Optimal. Leaf size=203 \[ -\frac{(103 A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(63 A+13 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(5 A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}} \]
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Rubi [A] time = 0.591436, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2978, 12, 2782, 205} \[ -\frac{(103 A+5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(63 A+13 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(5 A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 2978
Rule 12
Rule 2782
Rule 205
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{7/2}} \, dx &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{\int \frac{\frac{1}{2} a (11 A+B)-2 a (A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(5 A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{4} a^2 (73 A+11 B)-\frac{3}{2} a^2 (5 A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(5 A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(103 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{3 a^3 (63 A+13 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(5 A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(103 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(63 A+13 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(5 A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(103 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{(63 A+13 B) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{64 a^2 d}\\ &=\frac{(63 A+13 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(5 A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(103 A+5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 2.01223, size = 216, normalized size = 1.06 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{1}{8} \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) ((532 A-4 B) \cos (c+d x)+(103 A+5 B) \cos (2 (c+d x))+493 A-73 B)+\frac{3 i (63 A+13 B) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{24 d (a (\cos (c+d x)+1))^{7/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.606, size = 549, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.686, size = 716, normalized size = 3.53 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (63 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 63 \, A + 13 \, B\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (103 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (133 \, A - B\right )} \cos \left (d x + c\right ) + 195 \, A - 39 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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