Integrand size = 35, antiderivative size = 229 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {2 a^2 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)} \]
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Time = 0.35 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3055, 3060, 2855, 69, 67} \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {2 a^2 \left (A \left (8 n^2+30 n+25\right )+2 B \left (4 n^2+13 n+9\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A (2 n+5)+2 B (n+3)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]
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Rule 67
Rule 69
Rule 2855
Rule 3055
Rule 3060
Rubi steps \begin{align*} \text {integral}& = -\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)}+\frac {2 \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a d \left (2 B (1+n)+2 A \left (\frac {5}{2}+n\right )\right )+\frac {1}{2} a d (2 B (3+n)+A (5+2 n)) \sin (e+f x)\right ) \, dx}{d (5+2 n)} \\ & = -\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)}+\frac {\left (a \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \sqrt {a+a \sin (e+f x)} \, dx}{(3+2 n) (5+2 n)} \\ & = -\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)}+\frac {\left (a^3 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)}+\frac {\left (a^3 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \text {Subst}\left (\int \frac {x^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {2 a^2 \left (2 B \left (9+13 n+4 n^2\right )+A \left (25+30 n+8 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (2 B (3+n)+A (5+2 n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (d \sin (e+f x))^{1+n} \sqrt {a+a \sin (e+f x)}}{d f (5+2 n)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(478\) vs. \(2(229)=458\).
Time = 20.87 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.09 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {2^{1+n} \sec \left (\frac {1}{2} (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n (a (1+\sin (e+f x)))^{3/2} \tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}\right )^n \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^n \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {7}{2}+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},\frac {7}{2}+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},\frac {7}{2}+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{3+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {2 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {4+n}{2},\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4+n}+\tan \left (\frac {1}{2} (e+f x)\right ) \left (\frac {(3 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{5+n}+\frac {A \operatorname {Hypergeometric2F1}\left (\frac {7}{2}+n,\frac {6+n}{2},\frac {8+n}{2},-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{6+n}\right )\right )\right )\right )\right )}{f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]
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\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \sin \left (f x +e \right )\right )d x\]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (d \sin {\left (e + f x \right )}\right )^{n} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]
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