Optimal. Leaf size=99 \[ -\frac{e^{2 i a} 2^{-m-4} x^m (-i b x)^{-m} \text{Gamma}(m+2,-2 i b x)}{b^2}-\frac{e^{-2 i a} 2^{-m-4} x^m (i b x)^{-m} \text{Gamma}(m+2,2 i b x)}{b^2}+\frac{x^{m+2}}{2 (m+2)} \]
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Rubi [A] time = 0.142854, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3312, 3307, 2181} \[ -\frac{e^{2 i a} 2^{-m-4} x^m (-i b x)^{-m} \text{Gamma}(m+2,-2 i b x)}{b^2}-\frac{e^{-2 i a} 2^{-m-4} x^m (i b x)^{-m} \text{Gamma}(m+2,2 i b x)}{b^2}+\frac{x^{m+2}}{2 (m+2)} \]
Antiderivative was successfully verified.
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Rule 3312
Rule 3307
Rule 2181
Rubi steps
\begin{align*} \int x^{1+m} \sin ^2(a+b x) \, dx &=\int \left (\frac{x^{1+m}}{2}-\frac{1}{2} x^{1+m} \cos (2 a+2 b x)\right ) \, dx\\ &=\frac{x^{2+m}}{2 (2+m)}-\frac{1}{2} \int x^{1+m} \cos (2 a+2 b x) \, dx\\ &=\frac{x^{2+m}}{2 (2+m)}-\frac{1}{4} \int e^{-i (2 a+2 b x)} x^{1+m} \, dx-\frac{1}{4} \int e^{i (2 a+2 b x)} x^{1+m} \, dx\\ &=\frac{x^{2+m}}{2 (2+m)}-\frac{2^{-4-m} e^{2 i a} x^m (-i b x)^{-m} \Gamma (2+m,-2 i b x)}{b^2}-\frac{2^{-4-m} e^{-2 i a} x^m (i b x)^{-m} \Gamma (2+m,2 i b x)}{b^2}\\ \end{align*}
Mathematica [A] time = 0.305028, size = 116, normalized size = 1.17 \[ \frac{2^{-m-4} x^m \left (b^2 x^2\right )^{-m} \left (-(m+2) (\cos (a)-i \sin (a))^2 (-i b x)^m \text{Gamma}(m+2,2 i b x)-(m+2) (\cos (a)+i \sin (a))^2 (i b x)^m \text{Gamma}(m+2,-2 i b x)+2^{m+3} \left (b^2 x^2\right )^{m+1}\right )}{b^2 (m+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.125, size = 0, normalized size = 0. \begin{align*} \int{x}^{1+m} \left ( \sin \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (m + 2\right )} \int x x^{m} \cos \left (2 \, b x + 2 \, a\right )\,{d x} - e^{\left (m \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{2 \,{\left (m + 2\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.74335, size = 235, normalized size = 2.37 \begin{align*} \frac{4 \, b x x^{m + 1} +{\left (-i \, m - 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (2 i \, b\right ) - 2 i \, a\right )} \Gamma \left (m + 2, 2 i \, b x\right ) +{\left (i \, m + 2 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (-2 i \, b\right ) + 2 i \, a\right )} \Gamma \left (m + 2, -2 i \, b x\right )}{8 \,{\left (b m + 2 \, b\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \sin ^{2}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m + 1} \sin \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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