Optimal. Leaf size=271 \[ -\frac {2^{-m-4} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-4} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b} \]
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Rubi [A] time = 0.33, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4406, 3308, 2181} \[ -\frac {2^{-m-4} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {2^{-m-4} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2181
Rule 3308
Rule 4406
Rubi steps
\begin {align*} \int (c+d x)^m \cos (a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac {1}{4} (c+d x)^m \sin (2 a+2 b x)-\frac {1}{8} (c+d x)^m \sin (4 a+4 b x)\right ) \, dx\\ &=-\left (\frac {1}{8} \int (c+d x)^m \sin (4 a+4 b x) \, dx\right )+\frac {1}{4} \int (c+d x)^m \sin (2 a+2 b x) \, dx\\ &=-\left (\frac {1}{16} i \int e^{-i (4 a+4 b x)} (c+d x)^m \, dx\right )+\frac {1}{16} i \int e^{i (4 a+4 b x)} (c+d x)^m \, dx+\frac {1}{8} i \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac {1}{8} i \int e^{i (2 a+2 b x)} (c+d x)^m \, dx\\ &=-\frac {2^{-4-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 {2^{-4-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 {4^{-3-m} e^{4 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 {4 i b (c+d x)}{d}\right )}{b}+\frac {4^{-3-m} e^{-4 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 {4 i b (c+d x)}{d}\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 246, normalized size = 0.91 \[ \frac {4^{-m-3} e^{-\frac {4 i (a d+b c)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-2^{m+2} e^{2 i \left (a+\frac {3 b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )-2^{m+2} e^{2 i \left (3 a+\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )+e^{8 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )+e^{\frac {8 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )\right )}{b} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 184, normalized size = 0.68 \[ \frac {e^{\left (-\frac {d m \log \left (\frac {4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {4 i \, b d x + 4 i \, b c}{d}\right ) - 4 \, e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, b d x + 2 i \, b c}{d}\right ) - 4 \, e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, b d x - 2 i \, b c}{d}\right ) + e^{\left (-\frac {d m \log \left (-\frac {4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-4 i \, b d x - 4 i \, b c}{d}\right )}{64 \, b} \]
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 \[ \int {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \cos \left (b x +a \right ) \left (\sin ^{3}\left (b x +a \right )\right )\, dx \]
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 \[ \int {\left (d x + c\right )}^{m} \cos \left (b x + a\right ) \sin \left (b x + a\right )^{3}\,{d x} \]
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 \[ \int \cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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