Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \]
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Time = 0.04 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 276} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )} \]
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Rule 276
Rule 1369
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (a b+b^2 x^3\right )^5}{x^{23}} \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^5 b^5}{x^{23}}+\frac {5 a^4 b^6}{x^{20}}+\frac {10 a^3 b^7}{x^{17}}+\frac {10 a^2 b^8}{x^{14}}+\frac {5 a b^9}{x^{11}}+\frac {b^{10}}{x^8}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = -\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{22 x^{22} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 x^{19} \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^{16} \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 x^{13} \left (a+b x^3\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^{10} \left (a+b x^3\right )}-\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (6916 a^5+40040 a^4 b x^3+95095 a^3 b^2 x^6+117040 a^2 b^3 x^9+76076 a b^4 x^{12}+21736 b^5 x^{15}\right )}{152152 x^{22} \left (a+b x^3\right )} \]
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Time = 46.67 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{22} a^{5}-\frac {5}{19} a^{4} b \,x^{3}-\frac {5}{8} a^{3} b^{2} x^{6}-\frac {10}{13} a^{2} b^{3} x^{9}-\frac {1}{2} a \,b^{4} x^{12}-\frac {1}{7} b^{5} x^{15}\right )}{\left (b \,x^{3}+a \right ) x^{22}}\) | \(79\) |
gosper | \(-\frac {\left (21736 b^{5} x^{15}+76076 a \,b^{4} x^{12}+117040 a^{2} b^{3} x^{9}+95095 a^{3} b^{2} x^{6}+40040 a^{4} b \,x^{3}+6916 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{152152 x^{22} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (21736 b^{5} x^{15}+76076 a \,b^{4} x^{12}+117040 a^{2} b^{3} x^{9}+95095 a^{3} b^{2} x^{6}+40040 a^{4} b \,x^{3}+6916 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{152152 x^{22} \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
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Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {21736 \, b^{5} x^{15} + 76076 \, a b^{4} x^{12} + 117040 \, a^{2} b^{3} x^{9} + 95095 \, a^{3} b^{2} x^{6} + 40040 \, a^{4} b x^{3} + 6916 \, a^{5}}{152152 \, x^{22}} \]
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\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{23}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {21736 \, b^{5} x^{15} + 76076 \, a b^{4} x^{12} + 117040 \, a^{2} b^{3} x^{9} + 95095 \, a^{3} b^{2} x^{6} + 40040 \, a^{4} b x^{3} + 6916 \, a^{5}}{152152 \, x^{22}} \]
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Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {21736 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 76076 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 117040 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 95095 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 40040 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 6916 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{152152 \, x^{22}} \]
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Time = 8.34 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{22\,x^{22}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{7\,x^7\,\left (b\,x^3+a\right )}-\frac {a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{2\,x^{10}\,\left (b\,x^3+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{19\,x^{19}\,\left (b\,x^3+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{13\,x^{13}\,\left (b\,x^3+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{8\,x^{16}\,\left (b\,x^3+a\right )} \]
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