Optimal. Leaf size=135 \[ \frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {2^{-\frac {3}{2}+m} (A (3-m)-B m) \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a f (3-m)} \]
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Rubi [A]
time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2939, 2768, 72,
71} \begin {gather*} \frac {2^{m-\frac {3}{2}} (A (3-m)-B m) \sec ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{3 a f (3-m)}+\frac {B \sec ^3(e+f x) (a \sin (e+f x)+a)^m}{f (3-m)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rubi steps
\begin {align*} \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\left (A-\frac {B m}{3-m}\right ) \int \sec ^4(e+f x) (a+a \sin (e+f x))^m \, dx\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {\left (a^2 \left (A-\frac {B m}{3-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {\left (2^{-\frac {5}{2}+m} \left (A-\frac {B m}{3-m}\right ) \sec ^3(e+f x) (a-a \sin (e+f x))^{3/2} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {5}{2}+m}}{(a-a x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {B \sec ^3(e+f x) (a+a \sin (e+f x))^m}{f (3-m)}+\frac {2^{-\frac {3}{2}+m} \left (A-\frac {B m}{3-m}\right ) \, _2F_1\left (-\frac {3}{2},\frac {5}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sin (e+f x))\right ) \sec ^3(e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{3 a f}\\ \end {align*}
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Mathematica [F]
time = 1.46, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^4(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \left (\sec ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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