Integrand size = 32, antiderivative size = 112 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=-\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )}+\frac {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )} \]
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Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1369, 272, 45} \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\frac {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )}-\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )} \]
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Rule 45
Rule 272
Rule 1369
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x^{-1+2 n} \left (a b+b^2 x^n\right )^5 \, dx}{b^4 \left (a b+b^2 x^n\right )} \\ & = \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \text {Subst}\left (\int x \left (a b+b^2 x\right )^5 \, dx,x,x^n\right )}{b^4 n \left (a b+b^2 x^n\right )} \\ & = \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \text {Subst}\left (\int \left (-\frac {a \left (a b+b^2 x\right )^5}{b}+\frac {\left (a b+b^2 x\right )^6}{b^2}\right ) \, dx,x,x^n\right )}{b^4 n \left (a b+b^2 x^n\right )} \\ & = -\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )}+\frac {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\frac {x^{2 n} \left (\left (a+b x^n\right )^2\right )^{5/2} \left (21 a^5+70 a^4 b x^n+105 a^3 b^2 x^{2 n}+84 a^2 b^3 x^{3 n}+35 a b^4 x^{4 n}+6 b^5 x^{5 n}\right )}{42 n \left (a+b x^n\right )^5} \]
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Time = 0.11 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.86
method | result | size |
risch | \(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{5} x^{7 n}}{7 \left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{4} a \,x^{6 n}}{6 \left (a +b \,x^{n}\right ) n}+\frac {2 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b^{3} x^{5 n}}{\left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} b^{2} x^{4 n}}{2 \left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,a^{4} x^{3 n}}{3 \left (a +b \,x^{n}\right ) n}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{5} x^{2 n}}{2 \left (a +b \,x^{n}\right ) n}\) | \(208\) |
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none
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.66 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\frac {6 \, b^{5} x^{7 \, n} + 35 \, a b^{4} x^{6 \, n} + 84 \, a^{2} b^{3} x^{5 \, n} + 105 \, a^{3} b^{2} x^{4 \, n} + 70 \, a^{4} b x^{3 \, n} + 21 \, a^{5} x^{2 \, n}}{42 \, n} \]
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Timed out. \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\text {Timed out} \]
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none
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.66 \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\frac {6 \, b^{5} x^{7 \, n} + 35 \, a b^{4} x^{6 \, n} + 84 \, a^{2} b^{3} x^{5 \, n} + 105 \, a^{3} b^{2} x^{4 \, n} + 70 \, a^{4} b x^{3 \, n} + 21 \, a^{5} x^{2 \, n}}{42 \, n} \]
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\[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\int { {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {5}{2}} x^{2 \, n - 1} \,d x } \]
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Timed out. \[ \int x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx=\int x^{2\,n-1}\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{5/2} \,d x \]
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