Integrand size = 31, antiderivative size = 118 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \]
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Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {784, 77} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^{3/2} (a+b x)}-\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \]
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Rule 77
Rule 784
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right ) (A+B x)}{x^{7/2}} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a A b}{x^{7/2}}+\frac {b (A b+a B)}{x^{5/2}}+\frac {b^2 B}{x^{3/2}}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {2 a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a A+5 A b x+5 a B x+15 b B x^2\right )}{15 x^{5/2} (a+b x)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right )}{15 x^{\frac {5}{2}}}\) | \(34\) |
gosper | \(-\frac {2 \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 x^{\frac {5}{2}} \left (b x +a \right )}\) | \(44\) |
risch | \(-\frac {2 \left (15 B b \,x^{2}+5 A b x +5 a B x +3 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{15 x^{\frac {5}{2}} \left (b x +a \right )}\) | \(44\) |
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Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.23 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]
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Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (3 \, b x^{2} + a x\right )} B}{3 \, x^{\frac {5}{2}}} - \frac {2 \, {\left (5 \, b x^{2} + 3 \, a x\right )} A}{15 \, x^{\frac {7}{2}}} \]
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Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a x \mathrm {sgn}\left (b x + a\right ) + 5 \, A b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, x^{\frac {5}{2}}} \]
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Time = 10.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (2\,B\,x^2+\frac {2\,A\,a}{5\,b}+\frac {x\,\left (10\,A\,b+10\,B\,a\right )}{15\,b}\right )}{x^{7/2}+\frac {a\,x^{5/2}}{b}} \]
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