Optimal. Leaf size=158 \[ \frac{a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac{b \left (-8 a^2 b^2+24 a^4-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (2 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+a^5 x-\frac{a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.28609, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {4042, 3918, 4056, 4048, 3770, 3767, 8} \[ \frac{a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac{b \left (-8 a^2 b^2+24 a^4-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (2 a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+a^5 x-\frac{a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{4 d}-\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4042
Rule 3918
Rule 4056
Rule 4048
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx &=-\int (-a+b \sec (c+d x)) (a+b \sec (c+d x))^4 \, dx\\ &=-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{1}{4} \int (a+b \sec (c+d x))^2 \left (-4 a^3-b \left (4 a^2-3 b^2\right ) \sec (c+d x)+3 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=-\frac{a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{1}{12} \int (a+b \sec (c+d x)) \left (-12 a^4-3 a b \left (8 a^2-5 b^2\right ) \sec (c+d x)-3 b^2 \left (2 a^2-3 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-24 a^5-3 b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \sec (c+d x)-12 a b^2 \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^5 x+\frac{b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{2} \left (a b^2 \left (5 a^2-4 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (b \left (24 a^4-8 a^2 b^2-3 b^4\right )\right ) \int \sec (c+d x) \, dx\\ &=a^5 x+\frac{b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a b^2 \left (5 a^2-4 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^5 x+\frac{b \left (24 a^4-8 a^2 b^2-3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a b^2 \left (5 a^2-4 b^2\right ) \tan (c+d x)}{2 d}+\frac{b^3 \left (2 a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}-\frac{a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}-\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 6.40467, size = 1299, normalized size = 8.22 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.05, size = 205, normalized size = 1.3 \begin{align*}{a}^{5}x+{\frac{{a}^{5}c}{d}}+2\,{\frac{{a}^{3}{b}^{2}\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{4}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{a{b}^{4}\tan \left ( dx+c \right ) }{d}}-{\frac{a{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{b}^{5}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{b}^{5}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}-{\frac{3\,{b}^{5}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.974706, size = 259, normalized size = 1.64 \begin{align*} \frac{16 \,{\left (d x + c\right )} a^{5} - 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{4} + b^{5}{\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 )} + 8 \, a^{2} b^{3}{\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 )} + 48 \, a^{4} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 32 \, a^{3} b^{2} \tan \left (d x + c\right )}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.546008, size = 428, normalized size = 2.71 \begin{align*} \frac{16 \, a^{5} d x \cos \left (d x + c\right )^{4} +{\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (8 \, a b^{4} \cos \left (d x + c\right ) + 2 \, b^{5} - 16 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} +{\left (8 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a - b \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{4}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.26838, size = 512, normalized size = 3.24 \begin{align*} \frac{8 \,{\left (d x + c\right )} a^{5} +{\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (24 \, a^{4} b - 8 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (16 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 8 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 5 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 8 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 48 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, b^{5} \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}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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