Integrand size = 30, antiderivative size = 295 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )} \]
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Time = 0.05 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1126, 276} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )} \]
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Rule 276
Rule 1126
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{(d x)^{7/2}} \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5 b^5}{(d x)^{7/2}}+\frac {5 a^4 b^6}{d^2 (d x)^{3/2}}+\frac {10 a^3 b^7 \sqrt {d x}}{d^4}+\frac {10 a^2 b^8 (d x)^{5/2}}{d^6}+\frac {5 a b^9 (d x)^{9/2}}{d^8}+\frac {b^{10} (d x)^{13/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{5/2} \left (231 a^5+5775 a^4 b x^2-3850 a^3 b^2 x^4-1650 a^2 b^3 x^6-525 a b^4 x^8-77 b^5 x^{10}\right )}{1155 (d x)^{7/2} \left (a+b x^2\right )^5} \]
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Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28
method | result | size |
gosper | \(-\frac {2 x \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1155 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {7}{2}}}\) | \(83\) |
default | \(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right )}{1155 d \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {5}{2}}}\) | \(85\) |
risch | \(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right )}{1155 d^{3} \left (b \,x^{2}+a \right ) x^{2} \sqrt {d x}}\) | \(88\) |
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Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (77 \, b^{5} x^{10} + 525 \, a b^{4} x^{8} + 1650 \, a^{2} b^{3} x^{6} + 3850 \, a^{3} b^{2} x^{4} - 5775 \, a^{4} b x^{2} - 231 \, a^{5}\right )} \sqrt {d x}}{1155 \, d^{4} x^{3}} \]
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\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{\left (d x\right )^{\frac {7}{2}}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (7 \, {\left (11 \, b^{5} \sqrt {d} x^{3} + 15 \, a b^{4} \sqrt {d} x\right )} x^{\frac {9}{2}} + 60 \, {\left (7 \, a b^{4} \sqrt {d} x^{3} + 11 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {5}{2}} + 330 \, {\left (3 \, a^{2} b^{3} \sqrt {d} x^{3} + 7 \, a^{3} b^{2} \sqrt {d} x\right )} \sqrt {x} + \frac {1540 \, {\left (a^{3} b^{2} \sqrt {d} x^{3} - 3 \, a^{4} b \sqrt {d} x\right )}}{x^{\frac {3}{2}}} - \frac {231 \, {\left (5 \, a^{4} b \sqrt {d} x^{3} + a^{5} \sqrt {d} x\right )}}{x^{\frac {7}{2}}}\right )}}{1155 \, d^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {231 \, {\left (25 \, a^{4} b d^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} d^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{\sqrt {d x} d^{2} x^{2}} - \frac {77 \, \sqrt {d x} b^{5} d^{105} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 525 \, \sqrt {d x} a b^{4} d^{105} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 1650 \, \sqrt {d x} a^{2} b^{3} d^{105} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 3850 \, \sqrt {d x} a^{3} b^{2} d^{105} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{105}}\right )}}{1155 \, d^{4}} \]
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Time = 13.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,b^4\,x^{10}}{15\,d^3}-\frac {10\,a^4\,x^2}{d^3}-\frac {2\,a^5}{5\,b\,d^3}+\frac {20\,a^3\,b\,x^4}{3\,d^3}+\frac {10\,a\,b^3\,x^8}{11\,d^3}+\frac {20\,a^2\,b^2\,x^6}{7\,d^3}\right )}{x^4\,\sqrt {d\,x}+\frac {a\,x^2\,\sqrt {d\,x}}{b}} \]
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