Optimal. Leaf size=120 \[ \frac {(m+1) x^{m+1} \cos ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+(m+1)^2}+\frac {2 b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+(m+1)^2}+\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left (4 b^2 n^2+(m+1)^2\right )} \]
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Rubi [A] time = 0.03, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4488, 30} \[ \frac {(m+1) x^{m+1} \cos ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+(m+1)^2}+\frac {2 b n x^{m+1} \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+(m+1)^2}+\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left (4 b^2 n^2+(m+1)^2\right )} \]
Antiderivative was successfully verified.
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Rule 30
Rule 4488
Rubi steps
\begin {align*} \int x^m \cos ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {(1+m) x^{1+m} \cos ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+4 b^2 n^2}+\frac {2 b n x^{1+m} \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+4 b^2 n^2}+\frac {\left (2 b^2 n^2\right ) \int x^m \, dx}{(1+m)^2+4 b^2 n^2}\\ &=\frac {2 b^2 n^2 x^{1+m}}{(1+m) \left ((1+m)^2+4 b^2 n^2\right )}+\frac {(1+m) x^{1+m} \cos ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+4 b^2 n^2}+\frac {2 b n x^{1+m} \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2+4 b^2 n^2}\\ \end {align*}
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Mathematica [C] time = 0.34, size = 91, normalized size = 0.76 \[ \frac {x^{m+1} \left (2 b (m+1) n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+(m+1)^2 \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+4 b^2 n^2+m^2+2 m+1\right )}{2 (m+1) (-2 i b n+m+1) (2 i b n+m+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 105, normalized size = 0.88 \[ \frac {2 \, {\left (b m + b\right )} n x x^{m} \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sin \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + {\left (2 \, b^{2} n^{2} x + {\left (m^{2} + 2 \, m + 1\right )} x \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2}\right )} x^{m}}{m^{3} + 4 \, {\left (b^{2} m + b^{2}\right )} n^{2} + 3 \, m^{2} + 3 \, m + 1} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.07, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\cos ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 646, normalized size = 5.38 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.79, size = 82, normalized size = 0.68 \[ \frac {x\,x^m}{2\,m+2}+\frac {x\,x^m\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}{4\,m+4+b\,n\,8{}\mathrm {i}}+\frac {x\,x^m\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}\,1{}\mathrm {i}}{m\,4{}\mathrm {i}+8\,b\,n+4{}\mathrm {i}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \log {\relax (x )} \cos ^{2}{\relax (a )} & \text {for}\: b = 0 \wedge m = -1 \\\int x^{m} \cos ^{2}{\left (- a + \frac {i m \log {\left (c x^{n} \right )}}{2 n} + \frac {i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {i \left (m + 1\right )}{2 n} \\\int x^{m} \cos ^{2}{\left (a + \frac {i m \log {\left (c x^{n} \right )}}{2 n} + \frac {i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {i \left (m + 1\right )}{2 n} \\\frac {\begin {cases} \log {\relax (x )} \cos {\left (2 a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\relax (x )} \cos {\left (2 a + 2 b \log {\relax (c )} \right )} & \text {for}\: n = 0 \\\frac {\sin {\left (2 a + 2 b n \log {\relax (x )} + 2 b \log {\relax (c )} \right )}}{2 b n} & \text {otherwise} \end {cases}}{2} + \frac {\log {\relax (x )}}{2} & \text {for}\: m = -1 \\\frac {2 b^{2} n^{2} x x^{m} \sin ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 b^{2} n^{2} x x^{m} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 b m n x x^{m} \sin {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cos {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 b n x x^{m} \sin {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cos {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {m^{2} x x^{m} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 m x x^{m} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} + \frac {x x^{m} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{4 b^{2} m n^{2} + 4 b^{2} n^{2} + m^{3} + 3 m^{2} + 3 m + 1} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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