Optimal. Leaf size=177 \[ \frac{a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac{a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(5 A+7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac{(8 A+7 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+a^4 A x+\frac{a C \tan (c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]
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Rubi [A] time = 0.294305, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4055, 3917, 3914, 3767, 8, 3770} \[ \frac{a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac{a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(5 A+7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac{(8 A+7 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+a^4 A x+\frac{a C \tan (c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d} \]
Antiderivative was successfully verified.
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Rule 4055
Rule 3917
Rule 3914
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{\int (a+a \sec (c+d x))^4 (5 a A+4 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{\int (a+a \sec (c+d x))^3 \left (20 a^2 A+4 a^2 (5 A+7 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac{\int (a+a \sec (c+d x))^2 \left (60 a^3 A+20 a^3 (8 A+7 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac{(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{\int (a+a \sec (c+d x)) \left (120 a^4 A+60 a^4 (10 A+7 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=a^4 A x+\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac{(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}+\frac{1}{2} \left (a^4 (10 A+7 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (a^4 (12 A+7 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 A x+\frac{a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac{(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}-\frac{\left (a^4 (10 A+7 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^4 A x+\frac{a^4 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 (10 A+7 C) \tan (c+d x)}{2 d}+\frac{a C (a+a \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac{(5 A+7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{15 d}+\frac{(8 A+7 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{6 d}\\ \end{align*}
Mathematica [B] time = 2.9184, size = 418, normalized size = 2.36 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (\sec (c) (-780 A \sin (2 c+d x)+120 A \sin (c+2 d x)+120 A \sin (3 c+2 d x)+820 A \sin (2 c+3 d x)-180 A \sin (4 c+3 d x)+60 A \sin (3 c+4 d x)+60 A \sin (5 c+4 d x)+200 A \sin (4 c+5 d x)+150 A d x \cos (2 c+d x)+75 A d x \cos (2 c+3 d x)+75 A d x \cos (4 c+3 d x)+15 A d x \cos (4 c+5 d x)+15 A d x \cos (6 c+5 d x)+1220 A \sin (d x)+150 A d x \cos (d x)-480 C \sin (2 c+d x)+330 C \sin (c+2 d x)+330 C \sin (3 c+2 d x)+800 C \sin (2 c+3 d x)-30 C \sin (4 c+3 d x)+105 C \sin (3 c+4 d x)+105 C \sin (5 c+4 d x)+166 C \sin (4 c+5 d x)+1180 C \sin (d x))-240 (12 A+7 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.066, size = 226, normalized size = 1.3 \begin{align*}{a}^{4}Ax+{\frac{A{a}^{4}c}{d}}+{\frac{83\,{a}^{4}C\tan \left ( dx+c \right ) }{15\,d}}+6\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{7\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{20\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{34\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+2\,{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.955568, size = 416, normalized size = 2.35 \begin{align*} \frac{20 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 60 \,{\left (d x + c\right )} A a^{4} + 4 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{4} + 120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 15 \, C a^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 360 \, A a^{4} \tan \left (d x + c\right ) + 60 \, C a^{4} \tan \left (d x + c\right )}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.539463, size = 452, normalized size = 2.55 \begin{align*} \frac{60 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \,{\left (12 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (12 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (100 \, A + 83 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (5 \, A + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, C a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int A\, dx + \int 4 A \sec{\left (c + d x \right )}\, dx + \int 6 A \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.25417, size = 347, normalized size = 1.96 \begin{align*} \frac{30 \,{\left (d x + c\right )} A a^{4} + 15 \,{\left (12 \, A a^{4} + 7 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (12 \, A a^{4} + 7 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (150 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 105 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 680 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 490 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1180 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 896 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 920 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 790 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 270 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 375 \, C a^{4} \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 )}^{5}}}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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