Integrand size = 17, antiderivative size = 73 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {3 \log (x)}{8}-\frac {3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{8 b n}-\frac {\cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4 b n} \]
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Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2715, 8} \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {\sin ^3\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b n}-\frac {3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{8 b n}+\frac {3 \log (x)}{8} \]
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Rule 8
Rule 2715
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \sin ^4(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = -\frac {\cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4 b n}+\frac {3 \text {Subst}\left (\int \sin ^2(a+b x) \, dx,x,\log \left (c x^n\right )\right )}{4 n} \\ & = -\frac {3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{8 b n}-\frac {\cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4 b n}+\frac {3 \text {Subst}\left (\int 1 \, dx,x,\log \left (c x^n\right )\right )}{8 n} \\ & = \frac {3 \log (x)}{8}-\frac {3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{8 b n}-\frac {\cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4 b n} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {12 \left (a+b \log \left (c x^n\right )\right )-8 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )}{32 b n} \]
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Time = 6.52 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {12 \ln \left (x \right ) b n +\sin \left (4 b \ln \left (c \,x^{n}\right )+4 a \right )-8 \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )}{32 b n}\) | \(46\) |
derivativedivides | \(\frac {-\frac {\left ({\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}+\frac {3 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )}{2}\right ) \cos \left (a +b \ln \left (c \,x^{n}\right )\right )}{4}+\frac {3 b \ln \left (c \,x^{n}\right )}{8}+\frac {3 a}{8}}{n b}\) | \(61\) |
default | \(\frac {-\frac {\left ({\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}+\frac {3 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )}{2}\right ) \cos \left (a +b \ln \left (c \,x^{n}\right )\right )}{4}+\frac {3 b \ln \left (c \,x^{n}\right )}{8}+\frac {3 a}{8}}{n b}\) | \(61\) |
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Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {3 \, b n \log \left (x\right ) + {\left (2 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 5 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{8 \, b n} \]
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Time = 11.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.37 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=- \frac {\begin {cases} \log {\left (x \right )} \cos {\left (2 a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\left (x \right )} \cos {\left (2 a + 2 b \log {\left (c \right )} \right )} & \text {for}\: n = 0 \\\frac {\sin {\left (2 a + 2 b \log {\left (c x^{n} \right )} \right )}}{2 b n} & \text {otherwise} \end {cases}}{2} + \frac {\begin {cases} \log {\left (x \right )} \cos {\left (4 a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = 0\right ) \\\log {\left (x \right )} \cos {\left (4 a + 4 b \log {\left (c \right )} \right )} & \text {for}\: n = 0 \\\frac {\sin {\left (4 a + 4 b \log {\left (c x^{n} \right )} \right )}}{4 b n} & \text {otherwise} \end {cases}}{8} + \frac {3 \log {\left (x \right )}}{8} \]
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Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {12 \, b n \log \left (x\right ) + \cos \left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right ) \sin \left (4 \, b \log \left (c\right )\right ) - 8 \, \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) \sin \left (2 \, b \log \left (c\right )\right ) + \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (x^{n}\right ) + 4 \, a\right ) - 8 \, \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}{32 \, b n} \]
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\[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x} \,d x } \]
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Time = 29.63 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^4\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {3\,\ln \left (x^n\right )}{8\,n}-\frac {\frac {\sin \left (2\,a+2\,b\,\ln \left (c\,x^n\right )\right )}{4}-\frac {\sin \left (4\,a+4\,b\,\ln \left (c\,x^n\right )\right )}{32}}{b\,n} \]
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