Optimal. Leaf size=199 \[ \frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(115 A+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{5 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.698175, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3042, 2978, 2984, 2985, 2649, 206, 2773} \[ \frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(115 A+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{5 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3042
Rule 2978
Rule 2984
Rule 2985
Rule 2649
Rule 206
Rule 2773
Rubi steps
\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\left (a (5 A+C)-\frac{1}{2} a (5 A-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (\frac{1}{2} a^2 (35 A+3 C)-\frac{3}{4} a^2 (15 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-20 a^3 A+\frac{1}{4} a^3 (35 A+3 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3}+\frac{(115 A+3 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^2 d}-\frac{(115 A+3 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 )}{16 a^2 d}\\ &=-\frac{5 A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac{(115 A+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 4.44601, size = 185, normalized size = 0.93 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left ((230 A+6 C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\frac{1}{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sec ^3\left (\frac{1}{2} (c+d x)\right ) (2 (55 A+7 C) \cos (c+d x)+(35 A+3 C) \cos (2 (c+d x))+67 A+3 C)-160 \sqrt{2} A \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d (a (\cos (c+d x)+1))^{5/2} (2 A+C \cos (2 (c+d x))+C)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.127, size = 815, normalized size = 4.1 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.00874, size = 1002, normalized size = 5.04 \begin{align*} \frac{\sqrt{2}{\left ({\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (115 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 80 \,{\left (A \cos \left (d x + c\right )^{4} + 3 \, A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left ({\left (35 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (55 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\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 [B] time = 3.8807, size = 518, normalized size = 2.6 \begin{align*} \frac{2 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} + C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} + \frac{\sqrt{2}{\left (21 \, A a^{5} + 5 \, C a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{\sqrt{2}{\left (115 \, A \sqrt{a} + 3 \, C \sqrt{a}\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 )}^{2}\right )}{a^{3}} - \frac{160 \, A \log \left ({\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 )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{a^{\frac{5}{2}}} + \frac{160 \, A \log \left ({\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 )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{a^{\frac{5}{2}}} + \frac{128 \, \sqrt{2}{\left (3 \,{\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 )}^{2} A \sqrt{a} - A a^{\frac{3}{2}}\right )}}{{\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 )}^{4} - 6 \,{\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 )}^{2} a + a^{2}\right )} a^{2}}}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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