Integrand size = 14, antiderivative size = 141 \[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\frac {x^{1+m}}{2 (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 i a} x^{1+m} \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 i b x^n\right )}{n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 i a} x^{1+m} \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 i b x^n\right )}{n} \]
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Time = 0.24 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3507, 3505, 2250} \[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=-\frac {e^{2 i a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (-i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 i b x^n\right )}{n}-\frac {e^{-2 i a} 2^{-\frac {m+2 n+1}{n}} x^{m+1} \left (i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 i b x^n\right )}{n}+\frac {x^{m+1}}{2 (m+1)} \]
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Rule 2250
Rule 3505
Rule 3507
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x^m}{2}+\frac {1}{2} x^m \cos \left (2 a+2 b x^n\right )\right ) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{2} \int x^m \cos \left (2 a+2 b x^n\right ) \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}+\frac {1}{4} \int e^{-2 i a-2 i b x^n} x^m \, dx+\frac {1}{4} \int e^{2 i a+2 i b x^n} x^m \, dx \\ & = \frac {x^{1+m}}{2 (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 i a} x^{1+m} \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 i b x^n\right )}{n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 i a} x^{1+m} \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 i b x^n\right )}{n} \\ \end{align*}
Time = 0.57 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=-\frac {x^{1+m} \left (-2 n+2^{-\frac {1+m}{n}} e^{2 i a} (1+m) \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 i b x^n\right )+2^{-\frac {1+m}{n}} e^{-2 i a} (1+m) \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 i b x^n\right )\right )}{4 (1+m) n} \]
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\[\int x^{m} \left (\cos ^{2}\left (a +b \,x^{n}\right )\right )d x\]
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\[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\int { x^{m} \cos \left (b x^{n} + a\right )^{2} \,d x } \]
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\[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\int x^{m} \cos ^{2}{\left (a + b x^{n} \right )}\, dx \]
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\[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\int { x^{m} \cos \left (b x^{n} + a\right )^{2} \,d x } \]
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\[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\int { x^{m} \cos \left (b x^{n} + a\right )^{2} \,d x } \]
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Timed out. \[ \int x^m \cos ^2\left (a+b x^n\right ) \, dx=\int x^m\,{\cos \left (a+b\,x^n\right )}^2 \,d x \]
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