Optimal. Leaf size=140 \[ \frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 a d}+\frac {28 \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d} \]
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Rubi [A] time = 0.24, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2778, 2968, 3023, 2751, 2649, 206} \[ \frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 a d}+\frac {28 \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 2649
Rule 2751
Rule 2778
Rule 2968
Rule 3023
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {\int \frac {\cos (c+d x) (-4 a+a \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {\int \frac {-4 a \cos (c+d x)+a \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}-\frac {2 \int \frac {\frac {a^2}{2}-7 a^2 \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac {28 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}-\int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {28 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac {2 \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} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {28 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 118, normalized size = 0.84 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (60 \sin \left (\frac {1}{2} (c+d x)\right )-5 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )+30 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )-30 \log \left (\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )\right )\right )}{15 d \sqrt {a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 143, normalized size = 1.02 \[ \frac {4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.29, size = 116, normalized size = 0.83 \[ \frac {\sqrt {2} {\left (\frac {15 \, \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (17 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {5}{2}}}\right )}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.39, size = 183, normalized size = 1.31 \[ -\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -30 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right )}{15 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]
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 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]
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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