Optimal. Leaf size=120 \[ \frac{a^2 \tan ^3(c+d x)}{3 d (a+b)^3}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{9/2}}-\frac{a^3 \tan (c+d x)}{d (a+b)^4}+\frac{\tan ^7(c+d x)}{7 d (a+b)}-\frac{a \tan ^5(c+d x)}{5 d (a+b)^2} \]
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Rubi [A] time = 0.12543, antiderivative size = 120, 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 = {3195, 302, 205} \[ \frac{a^2 \tan ^3(c+d x)}{3 d (a+b)^3}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{9/2}}-\frac{a^3 \tan (c+d x)}{d (a+b)^4}+\frac{\tan ^7(c+d x)}{7 d (a+b)}-\frac{a \tan ^5(c+d x)}{5 d (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 3195
Rule 302
Rule 205
Rubi steps
\begin{align*} \int \frac{\tan ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a^3}{(a+b)^4}+\frac{a^2 x^2}{(a+b)^3}-\frac{a x^4}{(a+b)^2}+\frac{x^6}{a+b}+\frac{a^4}{(a+b)^4 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{a^3 \tan (c+d x)}{(a+b)^4 d}+\frac{a^2 \tan ^3(c+d x)}{3 (a+b)^3 d}-\frac{a \tan ^5(c+d x)}{5 (a+b)^2 d}+\frac{\tan ^7(c+d x)}{7 (a+b) d}+\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{(a+b)^4 d}\\ &=\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{9/2} d}-\frac{a^3 \tan (c+d x)}{(a+b)^4 d}+\frac{a^2 \tan ^3(c+d x)}{3 (a+b)^3 d}-\frac{a \tan ^5(c+d x)}{5 (a+b)^2 d}+\frac{\tan ^7(c+d x)}{7 (a+b) d}\\ \end{align*}
Mathematica [A] time = 2.40066, size = 147, normalized size = 1.22 \[ \frac{\tan (c+d x) \left (\left (254 a^2 b+122 a^3+177 a b^2+45 b^3\right ) \sec ^2(c+d x)-122 a^2 b-176 a^3-66 a b^2+15 (a+b)^3 \sec ^6(c+d x)-3 (a+b)^2 (22 a+15 b) \sec ^4(c+d x)-15 b^3\right )}{105 d (a+b)^4}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{d (a+b)^{9/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.118, size = 252, normalized size = 2.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{7}{a}^{3}}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{3\, \left ( \tan \left ( dx+c \right ) \right ) ^{7}{a}^{2}b}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( a+b \right ) ^{4}}}+{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( a+b \right ) ^{4}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{3}}{5\,d \left ( a+b \right ) ^{4}}}-{\frac{2\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{2}b}{5\,d \left ( a+b \right ) ^{4}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}a{b}^{2}}{5\,d \left ( a+b \right ) ^{4}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{3}}{3\,d \left ( a+b \right ) ^{4}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{2}b}{3\,d \left ( a+b \right ) ^{4}}}-{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d \left ( a+b \right ) ^{4}}}+{\frac{{a}^{4}}{d \left ( a+b \right ) ^{4}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \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 = 2.28648, size = 1453, normalized size = 12.11 \begin{align*} \left [\frac{105 \, a^{3} \sqrt{-\frac{a}{a + b}} \cos \left (d x + c\right )^{7} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \,{\left ({\left (176 \, a^{3} + 122 \, a^{2} b + 66 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{6} -{\left (122 \, a^{3} + 254 \, a^{2} b + 177 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 15 \, a^{3} - 45 \, a^{2} b - 45 \, a b^{2} - 15 \, b^{3} + 3 \,{\left (22 \, a^{3} + 59 \, a^{2} b + 52 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{420 \,{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{7}}, -\frac{105 \, a^{3} \sqrt{\frac{a}{a + b}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{7} + 2 \,{\left ({\left (176 \, a^{3} + 122 \, a^{2} b + 66 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{6} -{\left (122 \, a^{3} + 254 \, a^{2} b + 177 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 15 \, a^{3} - 45 \, a^{2} b - 45 \, a b^{2} - 15 \, b^{3} + 3 \,{\left (22 \, a^{3} + 59 \, a^{2} b + 52 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{210 \,{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{7}}\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 [B] time = 9.94162, size = 637, normalized size = 5.31 \begin{align*} \frac{\frac{105 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} a^{4}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt{a^{2} + a b}} + \frac{15 \, a^{6} \tan \left (d x + c\right )^{7} + 90 \, a^{5} b \tan \left (d x + c\right )^{7} + 225 \, a^{4} b^{2} \tan \left (d x + c\right )^{7} + 300 \, a^{3} b^{3} \tan \left (d x + c\right )^{7} + 225 \, a^{2} b^{4} \tan \left (d x + c\right )^{7} + 90 \, a b^{5} \tan \left (d x + c\right )^{7} + 15 \, b^{6} \tan \left (d x + c\right )^{7} - 21 \, a^{6} \tan \left (d x + c\right )^{5} - 105 \, a^{5} b \tan \left (d x + c\right )^{5} - 210 \, a^{4} b^{2} \tan \left (d x + c\right )^{5} - 210 \, a^{3} b^{3} \tan \left (d x + c\right )^{5} - 105 \, a^{2} b^{4} \tan \left (d x + c\right )^{5} - 21 \, a b^{5} \tan \left (d x + c\right )^{5} + 35 \, a^{6} \tan \left (d x + c\right )^{3} + 140 \, a^{5} b \tan \left (d x + c\right )^{3} + 210 \, a^{4} b^{2} \tan \left (d x + c\right )^{3} + 140 \, a^{3} b^{3} \tan \left (d x + c\right )^{3} + 35 \, a^{2} b^{4} \tan \left (d x + c\right )^{3} - 105 \, a^{6} \tan \left (d x + c\right ) - 315 \, a^{5} b \tan \left (d x + c\right ) - 315 \, a^{4} b^{2} \tan \left (d x + c\right ) - 105 \, a^{3} b^{3} \tan \left (d x + c\right )}{a^{7} + 7 \, a^{6} b + 21 \, a^{5} b^{2} + 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 21 \, a^{2} b^{5} + 7 \, a b^{6} + b^{7}}}{105 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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