Integrand size = 22, antiderivative size = 247 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5} \]
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Time = 0.03 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1357, 200} \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5} \]
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Rule 200
Rule 1357
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \int \left (2 a b+2 b^2 x^3\right )^5 \, dx}{\left (2 a b+2 b^2 x^3\right )^5} \\ & = \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \int \left (32 a^5 b^5+160 a^4 b^6 x^3+320 a^3 b^7 x^6+320 a^2 b^8 x^9+160 a b^9 x^{12}+32 b^{10} x^{15}\right ) \, dx}{\left (2 a b+2 b^2 x^3\right )^5} \\ & = \frac {a^5 x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a^4 b x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{4 \left (a+b x^3\right )^5}+\frac {10 a^3 b^2 x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{7 \left (a+b x^3\right )^5}+\frac {a^2 b^3 x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{\left (a+b x^3\right )^5}+\frac {5 a b^4 x^{13} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{13 \left (a+b x^3\right )^5}+\frac {b^5 x^{16} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{16 \left (a+b x^3\right )^5} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.33 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x \sqrt {\left (a+b x^3\right )^2} \left (1456 a^5+1820 a^4 b x^3+2080 a^3 b^2 x^6+1456 a^2 b^3 x^9+560 a b^4 x^{12}+91 b^5 x^{15}\right )}{1456 \left (a+b x^3\right )} \]
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Time = 1.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(\frac {x \left (91 b^{5} x^{15}+560 a \,b^{4} x^{12}+1456 a^{2} b^{3} x^{9}+2080 a^{3} b^{2} x^{6}+1820 a^{4} b \,x^{3}+1456 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1456 \left (b \,x^{3}+a \right )^{5}}\) | \(78\) |
default | \(\frac {x \left (91 b^{5} x^{15}+560 a \,b^{4} x^{12}+1456 a^{2} b^{3} x^{9}+2080 a^{3} b^{2} x^{6}+1820 a^{4} b \,x^{3}+1456 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1456 \left (b \,x^{3}+a \right )^{5}}\) | \(78\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x}{b \,x^{3}+a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \,a^{4} x^{4}}{4 \left (b \,x^{3}+a \right )}+\frac {10 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{7}}{7 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{10}}{b \,x^{3}+a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} a \,x^{13}}{13 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{16}}{16 b \,x^{3}+16 a}\) | \(174\) |
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Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{13} \, a b^{4} x^{13} + a^{2} b^{3} x^{10} + \frac {10}{7} \, a^{3} b^{2} x^{7} + \frac {5}{4} \, a^{4} b x^{4} + a^{5} x \]
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\[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int \left (a^{2} + 2 a b x^{3} + b^{2} x^{6}\right )^{\frac {5}{2}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{13} \, a b^{4} x^{13} + a^{2} b^{3} x^{10} + \frac {10}{7} \, a^{3} b^{2} x^{7} + \frac {5}{4} \, a^{4} b x^{4} + a^{5} x \]
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Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.41 \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{13} \, a b^{4} x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{2} b^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{7} \, a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{4} \, a^{4} b x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} x \mathrm {sgn}\left (b x^{3} + a\right ) \]
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Timed out. \[ \int \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int {\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]
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