Optimal. Leaf size=311 \[ \frac {(e \cos (c+d x))^{-m} \sec ^4(c+d x) (-1+\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) d e^3 (2+m)}+\frac {(-2 b+a (2+m)) (e \cos (c+d x))^{-m} \sec ^4(c+d x) (-1+\sin (c+d x)) (1+\sin (c+d x))^2 (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 m (2+m)}-\frac {\left (-b^2+a^2 (1+m)\right ) (e \cos (c+d x))^{-m} \, _2F_1\left (\frac {m}{2},1+m;2+m;-\frac {2 (a+b \sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right ) \sec ^4(c+d x) (1+\sin (c+d x))^3 \left (\frac {(a+b) (1+\sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right )^{\frac {1}{2} (-2+m)} (a+b \sin (c+d x))^{1+m}}{(a-b)^3 d e^3 m (1+m)} \]
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Rubi [A]
time = 0.35, antiderivative size = 420, normalized size of antiderivative = 1.35, number of steps
used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2779, 2777,
2999, 98, 134} \begin {gather*} -\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac {m+2}{2};m+2;\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) \left (a^2-b^2\right )}+\frac {a (\sin (c+d x)+1) (e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) \left (a^2-b^2\right )}+\frac {a 2^{-m/2} (a m+a+b) (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+2}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (-\frac {m}{2},\frac {m+2}{2};\frac {2-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e m (m+2) (a-b) (a+b)^2}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)} \end {gather*}
Antiderivative was successfully verified.
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Rule 98
Rule 134
Rule 2777
Rule 2779
Rule 2999
Rubi steps
\begin {align*} \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}+\frac {a \int \frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}-\frac {b \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx}{(a-b) e^2 (2+m)}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {\left (a (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac {2+m}{2}} (1+\sin (c+d x))^{\frac {2+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-2-m)} (1+x)^{\frac {1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac {\left (a (a+b+a m) (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac {2+m}{2}} (1+\sin (c+d x))^{\frac {2+m}{2}}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-2-m)} (1+x)^{\frac {1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) (a+b) d e (2+m)}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac {2^{-m/2} a (a+b+a m) (e \cos (c+d x))^{-2-m} \, _2F_1\left (-\frac {m}{2},\frac {2+m}{2};\frac {2-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x)) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {2+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e m (2+m)}\\ \end {align*}
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Mathematica [A]
time = 3.25, size = 319, normalized size = 1.03 \begin {gather*} \frac {(e \cos (c+d x))^{-m} \sec ^2(c+d x) (a+b \sin (c+d x))^m \left (-a-b \sin (c+d x)+\frac {b \, _2F_1\left (1+m,\frac {2+m}{2};2+m;-\frac {2 (a+b \sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right ) (1+\sin (c+d x)) \left (\frac {(a+b) (1+\sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))}{(a-b) (1+m)}+\frac {a (1-\sin (c+d x)) (1+\sin (c+d x)) \left (2^{-m/2} (a+b+a m) \, _2F_1\left (-\frac {m}{2},\frac {2+m}{2};1-\frac {m}{2};-\frac {(a-b) (-1+\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{m/2}-\frac {m (a+b \sin (c+d x))}{-1+\sin (c+d x)}\right )}{(a+b) m}\right )}{(a-b) d e^3 (2+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{-3-m} \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx\]
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 = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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