Integrand size = 18, antiderivative size = 130 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {b p}{15 a x^{5/2}}+\frac {b^2 p}{12 a^2 x^2}-\frac {b^3 p}{9 a^3 x^{3/2}}+\frac {b^4 p}{6 a^4 x}-\frac {b^5 p}{3 a^5 \sqrt {x}}+\frac {b^6 p \log \left (a+b \sqrt {x}\right )}{3 a^6}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}-\frac {b^6 p \log (x)}{6 a^6} \]
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Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2504, 2442, 46} \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {b^6 p \log \left (a+b \sqrt {x}\right )}{3 a^6}-\frac {b^6 p \log (x)}{6 a^6}-\frac {b^5 p}{3 a^5 \sqrt {x}}+\frac {b^4 p}{6 a^4 x}-\frac {b^3 p}{9 a^3 x^{3/2}}+\frac {b^2 p}{12 a^2 x^2}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}-\frac {b p}{15 a x^{5/2}} \]
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Rule 46
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x^7} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}+\frac {1}{3} (b p) \text {Subst}\left (\int \frac {1}{x^6 (a+b x)} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}+\frac {1}{3} (b p) \text {Subst}\left (\int \left (\frac {1}{a x^6}-\frac {b}{a^2 x^5}+\frac {b^2}{a^3 x^4}-\frac {b^3}{a^4 x^3}+\frac {b^4}{a^5 x^2}-\frac {b^5}{a^6 x}+\frac {b^6}{a^6 (a+b x)}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {b p}{15 a x^{5/2}}+\frac {b^2 p}{12 a^2 x^2}-\frac {b^3 p}{9 a^3 x^{3/2}}+\frac {b^4 p}{6 a^4 x}-\frac {b^5 p}{3 a^5 \sqrt {x}}+\frac {b^6 p \log \left (a+b \sqrt {x}\right )}{3 a^6}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}-\frac {b^6 p \log (x)}{6 a^6} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {a b p \sqrt {x} \left (-12 a^4+15 a^3 b \sqrt {x}-20 a^2 b^2 x+30 a b^3 x^{3/2}-60 b^4 x^2\right )+60 b^6 p x^3 \log \left (a+b \sqrt {x}\right )-60 a^6 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-30 b^6 p x^3 \log (x)}{180 a^6 x^3} \]
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Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.76
method | result | size |
parts | \(-\frac {\ln \left (c \left (a +b \sqrt {x}\right )^{p}\right )}{3 x^{3}}+\frac {p b \left (\frac {2 b^{5} \ln \left (a +b \sqrt {x}\right )}{a^{6}}-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{4}}{a^{5} \sqrt {x}}-\frac {2 b^{2}}{3 a^{3} x^{\frac {3}{2}}}-\frac {b^{5} \ln \left (x \right )}{a^{6}}+\frac {b^{3}}{a^{4} x}+\frac {b}{2 a^{2} x^{2}}\right )}{6}\) | \(99\) |
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Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {60 \, b^{6} p x^{3} \log \left (\sqrt {x}\right ) - 30 \, a^{2} b^{4} p x^{2} - 15 \, a^{4} b^{2} p x + 60 \, a^{6} \log \left (c\right ) - 60 \, {\left (b^{6} p x^{3} - a^{6} p\right )} \log \left (b \sqrt {x} + a\right ) + 4 \, {\left (15 \, a b^{5} p x^{2} + 5 \, a^{3} b^{3} p x + 3 \, a^{5} b p\right )} \sqrt {x}}{180 \, a^{6} x^{3}} \]
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Timed out. \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {1}{180} \, b p {\left (\frac {60 \, b^{5} \log \left (b \sqrt {x} + a\right )}{a^{6}} - \frac {30 \, b^{5} \log \left (x\right )}{a^{6}} - \frac {60 \, b^{4} x^{2} - 30 \, a b^{3} x^{\frac {3}{2}} + 20 \, a^{2} b^{2} x - 15 \, a^{3} b \sqrt {x} + 12 \, a^{4}}{a^{5} x^{\frac {5}{2}}}\right )} - \frac {\log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right )}{3 \, x^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 324 vs. \(2 (104) = 208\).
Time = 0.31 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.49 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {\frac {60 \, b^{7} p \log \left (b \sqrt {x} + a\right )}{{\left (b \sqrt {x} + a\right )}^{6} - 6 \, {\left (b \sqrt {x} + a\right )}^{5} a + 15 \, {\left (b \sqrt {x} + a\right )}^{4} a^{2} - 20 \, {\left (b \sqrt {x} + a\right )}^{3} a^{3} + 15 \, {\left (b \sqrt {x} + a\right )}^{2} a^{4} - 6 \, {\left (b \sqrt {x} + a\right )} a^{5} + a^{6}} - \frac {60 \, b^{7} p \log \left (b \sqrt {x} + a\right )}{a^{6}} + \frac {60 \, b^{7} p \log \left (b \sqrt {x}\right )}{a^{6}} + \frac {60 \, {\left (b \sqrt {x} + a\right )}^{5} b^{7} p - 330 \, {\left (b \sqrt {x} + a\right )}^{4} a b^{7} p + 740 \, {\left (b \sqrt {x} + a\right )}^{3} a^{2} b^{7} p - 855 \, {\left (b \sqrt {x} + a\right )}^{2} a^{3} b^{7} p + 522 \, {\left (b \sqrt {x} + a\right )} a^{4} b^{7} p - 137 \, a^{5} b^{7} p + 60 \, a^{5} b^{7} \log \left (c\right )}{{\left (b \sqrt {x} + a\right )}^{6} a^{5} - 6 \, {\left (b \sqrt {x} + a\right )}^{5} a^{6} + 15 \, {\left (b \sqrt {x} + a\right )}^{4} a^{7} - 20 \, {\left (b \sqrt {x} + a\right )}^{3} a^{8} + 15 \, {\left (b \sqrt {x} + a\right )}^{2} a^{9} - 6 \, {\left (b \sqrt {x} + a\right )} a^{10} + a^{11}}}{180 \, b} \]
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Time = 1.75 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {2\,b^6\,p\,\mathrm {atanh}\left (\frac {2\,b\,\sqrt {x}}{a}+1\right )}{3\,a^6}-\frac {\frac {b\,p}{5\,a}-\frac {b^2\,p\,\sqrt {x}}{4\,a^2}+\frac {b^5\,p\,x^2}{a^5}-\frac {b^4\,p\,x^{3/2}}{2\,a^4}+\frac {b^3\,p\,x}{3\,a^3}}{3\,x^{5/2}}-\frac {\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{3\,x^3} \]
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