Integrand size = 26, antiderivative size = 255 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^5 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )} \]
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Time = 0.12 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {b^5 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a^5 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )} \]
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Rule 45
Rule 660
Rule 1125
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int x^5 \left (a b+b^2 x\right )^5 \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (a^5 b^5 x^5+5 a^4 b^6 x^6+10 a^3 b^7 x^7+10 a^2 b^8 x^8+5 a b^9 x^9+b^{10} x^{10}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {a^5 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^5 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )} \\ \end{align*}
Time = 0.90 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.53 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{12} \left (462 a^5+1980 a^4 b x^2+3465 a^3 b^2 x^4+3080 a^2 b^3 x^6+1386 a b^4 x^8+252 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{5544 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(\frac {x^{12} \operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {6}{11} x^{10} b^{5}+3 a \,x^{8} b^{4}+\frac {20}{3} a^{2} x^{6} b^{3}+\frac {15}{2} a^{3} x^{4} b^{2}+\frac {30}{7} x^{2} a^{4} b +a^{5}\right )}{12}\) | \(66\) |
gosper | \(\frac {x^{12} \left (252 x^{10} b^{5}+1386 a \,x^{8} b^{4}+3080 a^{2} x^{6} b^{3}+3465 a^{3} x^{4} b^{2}+1980 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{12} \left (252 x^{10} b^{5}+1386 a \,x^{8} b^{4}+3080 a^{2} x^{6} b^{3}+3465 a^{3} x^{4} b^{2}+1980 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{12} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 b \,x^{2}+12 a}+\frac {5 a^{4} b \,x^{14} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{14 \left (b \,x^{2}+a \right )}+\frac {5 a^{3} b^{2} x^{16} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 \left (b \,x^{2}+a \right )}+\frac {5 a^{2} b^{3} x^{18} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{9 \left (b \,x^{2}+a \right )}+\frac {a \,b^{4} x^{20} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b \,x^{2}+4 a}+\frac {b^{5} x^{22} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{22 b \,x^{2}+22 a}\) | \(178\) |
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Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {1}{4} \, a b^{4} x^{20} + \frac {5}{9} \, a^{2} b^{3} x^{18} + \frac {5}{8} \, a^{3} b^{2} x^{16} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{12} \, a^{5} x^{12} \]
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\[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{11} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {1}{4} \, a b^{4} x^{20} + \frac {5}{9} \, a^{2} b^{3} x^{18} + \frac {5}{8} \, a^{3} b^{2} x^{16} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{12} \, a^{5} x^{12} \]
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{4} \, a b^{4} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a^{2} b^{3} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{3} b^{2} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{14} \, a^{4} b x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{12} \, a^{5} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) \]
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Timed out. \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{11}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]
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