Optimal. Leaf size=623 \[ \frac {3 a \left (a^2 (n+6)+3 b^2 (n+1)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) (n+2) (n+4) (n+6) \sqrt {\cos ^2(c+d x)}}+\frac {3 b \left (3 a^2 (n+7)+b^2 (n+2)\right ) \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) (n+3) (n+5) (n+7) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+15 n+53\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (n+4) (n+5) (n+6) (n+7)}-\frac {\left (a^2 (n+2) (n+3)-b^2 (n+6) (n+8)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6) (n+7)}-\frac {3 a \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+16 n+58\right )+3 b^4 \left (n^2+12 n+35\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{b^2 d (n+2) (n+4) (n+5) (n+6) (n+7)}-\frac {3 \left (2 a^4 \left (n^2+5 n+6\right )-2 a^2 b^2 \left (n^2+16 n+57\right )+b^4 \left (n^2+10 n+24\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{b d (n+3) (n+4) (n+5) (n+6) (n+7)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (n+6) (n+7)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^4}{b d (n+7)} \]
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Rubi [A] time = 1.76, antiderivative size = 623, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2895, 3049, 3033, 3023, 2748, 2643} \[ \frac {3 a \left (a^2 (n+6)+3 b^2 (n+1)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{d (n+1) (n+2) (n+4) (n+6) \sqrt {\cos ^2(c+d x)}}+\frac {3 b \left (3 a^2 (n+7)+b^2 (n+2)\right ) \cos (c+d x) \sin ^{n+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{d (n+2) (n+3) (n+5) (n+7) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (n^2+5 n+6\right )-b^2 \left (n^2+15 n+53\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (n+4) (n+5) (n+6) (n+7)}-\frac {3 a \left (-2 a^2 b^2 \left (n^2+16 n+58\right )+2 a^4 \left (n^2+5 n+6\right )+3 b^4 \left (n^2+12 n+35\right )\right ) \cos (c+d x) \sin ^{n+1}(c+d x)}{b^2 d (n+2) (n+4) (n+5) (n+6) (n+7)}-\frac {3 \left (-2 a^2 b^2 \left (n^2+16 n+57\right )+2 a^4 \left (n^2+5 n+6\right )+b^4 \left (n^2+10 n+24\right )\right ) \cos (c+d x) \sin ^{n+2}(c+d x)}{b d (n+3) (n+4) (n+5) (n+6) (n+7)}-\frac {\left (a^2 (n+2) (n+3)-b^2 (n+6) (n+8)\right ) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (n+5) (n+6) (n+7)}+\frac {a (n+3) \cos (c+d x) \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (n+6) (n+7)}-\frac {\cos (c+d x) \sin ^{n+2}(c+d x) (a+b \sin (c+d x))^4}{b d (n+7)} \]
Antiderivative was successfully verified.
