Optimal. Leaf size=148 \[ -\frac{\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}-\frac{\left (3 a^2-36 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{5}{16} x \left (a^2-12 a b+8 b^2\right )+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac{a (a-12 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]
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Rubi [A] time = 0.176964, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4132, 463, 455, 1814, 1153, 203} \[ -\frac{\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}-\frac{\left (3 a^2-36 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{5}{16} x \left (a^2-12 a b+8 b^2\right )+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac{a (a-12 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 463
Rule 455
Rule 1814
Rule 1153
Rule 203
Rubi steps
\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^6(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (a+b+b x^2\right )^2}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (7 a^2-6 (a+b)^2-6 b^2 x^2\right )}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac{a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac{\operatorname{Subst}\left (\int \frac{-a (a-12 b)+4 a (a-12 b) x^2-4 a (a-12 b) x^4+24 b^2 x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{24 f}\\ &=-\frac{\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac{\operatorname{Subst}\left (\int \frac{-7 a^2+84 a b-24 b^2+8 \left (a^2-12 a b+6 b^2\right ) x^2-48 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac{\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}-\frac{\operatorname{Subst}\left (\int \left (8 \left (a^2-12 a b+12 b^2\right )-48 b^2 x^2-\frac{15 \left (a^2-12 a b+8 b^2\right )}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac{\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac{b^2 \tan ^3(e+f x)}{3 f}+\frac{\left (5 \left (a^2-12 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f}\\ &=\frac{5}{16} \left (a^2-12 a b+8 b^2\right ) x-\frac{\left (3 a^2-36 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a (a-12 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{\left (a^2-12 a b+12 b^2\right ) \tan (e+f x)}{6 f}+\frac{a^2 \sin ^6(e+f x) \tan (e+f x)}{6 f}+\frac{b^2 \tan ^3(e+f x)}{3 f}\\ \end{align*}
Mathematica [B] time = 1.50546, size = 499, normalized size = 3.37 \[ \frac{\sec (e) \sec ^3(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (360 f x \left (a^2-12 a b+8 b^2\right ) \cos (2 e+f x)+360 f x \left (a^2-12 a b+8 b^2\right ) \cos (f x)-81 a^2 \sin (2 e+f x)-109 a^2 \sin (2 e+3 f x)-109 a^2 \sin (4 e+3 f x)-21 a^2 \sin (4 e+5 f x)-21 a^2 \sin (6 e+5 f x)+6 a^2 \sin (6 e+7 f x)+6 a^2 \sin (8 e+7 f x)-a^2 \sin (8 e+9 f x)-a^2 \sin (10 e+9 f x)+120 a^2 f x \cos (2 e+3 f x)+120 a^2 f x \cos (4 e+3 f x)-81 a^2 \sin (f x)-1164 a b \sin (2 e+f x)+2076 a b \sin (2 e+3 f x)+540 a b \sin (4 e+3 f x)+156 a b \sin (4 e+5 f x)+156 a b \sin (6 e+5 f x)-12 a b \sin (6 e+7 f x)-12 a b \sin (8 e+7 f x)-1440 a b f x \cos (2 e+3 f x)-1440 a b f x \cos (4 e+3 f x)+3444 a b \sin (f x)+2208 b^2 \sin (2 e+f x)-1936 b^2 \sin (2 e+3 f x)-144 b^2 \sin (4 e+3 f x)-48 b^2 \sin (4 e+5 f x)-48 b^2 \sin (6 e+5 f x)+960 b^2 f x \cos (2 e+3 f x)+960 b^2 f x \cos (4 e+3 f x)-3168 b^2 \sin (f x)\right )}{768 f (a \cos (2 (e+f x))+a+2 b)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 199, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +2\,ab \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{\cos \left ( fx+e \right ) }}+ \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) \cos \left ( fx+e \right ) -{\frac{15\,fx}{8}}-{\frac{15\,e}{8}} \right ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{3\,\cos \left ( fx+e \right ) }}-{\frac{4\,\cos \left ( fx+e \right ) }{3} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{2}}+{\frac{5\,e}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.5586, size = 221, normalized size = 1.49 \begin{align*} \frac{16 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \,{\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )}{\left (f x + e\right )} + 96 \,{\left (a b - b^{2}\right )} \tan \left (f x + e\right ) - \frac{3 \,{\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \,{\left (5 \, a^{2} - 24 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (5 \, a^{2} - 28 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.550109, size = 320, normalized size = 2.16 \begin{align*} \frac{15 \,{\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} -{\left (8 \, a^{2} \cos \left (f x + e\right )^{8} - 2 \,{\left (13 \, a^{2} - 12 \, a b\right )} \cos \left (f x + e\right )^{6} + 3 \,{\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 16 \,{\left (6 \, a b - 7 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 16 \, b^{2}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{3}} \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 = 1.26573, size = 266, normalized size = 1.8 \begin{align*} \frac{16 \, b^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right ) - 96 \, b^{2} \tan \left (f x + e\right ) + 15 \,{\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )}{\left (f x + e\right )} - \frac{33 \, a^{2} \tan \left (f x + e\right )^{5} - 108 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 192 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) - 84 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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