Optimal. Leaf size=162 \[ \frac{i 2^{-m-3} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{i 2^{-m-3} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{2 d (m+1)} \]
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Rubi [A] time = 0.217265, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3312, 3307, 2181} \[ \frac{i 2^{-m-3} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{i 2^{-m-3} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{2 d (m+1)} \]
Antiderivative was successfully verified.
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Rule 3312
Rule 3307
Rule 2181
Rubi steps
\begin{align*} \int (c+d x)^m \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{2} (c+d x)^m-\frac{1}{2} (c+d x)^m \cos (2 a+2 b x)\right ) \, dx\\ &=\frac{(c+d x)^{1+m}}{2 d (1+m)}-\frac{1}{2} \int (c+d x)^m \cos (2 a+2 b x) \, dx\\ &=\frac{(c+d x)^{1+m}}{2 d (1+m)}-\frac{1}{4} \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac{1}{4} \int e^{i (2 a+2 b x)} (c+d x)^m \, dx\\ &=\frac{(c+d x)^{1+m}}{2 d (1+m)}+\frac{i 2^{-3-m} e^{2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{2 i b (c+d x)}{d}\right )}{b}-\frac{i 2^{-3-m} e^{-2 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{2 i b (c+d x)}{d}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.612256, size = 211, normalized size = 1.3 \[ \frac{2^{-m-3} (c+d x)^m \left (\frac{b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-i d (m+1) \left (-\frac{i b (c+d x)}{d}\right )^m \left (\cos \left (2 a-\frac{2 b c}{d}\right )-i \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \text{Gamma}\left (m+1,\frac{2 i b (c+d x)}{d}\right )+i d (m+1) \left (\frac{i b (c+d x)}{d}\right )^m \left (\cos \left (2 a-\frac{2 b c}{d}\right )+i \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \text{Gamma}\left (m+1,-\frac{2 i b (c+d x)}{d}\right )+b 2^{m+2} (c+d x) \left (\frac{b^2 (c+d x)^2}{d^2}\right )^m\right )}{b d (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.138, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{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 (d m + d\right )} \int{\left (d x + c\right )}^{m} \cos \left (2 \, b x + 2 \, a\right )\,{d x} - e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{2 \,{\left (d m + d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.81047, size = 340, normalized size = 2.1 \begin{align*} \frac{{\left (-i \, d m - i \, d\right )} e^{\left (-\frac{d m \log \left (\frac{2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{2 i \, b d x + 2 i \, b c}{d}\right ) +{\left (i \, d m + i \, d\right )} e^{\left (-\frac{d m \log \left (-\frac{2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{-2 i \, b d x - 2 i \, b c}{d}\right ) + 4 \,{\left (b d x + b c\right )}{\left (d x + c\right )}^{m}}{8 \,{\left (b d m + b d\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 \left (c + d x\right )^{m} \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{\left (d x + c\right )}^{m} \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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