Optimal. Leaf size=127 \[ \frac{(5 a+b) (a-b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac{(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}-\frac{(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d} \]
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Rubi [A] time = 0.141559, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 205} \[ \frac{(5 a+b) (a-b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac{(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}-\frac{(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d} \]
Antiderivative was successfully verified.
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Rule 3675
Rule 390
Rule 385
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{2 a-3 b}{b^3}+\frac{x^2}{b^2}+\frac{(a-b)^2 (2 a+b)+3 (a-b)^2 b x^2}{b^3 \left (a+b x^2\right )^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d}+\frac{\operatorname{Subst}\left (\int \frac{(a-b)^2 (2 a+b)+3 (a-b)^2 b x^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}\\ &=-\frac{(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d}-\frac{(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\left ((a-b)^2 (5 a+b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a b^3 d}\\ &=\frac{(a-b)^2 (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac{(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d}-\frac{(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.732435, size = 135, normalized size = 1.06 \[ \frac{\frac{3 (a-b)^2 (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}+4 \sqrt{b} (4 b-3 a) \tan (c+d x)+\frac{3 \sqrt{b} (b-a)^3 \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 b^{3/2} \tan (c+d x) \sec ^2(c+d x)}{6 b^{7/2} d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.093, size = 275, normalized size = 2.2 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{b}^{2}d}}-2\,{\frac{a\tan \left ( dx+c \right ) }{d{b}^{3}}}+3\,{\frac{\tan \left ( dx+c \right ) }{{b}^{2}d}}-{\frac{{a}^{2}\tan \left ( dx+c \right ) }{2\,d{b}^{3} \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{3\,a\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{3\,\tan \left ( dx+c \right ) }{2\,db \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( dx+c \right ) }{2\,ad \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{5\,{a}^{2}}{2\,d{b}^{3}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{9\,a}{2\,{b}^{2}d}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3}{2\,db}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{2\,ad}\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 [B] time = 1.85338, size = 1349, normalized size = 10.62 \begin{align*} \left [-\frac{3 \,{\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt{-a b} \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 (2 \, a^{2} b^{3} -{\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} +{\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}, -\frac{3 \,{\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} +{\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{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 ) - 2 \,{\left (2 \, a^{2} b^{3} -{\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} +{\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}\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 = 1.70501, size = 243, normalized size = 1.91 \begin{align*} \frac{\frac{3 \,{\left (5 \, a^{3} - 9 \, 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} a b^{3}} - \frac{3 \,{\left (a^{3} \tan \left (d x + c\right ) - 3 \, a^{2} b \tan \left (d x + c\right ) + 3 \, a b^{2} \tan \left (d x + c\right ) - b^{3} \tan \left (d x + c\right )\right )}}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{3}} + \frac{2 \,{\left (b^{4} \tan \left (d x + c\right )^{3} - 6 \, a b^{3} \tan \left (d x + c\right ) + 9 \, b^{4} \tan \left (d x + c\right )\right )}}{b^{6}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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