Optimal. Leaf size=170 \[ -\frac{4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac{4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac{(10 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(10 B-7 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{x (10 B-7 C)}{2 a^2}-\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.401685, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4072, 4020, 3787, 2633, 2635, 8} \[ -\frac{4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac{4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac{(10 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(10 B-7 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{x (10 B-7 C)}{2 a^2}-\frac{(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4020
Rule 3787
Rule 2633
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos ^3(c+d x) (3 a (2 B-C)-4 a (B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \cos ^3(c+d x) \left (12 a^2 (3 B-2 C)-3 a^2 (10 B-7 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(10 B-7 C) \int \cos ^2(c+d x) \, dx}{a^2}+\frac{(4 (3 B-2 C)) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac{(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(10 B-7 C) \int 1 \, dx}{2 a^2}-\frac{(4 (3 B-2 C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac{(10 B-7 C) x}{2 a^2}+\frac{4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac{(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end{align*}
Mathematica [B] time = 0.708838, size = 369, normalized size = 2.17 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-36 d x (10 B-7 C) \cos \left (c+\frac{d x}{2}\right )-156 B \sin \left (c+\frac{d x}{2}\right )+342 B \sin \left (c+\frac{3 d x}{2}\right )+118 B \sin \left (2 c+\frac{3 d x}{2}\right )+30 B \sin \left (2 c+\frac{5 d x}{2}\right )+30 B \sin \left (3 c+\frac{5 d x}{2}\right )-3 B \sin \left (3 c+\frac{7 d x}{2}\right )-3 B \sin \left (4 c+\frac{7 d x}{2}\right )+B \sin \left (4 c+\frac{9 d x}{2}\right )+B \sin \left (5 c+\frac{9 d x}{2}\right )-120 B d x \cos \left (c+\frac{3 d x}{2}\right )-120 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-36 d x (10 B-7 C) \cos \left (\frac{d x}{2}\right )+516 B \sin \left (\frac{d x}{2}\right )+147 C \sin \left (c+\frac{d x}{2}\right )-239 C \sin \left (c+\frac{3 d x}{2}\right )-63 C \sin \left (2 c+\frac{3 d x}{2}\right )-15 C \sin \left (2 c+\frac{5 d x}{2}\right )-15 C \sin \left (3 c+\frac{5 d x}{2}\right )+3 C \sin \left (3 c+\frac{7 d x}{2}\right )+3 C \sin \left (4 c+\frac{7 d x}{2}\right )+84 C d x \cos \left (c+\frac{3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.089, size = 322, normalized size = 1.9 \begin{align*} -{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{9\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+10\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}B}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-5\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{40\,B}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-8\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-3\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-10\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.45624, size = 502, normalized size = 2.95 \begin{align*} \frac{B{\left (\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{60 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - C{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.502698, size = 389, normalized size = 2.29 \begin{align*} -\frac{3 \,{\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (10 \, B - 7 \, C\right )} d x -{\left (2 \, B \cos \left (d x + c\right )^{4} -{\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, B - C\right )} \cos \left (d x + c\right )^{2} +{\left (66 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 48 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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 [A] time = 1.14133, size = 259, normalized size = 1.52 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}{\left (10 \, B - 7 \, C\right )}}{a^{2}} - \frac{2 \,{\left (30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac{B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 27 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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