Optimal. Leaf size=246 \[ -\frac{b^2 \left (3 a^2 (3 A+4 C)+2 b^2 (13 A-12 C)\right ) \sin (c+d x)}{24 d}+\frac{a b \left (a^2 (23 A+36 C)+12 A b^2\right ) \tan (c+d x)}{12 d}+\frac{\left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (a^2 (3 A+4 C)+4 A b^2\right ) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^4}{4 d}+\frac{A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+4 a b^3 C x \]
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Rubi [A] time = 0.934565, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3047, 3031, 3023, 2735, 3770} \[ -\frac{b^2 \left (3 a^2 (3 A+4 C)+2 b^2 (13 A-12 C)\right ) \sin (c+d x)}{24 d}+\frac{a b \left (a^2 (23 A+36 C)+12 A b^2\right ) \tan (c+d x)}{12 d}+\frac{\left (24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (a^2 (3 A+4 C)+4 A b^2\right ) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^4}{4 d}+\frac{A b \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+4 a b^3 C x \]
Antiderivative was successfully verified.
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Rule 3048
Rule 3047
Rule 3031
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^3 \left (4 A b+a (3 A+4 C) \cos (c+d x)-b (A-4 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x))^2 \left (3 \left (4 A b^2+a^2 (3 A+4 C)\right )+2 a b (7 A+12 C) \cos (c+d x)-b^2 (7 A-12 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{24} \int (a+b \cos (c+d x)) \left (2 \left (12 A b^3+\frac{1}{2} a^2 (46 A b+72 b C)\right )+a \left (3 a^2 (3 A+4 C)+2 b^2 (13 A+36 C)\right ) \cos (c+d x)-b \left (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (8 A b^4+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )-96 a b^3 C \cos (c+d x)+b^2 \left (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (8 A b^4+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )-96 a b^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=4 a b^3 C x-\frac{b^2 \left (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{8} \left (-8 A b^4-24 a^2 b^2 (A+2 C)-a^4 (3 A+4 C)\right ) \int \sec (c+d x) \, dx\\ &=4 a b^3 C x+\frac{\left (8 A b^4+24 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b^2 \left (2 b^2 (13 A-12 C)+3 a^2 (3 A+4 C)\right ) \sin (c+d x)}{24 d}+\frac{a b \left (12 A b^2+a^2 (23 A+36 C)\right ) \tan (c+d x)}{12 d}+\frac{\left (4 A b^2+a^2 (3 A+4 C)\right ) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A b (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+b \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 6.33435, size = 612, normalized size = 2.49 \[ \frac{4 \left (2 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )+3 a^3 b C \sin \left (\frac{1}{2} (c+d x)\right )+3 a A b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 \left (2 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )+3 a^3 b C \sin \left (\frac{1}{2} (c+d x)\right )+3 a A b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{72 a^2 A b^2+16 a^3 A b+9 a^4 A+12 a^4 C}{48 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{-72 a^2 A b^2-16 a^3 A b-9 a^4 A-12 a^4 C}{48 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\left (-24 a^2 A b^2-3 a^4 A-48 a^2 b^2 C-4 a^4 C-8 A b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{\left (24 a^2 A b^2+3 a^4 A+48 a^2 b^2 C+4 a^4 C+8 A b^4\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{2 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{a^4 A}{16 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{a^4 A}{16 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{4 a b^3 C (c+d x)}{d}+\frac{b^4 C \sin (c+d x)}{d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.071, size = 316, normalized size = 1.3 \begin{align*}{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C{b}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{aA{b}^{3}\tan \left ( dx+c \right ) }{d}}+4\,a{b}^{3}Cx+4\,{\frac{Ca{b}^{3}c}{d}}+3\,{\frac{{a}^{2}A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,A{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,A{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+4\,{\frac{{a}^{3}bC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.04228, size = 413, normalized size = 1.68 \begin{align*} \frac{64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} b + 192 \,{\left (d x + c\right )} C a b^{3} - 3 \, A 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 )} - 12 \, 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 )} - 72 \, A a^{2} b^{2}{\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 )} + 144 \, C a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C b^{4} \sin \left (d x + c\right ) + 192 \, C a^{3} b \tan \left (d x + c\right ) + 192 \, A a b^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.53185, size = 575, normalized size = 2.34 \begin{align*} \frac{192 \, C a b^{3} d x \cos \left (d x + c\right )^{4} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 24 \,{\left (A + 2 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 24 \,{\left (A + 2 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, C b^{4} \cos \left (d x + c\right )^{4} + 32 \, A a^{3} b \cos \left (d x + c\right ) + 6 \, A a^{4} + 32 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{3} b + 3 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{4} + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 = 1.53916, size = 797, normalized size = 3.24 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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