Integrand size = 16, antiderivative size = 78 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {b p}{12 a x^4}+\frac {b^2 p}{6 a^2 x^2}+\frac {b^3 p \log (x)}{3 a^3}-\frac {b^3 p \log \left (a+b x^2\right )}{6 a^3}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6} \]
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Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2442, 46} \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {b^3 p \log \left (a+b x^2\right )}{6 a^3}+\frac {b^3 p \log (x)}{3 a^3}+\frac {b^2 p}{6 a^2 x^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}-\frac {b p}{12 a x^4} \]
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Rule 46
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x^4} \, dx,x,x^2\right ) \\ & = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}+\frac {1}{6} (b p) \text {Subst}\left (\int \frac {1}{x^3 (a+b x)} \, dx,x,x^2\right ) \\ & = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6}+\frac {1}{6} (b p) \text {Subst}\left (\int \left (\frac {1}{a x^3}-\frac {b}{a^2 x^2}+\frac {b^2}{a^3 x}-\frac {b^3}{a^3 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b p}{12 a x^4}+\frac {b^2 p}{6 a^2 x^2}+\frac {b^3 p \log (x)}{3 a^3}-\frac {b^3 p \log \left (a+b x^2\right )}{6 a^3}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {1}{3} b p \left (-\frac {1}{4 a x^4}+\frac {b}{2 a^2 x^2}+\frac {b^2 \log (x)}{a^3}-\frac {b^2 \log \left (a+b x^2\right )}{2 a^3}\right )-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{6 x^6} \]
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Time = 0.77 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{6 x^{6}}+\frac {p b \left (-\frac {1}{4 a \,x^{4}}+\frac {b^{2} \ln \left (x \right )}{a^{3}}+\frac {b}{2 a^{2} x^{2}}-\frac {b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{3}}\right )}{3}\) | \(66\) |
parallelrisch | \(\frac {4 b^{3} p^{2} \ln \left (x \right ) x^{6}-2 x^{6} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{3} p -2 x^{6} b^{3} p^{2}+2 x^{4} a \,b^{2} p^{2}-x^{2} a^{2} b \,p^{2}-2 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{3} p}{12 x^{6} p \,a^{3}}\) | \(100\) |
risch | \(-\frac {\ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{6 x^{6}}-\frac {2 b^{3} p \ln \left (b \,x^{2}+a \right ) x^{6}-4 b^{3} p \ln \left (x \right ) x^{6}+i \pi \,a^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,a^{3} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi \,a^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi \,a^{3} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 a \,b^{2} p \,x^{4}+a^{2} b p \,x^{2}+2 \ln \left (c \right ) a^{3}}{12 a^{3} x^{6}}\) | \(206\) |
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Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {4 \, b^{3} p x^{6} \log \left (x\right ) + 2 \, a b^{2} p x^{4} - a^{2} b p x^{2} - 2 \, a^{3} \log \left (c\right ) - 2 \, {\left (b^{3} p x^{6} + a^{3} p\right )} \log \left (b x^{2} + a\right )}{12 \, a^{3} x^{6}} \]
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Time = 6.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.24 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 x^{6}} - \frac {b p}{12 a x^{4}} + \frac {b^{2} p}{6 a^{2} x^{2}} + \frac {b^{3} p \log {\left (x \right )}}{3 a^{3}} - \frac {b^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 a^{3}} & \text {for}\: a \neq 0 \\- \frac {p}{18 x^{6}} - \frac {\log {\left (c \left (b x^{2}\right )^{p} \right )}}{6 x^{6}} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {1}{12} \, b p {\left (\frac {2 \, b^{2} \log \left (b x^{2} + a\right )}{a^{3}} - \frac {2 \, b^{2} \log \left (x^{2}\right )}{a^{3}} - \frac {2 \, b x^{2} - a}{a^{2} x^{4}}\right )} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{6 \, x^{6}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (68) = 136\).
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.45 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=-\frac {\frac {2 \, b^{4} p \log \left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{3} - 3 \, {\left (b x^{2} + a\right )}^{2} a + 3 \, {\left (b x^{2} + a\right )} a^{2} - a^{3}} + \frac {2 \, b^{4} p \log \left (b x^{2} + a\right )}{a^{3}} - \frac {2 \, b^{4} p \log \left (b x^{2}\right )}{a^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{2} b^{4} p - 5 \, {\left (b x^{2} + a\right )} a b^{4} p + 3 \, a^{2} b^{4} p - 2 \, a^{2} b^{4} \log \left (c\right )}{{\left (b x^{2} + a\right )}^{3} a^{2} - 3 \, {\left (b x^{2} + a\right )}^{2} a^{3} + 3 \, {\left (b x^{2} + a\right )} a^{4} - a^{5}}}{12 \, b} \]
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Time = 1.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^7} \, dx=\frac {b^2\,p}{6\,a^2\,x^2}-\frac {b^3\,p\,\ln \left (b\,x^2+a\right )}{6\,a^3}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{6\,x^6}+\frac {b^3\,p\,\ln \left (x\right )}{3\,a^3}-\frac {b\,p}{12\,a\,x^4} \]
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