Integrand size = 26, antiderivative size = 255 \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {5 a^4 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^5 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )} \]
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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} \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a^5 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {5 a^4 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )} \]
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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^7 \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^7+5 a^4 b^6 x^{10}+10 a^3 b^7 x^{13}+10 a^2 b^8 x^{16}+5 a b^9 x^{19}+b^{10} x^{22}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {a^5 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {5 a^4 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^3 b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 \left (a+b x^3\right )}+\frac {10 a^2 b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {b^5 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^8 \sqrt {\left (a+b x^3\right )^2} \left (30107 a^5+109480 a^4 b x^3+172040 a^3 b^2 x^6+141680 a^2 b^3 x^9+60214 a b^4 x^{12}+10472 b^5 x^{15}\right )}{240856 \left (a+b x^3\right )} \]
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Time = 6.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x^{8} \left (10472 b^{5} x^{15}+60214 a \,b^{4} x^{12}+141680 a^{2} b^{3} x^{9}+172040 a^{3} b^{2} x^{6}+109480 a^{4} b \,x^{3}+30107 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{240856 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{8} \left (10472 b^{5} x^{15}+60214 a \,b^{4} x^{12}+141680 a^{2} b^{3} x^{9}+172040 a^{3} b^{2} x^{6}+109480 a^{4} b \,x^{3}+30107 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{240856 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{8} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{8 b \,x^{3}+8 a}+\frac {5 a^{4} b \,x^{11} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 \left (b \,x^{3}+a \right )}+\frac {5 a^{3} b^{2} x^{14} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{7 \left (b \,x^{3}+a \right )}+\frac {10 a^{2} b^{3} x^{17} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{17 \left (b \,x^{3}+a \right )}+\frac {a \,b^{4} x^{20} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{4 b \,x^{3}+4 a}+\frac {b^{5} x^{23} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{23 b \,x^{3}+23 a}\) | \(178\) |
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Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{23} \, b^{5} x^{23} + \frac {1}{4} \, a b^{4} x^{20} + \frac {10}{17} \, a^{2} b^{3} x^{17} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{11} \, a^{4} b x^{11} + \frac {1}{8} \, a^{5} x^{8} \]
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\[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{7} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{23} \, b^{5} x^{23} + \frac {1}{4} \, a b^{4} x^{20} + \frac {10}{17} \, a^{2} b^{3} x^{17} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{11} \, a^{4} b x^{11} + \frac {1}{8} \, a^{5} x^{8} \]
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Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{23} \, b^{5} x^{23} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{4} \, a b^{4} x^{20} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{17} \, a^{2} b^{3} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{7} \, a^{3} b^{2} x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{11} \, a^{4} b x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{8} \, a^{5} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) \]
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Timed out. \[ \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^7\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]
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