Optimal. Leaf size=105 \[ -i e^{-i x} \, _2F_1\left (1,-\frac {1}{2 n};1-\frac {1}{2 n};-e^{2 i n x}\right )-i e^{i x} \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-e^{2 i n x}\right )+\frac {1}{2} i e^{-i x}+\frac {1}{2} i e^{i x} \]
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Rubi [A] time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4557, 2194, 2251} \[ -i e^{-i x} \, _2F_1\left (1,-\frac {1}{2 n};1-\frac {1}{2 n};-e^{2 i n x}\right )-i e^{i x} \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-e^{2 i n x}\right )+\frac {1}{2} i e^{-i x}+\frac {1}{2} i e^{i x} \]
Antiderivative was successfully verified.
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Rule 2194
Rule 2251
Rule 4557
Rubi steps
\begin {align*} \int \sin (x) \tan (n x) \, dx &=\int \left (\frac {e^{-i x}}{2}-\frac {e^{i x}}{2}-\frac {e^{-i x}}{1+e^{2 i n x}}+\frac {e^{i x}}{1+e^{2 i n x}}\right ) \, dx\\ &=\frac {1}{2} \int e^{-i x} \, dx-\frac {1}{2} \int e^{i x} \, dx-\int \frac {e^{-i x}}{1+e^{2 i n x}} \, dx+\int \frac {e^{i x}}{1+e^{2 i n x}} \, dx\\ &=\frac {1}{2} i e^{-i x}+\frac {1}{2} i e^{i x}-i e^{-i x} \, _2F_1\left (1,-\frac {1}{2 n};1-\frac {1}{2 n};-e^{2 i n x}\right )-i e^{i x} \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-e^{2 i n x}\right )\\ \end {align*}
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Mathematica [A] time = 0.18, size = 200, normalized size = 1.90 \[ -\frac {i e^{-2 i x} \left ((2 n+1) e^{i (2 n x+x)} \, _2F_1\left (1,1-\frac {1}{2 n};2-\frac {1}{2 n};-e^{2 i n x}\right )+(2 n-1) \left ((2 n+1) e^{i x} \left (\, _2F_1\left (1,-\frac {1}{2 n};1-\frac {1}{2 n};-e^{2 i n x}\right )+e^{2 i x} \, _2F_1\left (1,\frac {1}{2 n};1+\frac {1}{2 n};-e^{2 i n x}\right )\right )-e^{i (2 n+3) x} \, _2F_1\left (1,1+\frac {1}{2 n};2+\frac {1}{2 n};-e^{2 i n x}\right )\right )\right )}{2 \left (4 n^2-1\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sin \relax (x) \tan \left (n x\right ), x\right ) \]
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 \[ \int \sin \relax (x) \tan \left (n x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.44, size = 0, normalized size = 0.00 \[ \int \sin \relax (x ) \tan \left (n x \right )\, dx \]
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 \[ \int \sin \relax (x) \tan \left (n x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {tan}\left (n\,x\right )\,\sin \relax (x) \,d x \]
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 \[ \int \sin {\relax (x )} \tan {\left (n x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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