Integrand size = 18, antiderivative size = 123 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {a^5 p \sqrt {x}}{3 b^5}-\frac {a^4 p x}{6 b^4}+\frac {a^3 p x^{3/2}}{9 b^3}-\frac {a^2 p x^2}{12 b^2}+\frac {a p x^{5/2}}{15 b}-\frac {p x^3}{18}-\frac {a^6 p \log \left (a+b \sqrt {x}\right )}{3 b^6}+\frac {1}{3} x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \]
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Time = 0.06 (sec) , antiderivative size = 123, 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, 45} \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=-\frac {a^6 p \log \left (a+b \sqrt {x}\right )}{3 b^6}+\frac {a^5 p \sqrt {x}}{3 b^5}-\frac {a^4 p x}{6 b^4}+\frac {a^3 p x^{3/2}}{9 b^3}-\frac {a^2 p x^2}{12 b^2}+\frac {1}{3} x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right )+\frac {a p x^{5/2}}{15 b}-\frac {p x^3}{18} \]
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Rule 45
Rule 2442
Rule 2504
Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^5 \log \left (c (a+b x)^p\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{3} x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-\frac {1}{3} (b p) \text {Subst}\left (\int \frac {x^6}{a+b x} \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{3} x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-\frac {1}{3} (b p) \text {Subst}\left (\int \left (-\frac {a^5}{b^6}+\frac {a^4 x}{b^5}-\frac {a^3 x^2}{b^4}+\frac {a^2 x^3}{b^3}-\frac {a x^4}{b^2}+\frac {x^5}{b}+\frac {a^6}{b^6 (a+b x)}\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a^5 p \sqrt {x}}{3 b^5}-\frac {a^4 p x}{6 b^4}+\frac {a^3 p x^{3/2}}{9 b^3}-\frac {a^2 p x^2}{12 b^2}+\frac {a p x^{5/2}}{15 b}-\frac {p x^3}{18}-\frac {a^6 p \log \left (a+b \sqrt {x}\right )}{3 b^6}+\frac {1}{3} x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.91 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {b p \sqrt {x} \left (60 a^5-30 a^4 b \sqrt {x}+20 a^3 b^2 x-15 a^2 b^3 x^{3/2}+12 a b^4 x^2-10 b^5 x^{5/2}\right )-60 a^6 p \log \left (a+b \sqrt {x}\right )+60 b^6 x^3 \log \left (c \left (a+b \sqrt {x}\right )^p\right )}{180 b^6} \]
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Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80
method | result | size |
parts | \(\frac {x^{3} \ln \left (c \left (a +b \sqrt {x}\right )^{p}\right )}{3}-\frac {p b \left (-\frac {2 \left (-\frac {x^{3} b^{5}}{6}+\frac {a \,x^{\frac {5}{2}} b^{4}}{5}-\frac {a^{2} b^{3} x^{2}}{4}+\frac {a^{3} x^{\frac {3}{2}} b^{2}}{3}-\frac {b \,a^{4} x}{2}+a^{5} \sqrt {x}\right )}{b^{6}}+\frac {2 a^{6} \ln \left (a +b \sqrt {x}\right )}{b^{7}}\right )}{6}\) | \(99\) |
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Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.85 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=-\frac {10 \, b^{6} p x^{3} - 60 \, b^{6} x^{3} \log \left (c\right ) + 15 \, a^{2} b^{4} p x^{2} + 30 \, a^{4} b^{2} p x - 60 \, {\left (b^{6} p x^{3} - a^{6} p\right )} \log \left (b \sqrt {x} + a\right ) - 4 \, {\left (3 \, a b^{5} p x^{2} + 5 \, a^{3} b^{3} p x + 15 \, a^{5} b p\right )} \sqrt {x}}{180 \, b^{6}} \]
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Time = 3.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.97 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=- \frac {b p \left (\frac {2 a^{6} \left (\begin {cases} \frac {\sqrt {x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{6}} - \frac {2 a^{5} \sqrt {x}}{b^{6}} + \frac {a^{4} x}{b^{5}} - \frac {2 a^{3} x^{\frac {3}{2}}}{3 b^{4}} + \frac {a^{2} x^{2}}{2 b^{3}} - \frac {2 a x^{\frac {5}{2}}}{5 b^{2}} + \frac {x^{3}}{3 b}\right )}{6} + \frac {x^{3} \log {\left (c \left (a + b \sqrt {x}\right )^{p} \right )}}{3} \]
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Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {1}{3} \, x^{3} \log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right ) - \frac {1}{180} \, b p {\left (\frac {60 \, a^{6} \log \left (b \sqrt {x} + a\right )}{b^{7}} + \frac {10 \, b^{5} x^{3} - 12 \, a b^{4} x^{\frac {5}{2}} + 15 \, a^{2} b^{3} x^{2} - 20 \, a^{3} b^{2} x^{\frac {3}{2}} + 30 \, a^{4} b x - 60 \, a^{5} \sqrt {x}}{b^{6}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (97) = 194\).
Time = 0.28 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.07 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {60 \, b x^{3} \log \left (c\right ) + {\left (\frac {60 \, {\left (b \sqrt {x} + a\right )}^{6} \log \left (b \sqrt {x} + a\right )}{b^{5}} - \frac {360 \, {\left (b \sqrt {x} + a\right )}^{5} a \log \left (b \sqrt {x} + a\right )}{b^{5}} + \frac {900 \, {\left (b \sqrt {x} + a\right )}^{4} a^{2} \log \left (b \sqrt {x} + a\right )}{b^{5}} - \frac {1200 \, {\left (b \sqrt {x} + a\right )}^{3} a^{3} \log \left (b \sqrt {x} + a\right )}{b^{5}} + \frac {900 \, {\left (b \sqrt {x} + a\right )}^{2} a^{4} \log \left (b \sqrt {x} + a\right )}{b^{5}} - \frac {360 \, {\left (b \sqrt {x} + a\right )} a^{5} \log \left (b \sqrt {x} + a\right )}{b^{5}} - \frac {10 \, {\left (b \sqrt {x} + a\right )}^{6}}{b^{5}} + \frac {72 \, {\left (b \sqrt {x} + a\right )}^{5} a}{b^{5}} - \frac {225 \, {\left (b \sqrt {x} + a\right )}^{4} a^{2}}{b^{5}} + \frac {400 \, {\left (b \sqrt {x} + a\right )}^{3} a^{3}}{b^{5}} - \frac {450 \, {\left (b \sqrt {x} + a\right )}^{2} a^{4}}{b^{5}} + \frac {360 \, {\left (b \sqrt {x} + a\right )} a^{5}}{b^{5}}\right )} p}{180 \, b} \]
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Time = 1.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.79 \[ \int x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {x^3\,\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{3}-\frac {p\,x^3}{18}-\frac {a^6\,p\,\ln \left (a+b\,\sqrt {x}\right )}{3\,b^6}-\frac {a^2\,p\,x^2}{12\,b^2}+\frac {a^3\,p\,x^{3/2}}{9\,b^3}+\frac {a^5\,p\,\sqrt {x}}{3\,b^5}+\frac {a\,p\,x^{5/2}}{15\,b}-\frac {a^4\,p\,x}{6\,b^4} \]
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