Integrand size = 22, antiderivative size = 100 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac {(3 b B+A c) \sqrt {x}}{4 b c^2 (b+c x)}+\frac {(3 b B+A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{3/2} c^{5/2}} \]
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Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {795, 79, 43, 65, 211} \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {(A c+3 b B) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{3/2} c^{5/2}}-\frac {\sqrt {x} (A c+3 b B)}{4 b c^2 (b+c x)}-\frac {x^{3/2} (b B-A c)}{2 b c (b+c x)^2} \]
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Rule 43
Rule 65
Rule 79
Rule 211
Rule 795
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {x} (A+B x)}{(b+c x)^3} \, dx \\ & = -\frac {(b B-A c) x^{3/2}}{2 b c (b+c x)^2}+\frac {(3 b B+A c) \int \frac {\sqrt {x}}{(b+c x)^2} \, dx}{4 b c} \\ & = -\frac {(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac {(3 b B+A c) \sqrt {x}}{4 b c^2 (b+c x)}+\frac {(3 b B+A c) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b c^2} \\ & = -\frac {(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac {(3 b B+A c) \sqrt {x}}{4 b c^2 (b+c x)}+\frac {(3 b B+A c) \text {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b c^2} \\ & = -\frac {(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac {(3 b B+A c) \sqrt {x}}{4 b c^2 (b+c x)}+\frac {(3 b B+A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{3/2} c^{5/2}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {\sqrt {x} \left (3 b^2 B+A b c+5 b B c x-A c^2 x\right )}{4 b c^2 (b+c x)^2}+\frac {(3 b B+A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{3/2} c^{5/2}} \]
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Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {\frac {\left (A c -5 B b \right ) x^{\frac {3}{2}}}{4 b c}-\frac {\left (A c +3 B b \right ) \sqrt {x}}{4 c^{2}}}{\left (c x +b \right )^{2}}+\frac {\left (A c +3 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 c^{2} b \sqrt {b c}}\) | \(79\) |
default | \(\frac {\frac {\left (A c -5 B b \right ) x^{\frac {3}{2}}}{4 b c}-\frac {\left (A c +3 B b \right ) \sqrt {x}}{4 c^{2}}}{\left (c x +b \right )^{2}}+\frac {\left (A c +3 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 c^{2} b \sqrt {b c}}\) | \(79\) |
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Time = 0.32 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.91 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\left [-\frac {{\left (3 \, B b^{3} + A b^{2} c + {\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b - 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) + 2 \, {\left (3 \, B b^{3} c + A b^{2} c^{2} + {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (b^{2} c^{5} x^{2} + 2 \, b^{3} c^{4} x + b^{4} c^{3}\right )}}, -\frac {{\left (3 \, B b^{3} + A b^{2} c + {\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (3 \, B b^{3} c + A b^{2} c^{2} + {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (b^{2} c^{5} x^{2} + 2 \, b^{3} c^{4} x + b^{4} c^{3}\right )}}\right ] \]
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Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\text {Timed out} \]
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Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {{\left (5 \, B b c - A c^{2}\right )} x^{\frac {3}{2}} + {\left (3 \, B b^{2} + A b c\right )} \sqrt {x}}{4 \, {\left (b c^{4} x^{2} + 2 \, b^{2} c^{3} x + b^{3} c^{2}\right )}} + \frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b c^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b c^{2}} - \frac {5 \, B b c x^{\frac {3}{2}} - A c^{2} x^{\frac {3}{2}} + 3 \, B b^{2} \sqrt {x} + A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b c^{2}} \]
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Time = 10.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int \frac {x^{7/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,b^{3/2}\,c^{5/2}}-\frac {\frac {\sqrt {x}\,\left (A\,c+3\,B\,b\right )}{4\,c^2}-\frac {x^{3/2}\,\left (A\,c-5\,B\,b\right )}{4\,b\,c}}{b^2+2\,b\,c\,x+c^2\,x^2} \]
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