Optimal. Leaf size=311 \[ -\frac {a^2 \sin ^2(e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac {a^2 \sin (e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f (1-m) (a-a \sin (e+f x))}+\frac {2^{m-\frac {3}{2}} \left (m^4+6 m^3-7 m^2-12 m+9\right ) (1-\sin (e+f x)) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {5}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{3 f (1-m) m}-\frac {\sec (e+f x) (a \sin (e+f x)+a)^{m-1} \left (a \left (-m^3-7 m^2-m+6\right )-a \left (-m^3-8 m^2-6 m+9\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))} \]
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Rubi [A] time = 0.36, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2719, 100, 153, 145, 70, 69} \[ -\frac {a^2 \sin ^2(e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac {a^2 \sin (e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f (1-m) (a-a \sin (e+f x))}+\frac {2^{m-\frac {3}{2}} \left (m^4+6 m^3-7 m^2-12 m+9\right ) (1-\sin (e+f x)) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {5}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{3 f (1-m) m}-\frac {\sec (e+f x) (a \sin (e+f x)+a)^{m-1} \left (a \left (-m^3-7 m^2-m+6\right )-a \left (-m^3-8 m^2-6 m+9\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 100
Rule 145
Rule 153
Rule 2719
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \tan ^4(e+f x) \, dx &=\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^4 (a+x)^{-\frac {5}{2}+m}}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f}\\ &=-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^2 (a+x)^{-\frac {5}{2}+m} \left (-3 a^2-a m x\right )}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f m}\\ &=\frac {a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x (a+x)^{-\frac {5}{2}+m} \left (2 a^3 m-a^2 \left (3-3 m-m^2\right ) x\right )}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f (1-m) m}\\ &=-\frac {\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac {a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac {\left (a \left (9-12 m-7 m^2+6 m^3+m^4\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-\frac {5}{2}+m}}{\sqrt {a-x}} \, dx,x,a \sin (e+f x)\right )}{3 f (1-m) m}\\ &=-\frac {\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac {a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac {\left (2^{-\frac {5}{2}+m} \left (9-12 m-7 m^2+6 m^3+m^4\right ) \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^m \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2 a}\right )^{-\frac {5}{2}+m}}{\sqrt {a-x}} \, dx,x,a \sin (e+f x)\right )}{3 a f (1-m) m}\\ &=\frac {2^{-\frac {3}{2}+m} \left (9-12 m-7 m^2+6 m^3+m^4\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{3 f (1-m) m}-\frac {\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac {a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}\\ \end {align*}
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Mathematica [F] time = 1.12, size = 0, normalized size = 0.00 \[ \int (a+a \sin (e+f x))^m \tan ^4(e+f x) \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{4}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.26, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (\tan ^{4}\left (f x +e \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
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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