Optimal. Leaf size=108 \[ \frac{\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac{(a-3 b) \tan ^3(c+d x)}{3 b^2 d}-\frac{(a-b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2} d}+\frac{\tan ^5(c+d x)}{5 b d} \]
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Rubi [A] time = 0.109796, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3675, 390, 205} \[ \frac{\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac{(a-3 b) \tan ^3(c+d x)}{3 b^2 d}-\frac{(a-b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2} d}+\frac{\tan ^5(c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 3675
Rule 390
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2-3 a b+3 b^2}{b^3}-\frac{(a-3 b) x^2}{b^2}+\frac{x^4}{b}+\frac{-a^3+3 a^2 b-3 a b^2+b^3}{b^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac{(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac{\tan ^5(c+d x)}{5 b d}-\frac{(a-b)^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}\\ &=-\frac{(a-b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2} d}+\frac{\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac{(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac{\tan ^5(c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 0.922971, size = 103, normalized size = 0.95 \[ \frac{\sqrt{b} \tan (c+d x) \left (15 a^2-b (5 a-9 b) \sec ^2(c+d x)-40 a b+3 b^2 \sec ^4(c+d x)+33 b^2\right )-\frac{15 (a-b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a}}}{15 b^{7/2} d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.073, size = 206, normalized size = 1.9 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,bd}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}a}{3\,d{b}^{2}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{bd}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d{b}^{3}}}-3\,{\frac{a\tan \left ( dx+c \right ) }{d{b}^{2}}}+3\,{\frac{\tan \left ( dx+c \right ) }{bd}}-{\frac{{a}^{3}}{d{b}^{3}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+3\,{\frac{{a}^{2}}{d{b}^{2}\sqrt{ab}}\arctan \left ({\frac{b\tan \left ( dx+c \right ) }{\sqrt{ab}}} \right ) }-3\,{\frac{a}{bd\sqrt{ab}}\arctan \left ({\frac{b\tan \left ( dx+c \right ) }{\sqrt{ab}}} \right ) }+{\frac{1}{d}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.78256, size = 992, normalized size = 9.19 \begin{align*} \left [\frac{15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{-a b} \cos \left (d x + c\right )^{5} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt{-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \,{\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} -{\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, a b^{4} d \cos \left (d x + c\right )^{5}}, \frac{15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt{a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \,{\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} -{\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, a b^{4} d \cos \left (d x + c\right )^{5}}\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 = 1.63043, size = 204, normalized size = 1.89 \begin{align*} -\frac{\frac{15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (d x + c\right )}{\sqrt{a b}}\right )\right )}}{\sqrt{a b} b^{3}} - \frac{3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, b^{4} \tan \left (d x + c\right )^{3} + 15 \, a^{2} b^{2} \tan \left (d x + c\right ) - 45 \, a b^{3} \tan \left (d x + c\right ) + 45 \, b^{4} \tan \left (d x + c\right )}{b^{5}}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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