Integrand size = 31, antiderivative size = 316 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=-\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {10 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {20 a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {10 a b^3 (A b+2 a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^4 (A b+5 a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^5 B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]
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Time = 0.08 (sec) , antiderivative size = 316, 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} \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=-\frac {20 a^2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt {x} (a+b x)}+\frac {2 b^4 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{3 (a+b x)}+\frac {10 a b^3 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{a+b x}+\frac {2 b^5 B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}-\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 a^4 \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{5 x^{5/2} (a+b x)}-\frac {10 a^3 b \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{3 x^{3/2} (a+b x)} \]
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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 )^5 (A+B x)}{x^{9/2}} \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5 A b^5}{x^{9/2}}+\frac {a^4 b^5 (5 A b+a B)}{x^{7/2}}+\frac {5 a^3 b^6 (2 A b+a B)}{x^{5/2}}+\frac {10 a^2 b^7 (A b+a B)}{x^{3/2}}+\frac {5 a b^8 (A b+2 a B)}{\sqrt {x}}+b^9 (A b+5 a B) \sqrt {x}+b^{10} B x^{3/2}\right ) \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = -\frac {2 a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^{7/2} (a+b x)}-\frac {2 a^4 (5 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac {10 a^3 b (2 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {20 a^2 b^2 (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {10 a b^3 (A b+2 a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^4 (A b+5 a B) x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {2 b^5 B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (1050 a^2 b^3 x^3 (A-B x)-175 a b^4 x^4 (3 A+B x)+350 a^3 b^2 x^2 (A+3 B x)-7 b^5 x^5 (5 A+3 B x)+35 a^4 b x (3 A+5 B x)+3 a^5 (5 A+7 B x)\right )}{105 x^{7/2} (a+b x)} \]
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Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(-\frac {2 \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 a^{5} B x +15 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )^{5}}\) | \(140\) |
default | \(-\frac {2 \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 a^{5} B x +15 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{105 x^{\frac {7}{2}} \left (b x +a \right )^{5}}\) | \(140\) |
risch | \(-\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (-21 B \,b^{5} x^{6}-35 A \,b^{5} x^{5}-175 B a \,b^{4} x^{5}-525 A a \,b^{4} x^{4}-1050 B \,a^{2} b^{3} x^{4}+1050 A \,a^{2} b^{3} x^{3}+1050 B \,a^{3} b^{2} x^{3}+350 A \,a^{3} b^{2} x^{2}+175 B \,a^{4} b \,x^{2}+105 A \,a^{4} b x +21 a^{5} B x +15 A \,a^{5}\right )}{105 \left (b x +a \right ) x^{\frac {7}{2}}}\) | \(140\) |
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Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2 \, {\left (21 \, B b^{5} x^{6} - 15 \, A a^{5} + 35 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 525 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 1050 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 175 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 21 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )}}{105 \, x^{\frac {7}{2}}} \]
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\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{\frac {9}{2}}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2}{15} \, {\left ({\left (3 \, b^{5} x^{2} + 5 \, a b^{4} x\right )} \sqrt {x} + \frac {20 \, {\left (a b^{4} x^{2} + 3 \, a^{2} b^{3} x\right )}}{\sqrt {x}} + \frac {90 \, {\left (a^{2} b^{3} x^{2} - a^{3} b^{2} x\right )}}{x^{\frac {3}{2}}} - \frac {20 \, {\left (3 \, a^{3} b^{2} x^{2} + a^{4} b x\right )}}{x^{\frac {5}{2}}} - \frac {5 \, a^{4} b x^{2} + 3 \, a^{5} x}{x^{\frac {7}{2}}}\right )} B + \frac {2}{105} \, A {\left (\frac {35 \, {\left (b^{5} x^{2} + 3 \, a b^{4} x\right )}}{\sqrt {x}} + \frac {420 \, {\left (a b^{4} x^{2} - a^{2} b^{3} x\right )}}{x^{\frac {3}{2}}} - \frac {210 \, {\left (3 \, a^{2} b^{3} x^{2} + a^{3} b^{2} x\right )}}{x^{\frac {5}{2}}} - \frac {28 \, {\left (5 \, a^{3} b^{2} x^{2} + 3 \, a^{4} b x\right )}}{x^{\frac {7}{2}}} - \frac {3 \, {\left (7 \, a^{4} b x^{2} + 5 \, a^{5} x\right )}}{x^{\frac {9}{2}}}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\frac {2}{5} \, B b^{5} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, B a b^{4} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A b^{5} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 20 \, B a^{2} b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 10 \, A a b^{4} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (1050 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1050 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 175 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 350 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 21 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{5} \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, x^{\frac {7}{2}}} \]
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Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{9/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^{9/2}} \,d x \]
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