Integrand size = 29, antiderivative size = 153 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {b^2 (9 A b-7 a B)}{a^5 \sqrt {x}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}+\frac {b^{5/2} (9 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \]
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Time = 0.05 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {27, 79, 53, 65, 211} \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {b^{5/2} (9 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}}+\frac {b^2 (9 A b-7 a B)}{a^5 \sqrt {x}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {A b-a B}{a b x^{7/2} (a+b x)} \]
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Rule 27
Rule 53
Rule 65
Rule 79
Rule 211
Rubi steps \begin{align*} \text {integral}& = \int \frac {A+B x}{x^{9/2} (a+b x)^2} \, dx \\ & = \frac {A b-a B}{a b x^{7/2} (a+b x)}-\frac {\left (-\frac {9 A b}{2}+\frac {7 a B}{2}\right ) \int \frac {1}{x^{9/2} (a+b x)} \, dx}{a b} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}-\frac {(9 A b-7 a B) \int \frac {1}{x^{7/2} (a+b x)} \, dx}{2 a^2} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}+\frac {(b (9 A b-7 a B)) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{2 a^3} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}-\frac {\left (b^2 (9 A b-7 a B)\right ) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{2 a^4} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {b^2 (9 A b-7 a B)}{a^5 \sqrt {x}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}+\frac {\left (b^3 (9 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 a^5} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {b^2 (9 A b-7 a B)}{a^5 \sqrt {x}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}+\frac {\left (b^3 (9 A b-7 a B)\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{a^5} \\ & = -\frac {9 A b-7 a B}{7 a^2 b x^{7/2}}+\frac {9 A b-7 a B}{5 a^3 x^{5/2}}-\frac {b (9 A b-7 a B)}{3 a^4 x^{3/2}}+\frac {b^2 (9 A b-7 a B)}{a^5 \sqrt {x}}+\frac {A b-a B}{a b x^{7/2} (a+b x)}+\frac {b^{5/2} (9 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {945 A b^4 x^4+105 a b^3 x^3 (6 A-7 B x)-6 a^4 (5 A+7 B x)-14 a^2 b^2 x^2 (9 A+35 B x)+2 a^3 b x (27 A+49 B x)}{105 a^5 x^{7/2} (a+b x)}+\frac {b^{5/2} (9 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{11/2}} \]
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Time = 0.16 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {2 b^{3} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{2 \sqrt {b a}}\right )}{a^{5}}-\frac {2 A}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A b +B a \right )}{5 a^{3} x^{\frac {5}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{3 a^{4} x^{\frac {3}{2}}}+\frac {2 b^{2} \left (4 A b -3 B a \right )}{a^{5} \sqrt {x}}\) | \(121\) |
default | \(\frac {2 b^{3} \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{2 \sqrt {b a}}\right )}{a^{5}}-\frac {2 A}{7 a^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A b +B a \right )}{5 a^{3} x^{\frac {5}{2}}}-\frac {2 b \left (3 A b -2 B a \right )}{3 a^{4} x^{\frac {3}{2}}}+\frac {2 b^{2} \left (4 A b -3 B a \right )}{a^{5} \sqrt {x}}\) | \(121\) |
risch | \(-\frac {2 \left (-420 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+105 A a \,b^{2} x^{2}-70 B \,a^{2} b \,x^{2}-42 A \,a^{2} b x +21 a^{3} B x +15 A \,a^{3}\right )}{105 a^{5} x^{\frac {7}{2}}}+\frac {b^{3} \left (\frac {2 \left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (9 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{\sqrt {b a}}\right )}{a^{5}}\) | \(126\) |
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Time = 0.28 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.43 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [-\frac {105 \, {\left ({\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{210 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}, \frac {105 \, {\left ({\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{5} + {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x\right )} \sqrt {x}}{105 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}\right ] \]
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Timed out. \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Timed out} \]
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Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {30 \, A a^{4} + 105 \, {\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} x^{4} + 70 \, {\left (7 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{3} - 14 \, {\left (7 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{2} + 6 \, {\left (7 \, B a^{4} - 9 \, A a^{3} b\right )} x}{105 \, {\left (a^{5} b x^{\frac {9}{2}} + a^{6} x^{\frac {7}{2}}\right )}} - \frac {{\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} \]
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Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {{\left (7 \, B a b^{3} - 9 \, A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} - \frac {B a b^{3} \sqrt {x} - A b^{4} \sqrt {x}}{{\left (b x + a\right )} a^{5}} - \frac {2 \, {\left (315 \, B a b^{2} x^{3} - 420 \, A b^{3} x^{3} - 70 \, B a^{2} b x^{2} + 105 \, A a b^{2} x^{2} + 21 \, B a^{3} x - 42 \, A a^{2} b x + 15 \, A a^{3}\right )}}{105 \, a^{5} x^{\frac {7}{2}}} \]
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Time = 9.93 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{9/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\frac {2\,x\,\left (9\,A\,b-7\,B\,a\right )}{35\,a^2}-\frac {2\,A}{7\,a}+\frac {2\,b^2\,x^3\,\left (9\,A\,b-7\,B\,a\right )}{3\,a^4}+\frac {b^3\,x^4\,\left (9\,A\,b-7\,B\,a\right )}{a^5}-\frac {2\,b\,x^2\,\left (9\,A\,b-7\,B\,a\right )}{15\,a^3}}{a\,x^{7/2}+b\,x^{9/2}}+\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (9\,A\,b-7\,B\,a\right )}{a^{11/2}} \]
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