Optimal. Leaf size=197 \[ \frac {2 c (A (2 m+5)-B (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 c (2 B m+5 B+4 C m+2 C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)} \]
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Rubi [A] time = 0.63, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3039, 2971, 2738} \[ \frac {2 c (A (2 m+5)-B (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 c (2 B m+5 B+4 C m+2 C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)} \]
Antiderivative was successfully verified.
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Rule 2738
Rule 2971
Rule 3039
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}-\frac {2 \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (-\frac {1}{2} a c (C (3-2 m)+A (5+2 m))-\frac {1}{2} a c (5 B+2 C+2 B m+4 C m) \sin (e+f x)\right ) \, dx}{a c (5+2 m)}\\ &=\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}+\frac {(5 B+2 C+2 B m+4 C m) \int (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)} \, dx}{a (5+2 m)}+\frac {(C-6 C m+A (5+2 m)-B (5+2 m)) \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx}{5+2 m}\\ &=\frac {2 c (C-6 C m+A (5+2 m)-B (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 c (5 B+2 C+2 B m+4 C m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}\\ \end {align*}
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Mathematica [A] time = 1.06, size = 177, normalized size = 0.90 \[ \frac {\sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m \left (8 A m^2+32 A m+30 A+2 (2 m+1) (2 B m+5 B-4 C) \sin (e+f x)-8 B m-20 B-C \left (4 m^2+8 m+3\right ) \cos (2 (e+f x))+4 C m^2+8 C m+19 C\right )}{f (2 m+1) (2 m+3) (2 m+5) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 310, normalized size = 1.57 \[ -\frac {2 \, {\left ({\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 4 \, {\left (A + B + C\right )} m^{2} + {\left (4 \, {\left (B + C\right )} m^{2} + 12 \, B m + 5 \, B - C\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (2 \, A + B\right )} m - {\left (4 \, {\left (A + C\right )} m^{2} + 4 \, {\left (4 \, A - B + 2 \, C\right )} m + 15 \, A - 10 \, B + 11 \, C\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (A + B + C\right )} m^{2} - {\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (2 \, A + B\right )} m + {\left (4 \, B m^{2} + 4 \, {\left (3 \, B - 2 \, C\right )} m + 5 \, B - 4 \, C\right )} \cos \left (f x + e\right ) + 15 \, A - 5 \, B + 7 \, C\right )} \sin \left (f x + e\right ) - 15 \, A + 5 \, B - 7 \, C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \]
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 \[ \int {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 5.91, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}\, \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.69, size = 644, normalized size = 3.27 \[ -\frac {2 \, {\left (\frac {2 \, {\left (\frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - a^{m} \sqrt {c} - \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} B e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + \frac {{\left (4 \, m^{2} + 8 \, m + 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 3\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac {4 \, {\left (\frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2 \, a^{m} \sqrt {c} - \frac {2 \, a^{m} \sqrt {c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} C e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + \frac {2 \, {\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac {{\left (a^{m} \sqrt {c} + \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 19.42, size = 510, normalized size = 2.59 \[ -\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (30\,A-15\,B+15\,C+32\,A\,m+4\,B\,m+8\,A\,m^2+4\,B\,m^2+4\,C\,m^2\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (A\,30{}\mathrm {i}-B\,15{}\mathrm {i}+C\,15{}\mathrm {i}+A\,m\,32{}\mathrm {i}+B\,m\,4{}\mathrm {i}+A\,m^2\,8{}\mathrm {i}+B\,m^2\,4{}\mathrm {i}+C\,m^2\,4{}\mathrm {i}\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {C\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,4{}\mathrm {i}+m\,8{}\mathrm {i}+3{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {C\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (10\,B-5\,C+4\,B\,m+2\,C\,m\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (B\,10{}\mathrm {i}-C\,5{}\mathrm {i}+B\,m\,4{}\mathrm {i}+C\,m\,2{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}\right )}{{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (8\,m^3+36\,m^2+46\,m+15\right )}{m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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