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 = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 203
Rule 455
Rule 463
Rule 1153
Rule 1814
Rule 4132
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*}
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Mathematica [B] time = 1.35, 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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fricas [A] time = 0.56, size = 131, normalized size = 0.89 \[ \frac {15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} - {\left (8 \, a^{2} \cos \left (f x + e\right )^{8} - 2 \, {\left (13 \, a^{2} - 12 \, a b\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 16 \, {\left (6 \, a b - 7 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 16 \, b^{2}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.27, size = 197, normalized size = 1.33 \[ \frac {16 \, b^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right ) - 96 \, b^{2} \tan \left (f x + e\right ) + 15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} - 108 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 192 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) - 84 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.92, size = 199, normalized size = 1.34 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+2 a b \left (\frac {\sin ^{7}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )-\frac {15 f x}{8}-\frac {15 e}{8}\right )+b^{2} \left (\frac {\sin ^{7}\left (f x +e \right )}{3 \cos \left (f x +e \right )^{3}}-\frac {4 \left (\sin ^{7}\left (f x +e \right )\right )}{3 \cos \left (f x +e \right )}-\frac {4 \left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{3}+\frac {5 f x}{2}+\frac {5 e}{2}\right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 164, normalized size = 1.11 \[ \frac {16 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + 96 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right ) - \frac {3 \, {\left (11 \, a^{2} - 36 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} - 24 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} - 28 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.79, size = 163, normalized size = 1.10 \[ x\,\left (\frac {5\,a^2}{16}-\frac {15\,a\,b}{4}+\frac {5\,b^2}{2}\right )-\frac {\left (\frac {11\,a^2}{16}-\frac {9\,a\,b}{4}+\frac {b^2}{2}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a^2}{6}-4\,a\,b+b^2\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a^2}{16}-\frac {7\,a\,b}{4}+\frac {b^2}{2}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,b^2-2\,b\,\left (a+b\right )\right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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