Optimal. Leaf size=128 \[ \frac {x^2}{2}+\frac {1}{8} e^{2 x} x-\frac {3 e^{2 x}}{32}+\frac {1}{4} e^{2 x} x \sin ^2(x)-\frac {1}{16} e^{2 x} \sin ^2(x)+e^x x \sin (x)+\frac {1}{32} e^{2 x} \sin (2 x)-e^x x \cos (x)+e^x \cos (x)-\frac {1}{32} e^{2 x} \cos (2 x)-\frac {1}{4} e^{2 x} x \sin (x) \cos (x)+\frac {1}{16} e^{2 x} \sin (x) \cos (x) \]
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Rubi [A] time = 0.19, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6742, 4432, 4465, 4433, 4434, 2194, 4469, 12} \[ \frac {x^2}{2}+\frac {1}{8} e^{2 x} x-\frac {3 e^{2 x}}{32}+\frac {1}{4} e^{2 x} x \sin ^2(x)-\frac {1}{16} e^{2 x} \sin ^2(x)+e^x x \sin (x)+\frac {1}{32} e^{2 x} \sin (2 x)-e^x x \cos (x)+e^x \cos (x)-\frac {1}{32} e^{2 x} \cos (2 x)-\frac {1}{4} e^{2 x} x \sin (x) \cos (x)+\frac {1}{16} e^{2 x} \sin (x) \cos (x) \]
Antiderivative was successfully verified.
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Rule 12
Rule 2194
Rule 4432
Rule 4433
Rule 4434
Rule 4465
Rule 4469
Rule 6742
Rubi steps
\begin {align*} \int x \left (1+e^x \sin (x)\right )^2 \, dx &=\int \left (x+2 e^x x \sin (x)+e^{2 x} x \sin ^2(x)\right ) \, dx\\ &=\frac {x^2}{2}+2 \int e^x x \sin (x) \, dx+\int e^{2 x} x \sin ^2(x) \, dx\\ &=\frac {1}{8} e^{2 x} x+\frac {x^2}{2}-e^x x \cos (x)+e^x x \sin (x)-\frac {1}{4} e^{2 x} x \cos (x) \sin (x)+\frac {1}{4} e^{2 x} x \sin ^2(x)-2 \int \left (-\frac {1}{2} e^x \cos (x)+\frac {1}{2} e^x \sin (x)\right ) \, dx-\int \left (\frac {e^{2 x}}{8}-\frac {1}{4} e^{2 x} \cos (x) \sin (x)+\frac {1}{4} e^{2 x} \sin ^2(x)\right ) \, dx\\ &=\frac {1}{8} e^{2 x} x+\frac {x^2}{2}-e^x x \cos (x)+e^x x \sin (x)-\frac {1}{4} e^{2 x} x \cos (x) \sin (x)+\frac {1}{4} e^{2 x} x \sin ^2(x)-\frac {1}{8} \int e^{2 x} \, dx+\frac {1}{4} \int e^{2 x} \cos (x) \sin (x) \, dx-\frac {1}{4} \int e^{2 x} \sin ^2(x) \, dx+\int e^x \cos (x) \, dx-\int e^x \sin (x) \, dx\\ &=-\frac {e^{2 x}}{16}+\frac {1}{8} e^{2 x} x+\frac {x^2}{2}+e^x \cos (x)-e^x x \cos (x)+e^x x \sin (x)+\frac {1}{16} e^{2 x} \cos (x) \sin (x)-\frac {1}{4} e^{2 x} x \cos (x) \sin (x)-\frac {1}{16} e^{2 x} \sin ^2(x)+\frac {1}{4} e^{2 x} x \sin ^2(x)-\frac {1}{16} \int e^{2 x} \, dx+\frac {1}{4} \int \frac {1}{2} e^{2 x} \sin (2 x) \, dx\\ &=-\frac {3 e^{2 x}}{32}+\frac {1}{8} e^{2 x} x+\frac {x^2}{2}+e^x \cos (x)-e^x x \cos (x)+e^x x \sin (x)+\frac {1}{16} e^{2 x} \cos (x) \sin (x)-\frac {1}{4} e^{2 x} x \cos (x) \sin (x)-\frac {1}{16} e^{2 x} \sin ^2(x)+\frac {1}{4} e^{2 x} x \sin ^2(x)+\frac {1}{8} \int e^{2 x} \sin (2 x) \, dx\\ &=-\frac {3 e^{2 x}}{32}+\frac {1}{8} e^{2 x} x+\frac {x^2}{2}+e^x \cos (x)-e^x x \cos (x)-\frac {1}{32} e^{2 x} \cos (2 x)+e^x x \sin (x)+\frac {1}{16} e^{2 x} \cos (x) \sin (x)-\frac {1}{4} e^{2 x} x \cos (x) \sin (x)-\frac {1}{16} e^{2 x} \sin ^2(x)+\frac {1}{4} e^{2 x} x \sin ^2(x)+\frac {1}{32} e^{2 x} \sin (2 x)\\ \end {align*}
