Optimal. Leaf size=154 \[ \frac{4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac{\left (24 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{5 a^3 b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]
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Rubi [A] time = 0.335893, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2792, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac{\left (24 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{5 a^3 b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3031
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx &=\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x)) \left (10 a^2 b+3 a \left (a^2+4 b^2\right ) \cos (c+d x)+b \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{12} \int \left (-3 a^2 \left (3 a^2+22 b^2\right )-16 a b \left (2 a^2+3 b^2\right ) \cos (c+d x)-3 b^2 \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-32 a b \left (2 a^2+3 b^2\right )-3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{3} \left (4 a b \left (2 a^2+3 b^2\right )\right ) \int \sec ^2(c+d x) \, dx-\frac{1}{8} \left (-3 a^4-24 a^2 b^2-8 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (4 a b \left (2 a^2+3 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{\left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac{a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.498054, size = 101, normalized size = 0.66 \[ \frac{3 \left (24 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) \left (32 b \left (3 \left (a^2+b^2\right )+a^2 \tan ^2(c+d x)\right )+9 a \left (a^2+8 b^2\right ) \sec (c+d x)+6 a^3 \sec ^3(c+d x)\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.074, size = 188, normalized size = 1.2 \begin{align*}{\frac{{a}^{4} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,{a}^{3}b \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+3\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{a{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.989196, size = 252, normalized size = 1.64 \begin{align*} \frac{64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} b - 3 \, 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 )} - 72 \, 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 )} + 24 \, b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, 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 = 1.89799, size = 394, normalized size = 2.56 \begin{align*} \frac{3 \,{\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \, a^{3} b \cos \left (d x + c\right ) + 6 \, a^{4} + 32 \,{\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \,{\left (a^{4} + 8 \, 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 = 1.43225, size = 486, normalized size = 3.16 \begin{align*} \frac{3 \,{\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 96 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 160 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 288 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 160 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 288 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \, a b^{3} \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 )}^{4}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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