10 B+3 C \cos (2 (c+d x))+29 C)-15 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{15 d \sqrt {a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 171, normalized size = 1.04 \[ \frac {4 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + {\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 15 \, A - 5 \, B + 13 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.22, size = 189, normalized size = 1.15 \[ \frac {\frac {15 \, {\left (\sqrt {2} A - \sqrt {2} B + \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} + \frac {2 \, {\left (15 \, \sqrt {2} A a^{2} + 15 \, \sqrt {2} C a^{2} + {\left (30 \, \sqrt {2} A a^{2} - 10 \, \sqrt {2} B a^{2} + 20 \, \sqrt {2} C a^{2} + {\left (15 \, \sqrt {2} A a^{2} - 10 \, \sqrt {2} B a^{2} + 17 \, \sqrt {2} C a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.41, size = 318, normalized size = 1.94 \[ \frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C +30 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-15 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +15 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +30 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-15 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a C \right )}{15 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]
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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