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Mathematica [A] time = 0.17, size = 67, normalized size = 0.52 \[ \frac {1}{8} \left (4 x^2+e^{2 x} (2 x-1)+8 e^x x \sin (x)-e^{2 x} x \cos (2 x)-8 e^x (x-1) \cos (x)-e^{2 x} (2 x-1) \sin (x) \cos (x)\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.43, size = 55, normalized size = 0.43 \[ -{\left (x - 1\right )} \cos \relax (x) e^{x} + \frac {1}{2} \, x^{2} - \frac {1}{8} \, {\left (2 \, x \cos \relax (x)^{2} - 3 \, x + 1\right )} e^{\left (2 \, x\right )} - \frac {1}{8} \, {\left ({\left (2 \, x - 1\right )} \cos \relax (x) e^{\left (2 \, x\right )} - 8 \, x e^{x}\right )} \sin \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.10, size = 57, normalized size = 0.45 \[ \frac {1}{2} \, x^{2} - \frac {1}{16} \, {\left (2 \, x \cos \left (2 \, x\right ) + {\left (2 \, x - 1\right )} \sin \left (2 \, x\right )\right )} e^{\left (2 \, x\right )} + \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - {\left ({\left (x - 1\right )} \cos \relax (x) - x \sin \relax (x)\right )} e^{x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 63, normalized size = 0.49 \[ -\frac {x \cos \left (2 x \right ) {\mathrm e}^{2 x}}{8}+x \,{\mathrm e}^{x} \sin \relax (x )+\frac {x^{2}}{2}+\frac {x \,{\mathrm e}^{2 x}}{4}+2 \left (-\frac {x}{2}+\frac {1}{2}\right ) \cos \relax (x ) {\mathrm e}^{x}+\frac {\left (-\frac {x}{4}+\frac {1}{8}\right ) {\mathrm e}^{2 x} \sin \left (2 x \right )}{2}-\frac {{\mathrm e}^{2 x}}{8} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 58, normalized size = 0.45 \[ -\frac {1}{8} \, x \cos \left (2 \, x\right ) e^{\left (2 \, x\right )} - {\left (x - 1\right )} \cos \relax (x) e^{x} - \frac {1}{16} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \sin \left (2 \, x\right ) + x e^{x} \sin \relax (x) + \frac {1}{2} \, x^{2} + \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.28, size = 69, normalized size = 0.54 \[ \frac {3\,x\,{\mathrm {e}}^{2\,x}}{8}-\frac {{\mathrm {e}}^{2\,x}}{8}+{\mathrm {e}}^x\,\cos \relax (x)+\frac {x^2}{2}-\frac {x\,{\mathrm {e}}^{2\,x}\,{\cos \relax (x)}^2}{4}+\frac {{\mathrm {e}}^{2\,x}\,\cos \relax (x)\,\sin \relax (x)}{8}-x\,{\mathrm {e}}^x\,\cos \relax (x)+x\,{\mathrm {e}}^x\,\sin \relax (x)-\frac {x\,{\mathrm {e}}^{2\,x}\,\cos \relax (x)\,\sin \relax (x)}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.95, size = 109, normalized size = 0.85 \[ \frac {x^{2}}{2} + \frac {3 x e^{2 x} \sin ^{2}{\relax (x )}}{8} - \frac {x e^{2 x} \sin {\relax (x )} \cos {\relax (x )}}{4} + \frac {x e^{2 x} \cos ^{2}{\relax (x )}}{8} + x e^{x} \sin {\relax (x )} - x e^{x} \cos {\relax (x )} - \frac {e^{2 x} \sin ^{2}{\relax (x )}}{8} + \frac {e^{2 x} \sin {\relax (x )} \cos {\relax (x )}}{8} - \frac {e^{2 x} \cos ^{2}{\relax (x )}}{8} + e^{x} \cos {\relax (x )} \]
Verification of antiderivative is not currently implemented for this CAS.
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