Integrand size = 26, antiderivative size = 255 \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )} \]
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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} \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )} \]
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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 x^{12} \left (a b+b^2 x^3\right )^5 \, 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 (a^5 b^5 x^{12}+5 a^4 b^6 x^{15}+10 a^3 b^7 x^{18}+10 a^2 b^8 x^{21}+5 a b^9 x^{24}+b^{10} x^{27}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{13} \sqrt {\left (a+b x^3\right )^2} \left (117040 a^5+475475 a^4 b x^3+800800 a^3 b^2 x^6+691600 a^2 b^3 x^9+304304 a b^4 x^{12}+54340 b^5 x^{15}\right )}{1521520 \left (a+b x^3\right )} \]
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Time = 13.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x^{13} \left (54340 b^{5} x^{15}+304304 a \,b^{4} x^{12}+691600 a^{2} b^{3} x^{9}+800800 a^{3} b^{2} x^{6}+475475 a^{4} b \,x^{3}+117040 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1521520 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{13} \left (54340 b^{5} x^{15}+304304 a \,b^{4} x^{12}+691600 a^{2} b^{3} x^{9}+800800 a^{3} b^{2} x^{6}+475475 a^{4} b \,x^{3}+117040 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1521520 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{13} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{13 b \,x^{3}+13 a}+\frac {5 a^{4} b \,x^{16} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{16 \left (b \,x^{3}+a \right )}+\frac {10 a^{3} b^{2} x^{19} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{19 \left (b \,x^{3}+a \right )}+\frac {5 a^{2} b^{3} x^{22} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 \left (b \,x^{3}+a \right )}+\frac {a \,b^{4} x^{25} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{5 b \,x^{3}+5 a}+\frac {b^{5} x^{28} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{28 b \,x^{3}+28 a}\) | \(178\) |
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Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{28} \, b^{5} x^{28} + \frac {1}{5} \, a b^{4} x^{25} + \frac {5}{11} \, a^{2} b^{3} x^{22} + \frac {10}{19} \, a^{3} b^{2} x^{19} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{13} \, a^{5} x^{13} \]
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\[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{12} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{28} \, b^{5} x^{28} + \frac {1}{5} \, a b^{4} x^{25} + \frac {5}{11} \, a^{2} b^{3} x^{22} + \frac {10}{19} \, a^{3} b^{2} x^{19} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{13} \, a^{5} x^{13} \]
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Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{28} \, b^{5} x^{28} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a b^{4} x^{25} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{11} \, a^{2} b^{3} x^{22} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{19} \, a^{3} b^{2} x^{19} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{16} \, a^{4} b x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{13} \, a^{5} x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) \]
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Timed out. \[ \int x^{12} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{12}\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]
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