Optimal. Leaf size=166 \[ -\frac {\sqrt {b} (a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f}+\frac {(3 a+2 b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f}-\frac {\left (11 a^2+18 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac {x \left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right )}{16 a^4}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f} \]
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Rubi [A] time = 0.34, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4132, 470, 578, 527, 522, 203, 205} \[ -\frac {\left (11 a^2+18 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac {x \left (30 a^2 b+5 a^3+40 a b^2+16 b^3\right )}{16 a^4}-\frac {\sqrt {b} (a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f}+\frac {(3 a+2 b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f}+\frac {\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 470
Rule 522
Rule 527
Rule 578
Rule 4132
Rubi steps
\begin {align*} \int \frac {\sin ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b)-3 (2 a+b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac {(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}-\frac {\operatorname {Subst}\left (\int \frac {3 (a+b) (3 a+2 b)-3 \left (8 a^2+13 a b+6 b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac {\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}+\frac {\operatorname {Subst}\left (\int \frac {3 (a+b) (a+2 b) (5 a+4 b)-3 b \left (11 a^2+18 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=-\frac {\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}-\frac {\left (b (a+b)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^4 f}+\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^4 f}\\ &=\frac {\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) x}{16 a^4}-\frac {\sqrt {b} (a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f}-\frac {\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}\\ \end {align*}
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Mathematica [C] time = 4.09, size = 357, normalized size = 2.15 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt {b (\cos (e)-i \sin (e))^4} \left (3 a^3 (9 a+8 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )+2 \sqrt {b} \sqrt {a+b} \left (-a^3 \sin (6 (e+f x))-12 a^3 e+60 a^3 f x-3 a \left (15 a^2+32 a b+16 b^2\right ) \sin (2 (e+f x))+3 a^2 (3 a+2 b) \sin (4 (e+f x))+360 a^2 b f x+480 a b^2 f x+192 b^3 f x\right )\right )+3 \sqrt {b} \left (9 a^4+136 a^3 b+384 a^2 b^2+384 a b^3+128 b^4\right ) (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )\right )}{768 a^4 \sqrt {b} f \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.75, size = 428, normalized size = 2.58 \[ \left [\frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 12 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}, \frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 24 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 250, normalized size = 1.51 \[ \frac {\frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}} - \frac {48 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{4}} - \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} + 54 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 42 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.85, size = 460, normalized size = 2.77 \[ -\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f a \sqrt {\left (a +b \right ) b}}-\frac {3 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a +b \right ) b}}-\frac {3 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{3} \sqrt {\left (a +b \right ) b}}-\frac {b^{4} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{4} \sqrt {\left (a +b \right ) b}}-\frac {9 \left (\tan ^{5}\left (f x +e \right )\right ) b}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\left (\tan ^{5}\left (f x +e \right )\right ) b^{2}}{2 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {11 \left (\tan ^{5}\left (f x +e \right )\right )}{16 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {2 \left (\tan ^{3}\left (f x +e \right )\right ) b}{f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\left (\tan ^{3}\left (f x +e \right )\right ) b^{2}}{f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{6 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \tan \left (f x +e \right )}{16 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {7 \tan \left (f x +e \right ) b}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\tan \left (f x +e \right ) b^{2}}{2 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {15 \arctan \left (\tan \left (f x +e \right )\right ) b}{8 f \,a^{2}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{2 f \,a^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{3}}{f \,a^{4}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f a} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 209, normalized size = 1.26 \[ -\frac {\frac {3 \, {\left (11 \, a^{2} + 18 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 14 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{3} \tan \left (f x + e\right )^{6} + 3 \, a^{3} \tan \left (f x + e\right )^{4} + 3 \, a^{3} \tan \left (f x + e\right )^{2} + a^{3}} - \frac {3 \, {\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}} + \frac {48 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{4}}}{48 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.71, size = 1448, normalized size = 8.72 \[ \text {result too large to display} \]
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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