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Rule 2643
Rule 2748
Rule 2895
Rule 3023
Rule 3033
Rule 3049
Rubi steps
\begin {align*} \int \cos ^4(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^3 \left (a^2 (1+n) (3+n)-b^2 (6+n) (7+n)+3 a b \sin (c+d x)-\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (6+n) (7+n)}\\ &=-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x))^2 \left (3 a \left (a^2 \left (3+4 n+n^2\right )-b^2 \left (54+15 n+n^2\right )\right )+3 b \left (2 a^2-b^2 (6+n)\right ) \sin (c+d x)-3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (5+n) (6+n) (7+n)}\\ &=-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) (a+b \sin (c+d x)) \left (3 a^2 \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (163+46 n+3 n^2\right )\right )+3 a b \left (2 a^2-b^2 \left (81+26 n+2 n^2\right )\right ) \sin (c+d x)-3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (4+n) (5+n) (6+n) (7+n)}\\ &=-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) \left (3 a^3 (3+n) \left (2 a^2 \left (3+4 n+n^2\right )-b^2 \left (163+46 n+3 n^2\right )\right )-3 b^3 \left (24+10 n+n^2\right ) \left (b^2 (2+n)+3 a^2 (7+n)\right ) \sin (c+d x)-3 a (3+n) \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \sin ^2(c+d x)\right ) \, dx}{b^2 (3+n) (4+n) (5+n) (6+n) (7+n)}\\ &=-\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}-\frac {\int \sin ^n(c+d x) \left (-3 a b^2 (3+n) \left (35+12 n+n^2\right ) \left (3 b^2 (1+n)+a^2 (6+n)\right )-3 b^3 (2+n) \left (24+10 n+n^2\right ) \left (b^2 (2+n)+3 a^2 (7+n)\right ) \sin (c+d x)\right ) \, dx}{b^2 (2+n) (3+n) (4+n) (5+n) (6+n) (7+n)}\\ &=-\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}+\frac {\left (3 a \left (3 b^2 (1+n)+a^2 (6+n)\right )\right ) \int \sin ^n(c+d x) \, dx}{(2+n) (4+n) (6+n)}+\frac {\left (3 b \left (b^2 (2+n)+3 a^2 (7+n)\right )\right ) \int \sin ^{1+n}(c+d x) \, dx}{(3+n) (5+n) (7+n)}\\ &=-\frac {3 a \left (2 a^4 \left (6+5 n+n^2\right )+3 b^4 \left (35+12 n+n^2\right )-2 a^2 b^2 \left (58+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x)}{b^2 d (2+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 a \left (3 b^2 (1+n)+a^2 (6+n)\right ) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+n}{2};\frac {3+n}{2};\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) (2+n) (4+n) (6+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 \left (2 a^4 \left (6+5 n+n^2\right )+b^4 \left (24+10 n+n^2\right )-2 a^2 b^2 \left (57+16 n+n^2\right )\right ) \cos (c+d x) \sin ^{2+n}(c+d x)}{b d (3+n) (4+n) (5+n) (6+n) (7+n)}+\frac {3 b \left (b^2 (2+n)+3 a^2 (7+n)\right ) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2+n}{2};\frac {4+n}{2};\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) (3+n) (5+n) (7+n) \sqrt {\cos ^2(c+d x)}}-\frac {3 a \left (a^2 \left (6+5 n+n^2\right )-b^2 \left (53+15 n+n^2\right )\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^2}{b^2 d (4+n) (5+n) (6+n) (7+n)}-\frac {\left (a^2 (2+n) (3+n)-b^2 (6+n) (8+n)\right ) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^3}{b^2 d (5+n) (6+n) (7+n)}+\frac {a (3+n) \cos (c+d x) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^4}{b^2 d (6+n) (7+n)}-\frac {\cos (c+d x) \sin ^{2+n}(c+d x) (a+b \sin (c+d x))^4}{b d (7+n)}\\ \end {align*}
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Mathematica [A] time = 0.85, size = 195, normalized size = 0.31 \[ \frac {\sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{n+1}(c+d x) \left (\frac {a^3 \, _2F_1\left (-\frac {3}{2},\frac {n+1}{2};\frac {n+3}{2};\sin ^2(c+d x)\right )}{n+1}+b \sin (c+d x) \left (\frac {3 a^2 \, _2F_1\left (-\frac {3}{2},\frac {n+2}{2};\frac {n+4}{2};\sin ^2(c+d x)\right )}{n+2}+b \sin (c+d x) \left (\frac {3 a \, _2F_1\left (-\frac {3}{2},\frac {n+3}{2};\frac {n+5}{2};\sin ^2(c+d x)\right )}{n+3}+\frac {b \sin (c+d x) \, _2F_1\left (-\frac {3}{2},\frac {n+4}{2};\frac {n+6}{2};\sin ^2(c+d x)\right )}{n+4}\right )\right )\right )}{d} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (d x + c\right )^{6} - {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (b^{3} \cos \left (d x + c\right )^{6} - {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}, x\right ) \]
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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 28.10, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{3}\, dx \]
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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}\,{d x} \]
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 \[ \int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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