Integrand size = 26, antiderivative size = 255 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {5 a b^4 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )}+\frac {b^5 x^{26} \sqrt {a^2+2 a b x^3+b^2 x^6}}{26 \left (a+b x^3\right )} \]
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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} \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{26} \sqrt {a^2+2 a b x^3+b^2 x^6}}{26 \left (a+b x^3\right )}+\frac {5 a b^4 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {a^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )} \]
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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^{10} \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^{10}+5 a^4 b^6 x^{13}+10 a^3 b^7 x^{16}+10 a^2 b^8 x^{19}+5 a b^9 x^{22}+b^{10} x^{25}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {a^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {5 a b^4 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )}+\frac {b^5 x^{26} \sqrt {a^2+2 a b x^3+b^2 x^6}}{26 \left (a+b x^3\right )} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{11} \sqrt {\left (a+b x^3\right )^2} \left (71162 a^5+279565 a^4 b x^3+460460 a^3 b^2 x^6+391391 a^2 b^3 x^9+170170 a b^4 x^{12}+30107 b^5 x^{15}\right )}{782782 \left (a+b x^3\right )} \]
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Time = 9.72 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x^{11} \left (30107 b^{5} x^{15}+170170 a \,b^{4} x^{12}+391391 a^{2} b^{3} x^{9}+460460 a^{3} b^{2} x^{6}+279565 a^{4} b \,x^{3}+71162 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{782782 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{11} \left (30107 b^{5} x^{15}+170170 a \,b^{4} x^{12}+391391 a^{2} b^{3} x^{9}+460460 a^{3} b^{2} x^{6}+279565 a^{4} b \,x^{3}+71162 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{782782 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{11} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 b \,x^{3}+11 a}+\frac {5 a^{4} b \,x^{14} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{14 \left (b \,x^{3}+a \right )}+\frac {10 a^{3} b^{2} x^{17} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{17 \left (b \,x^{3}+a \right )}+\frac {a^{2} b^{3} x^{20} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 b \,x^{3}+2 a}+\frac {5 a \,b^{4} x^{23} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{23 \left (b \,x^{3}+a \right )}+\frac {b^{5} x^{26} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{26 b \,x^{3}+26 a}\) | \(178\) |
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Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} + \frac {5}{23} \, a b^{4} x^{23} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{11} \, a^{5} x^{11} \]
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\[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{10} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} + \frac {5}{23} \, a b^{4} x^{23} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{11} \, a^{5} x^{11} \]
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{23} \, a b^{4} x^{23} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a^{2} b^{3} x^{20} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{17} \, a^{3} b^{2} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{14} \, a^{4} b x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{11} \, a^{5} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) \]
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Timed out. \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{10}\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]
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