Integrand size = 25, antiderivative size = 311 \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=-\frac {d^2 \cos (e+f x) (3+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} (3+b) d (3 d-2 b c (2+m)) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{3+b}\right ) \cos (e+f x) (3+b \sin (e+f x))^m \left (\frac {3+b \sin (e+f x)}{3+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (3 d (3 d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{3+b}\right ) \cos (e+f x) (3+b \sin (e+f x))^m \left (\frac {3+b \sin (e+f x)}{3+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}} \]
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Time = 0.30 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2870, 2835, 2744, 144, 143} \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=-\frac {\sqrt {2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rule 2870
Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\int (a+b \sin (e+f x))^m \left (b \left (d^2 (1+m)+c^2 (2+m)\right )-d (a d-2 b c (2+m)) \sin (e+f x)\right ) \, dx}{b (2+m)} \\ & = -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac {(d (a d-2 b c (2+m))) \int (a+b \sin (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \int (a+b \sin (e+f x))^m \, dx}{b^2 (2+m)} \\ & = -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac {(d (a d-2 b c (2+m)) \cos (e+f x)) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\left ((-a-b) d (a d-2 b c (2+m)) \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} (a+b) d (a d-2 b c (2+m)) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}} \\ \end{align*}
\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx \]
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\[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{2}d x\]
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\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\text {Timed out} \]
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\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]
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