Integrand size = 16, antiderivative size = 67 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=-\frac {2 a (A b-a B) (a+b x)^{5/2}}{5 b^3}+\frac {2 (A b-2 a B) (a+b x)^{7/2}}{7 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \]
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Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 (a+b x)^{7/2} (A b-2 a B)}{7 b^3}-\frac {2 a (a+b x)^{5/2} (A b-a B)}{5 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a (-A b+a B) (a+b x)^{3/2}}{b^2}+\frac {(A b-2 a B) (a+b x)^{5/2}}{b^2}+\frac {B (a+b x)^{7/2}}{b^2}\right ) \, dx \\ & = -\frac {2 a (A b-a B) (a+b x)^{5/2}}{5 b^3}+\frac {2 (A b-2 a B) (a+b x)^{7/2}}{7 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 (a+b x)^{5/2} \left (8 a^2 B+5 b^2 x (9 A+7 B x)-2 a b (9 A+10 B x)\right )}{315 b^3} \]
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Time = 0.52 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {4 \left (-\frac {5 x \left (\frac {7 B x}{9}+A \right ) b^{2}}{2}+a \left (\frac {10 B x}{9}+A \right ) b -\frac {4 a^{2} B}{9}\right ) \left (b x +a \right )^{\frac {5}{2}}}{35 b^{3}}\) | \(41\) |
gosper | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-35 b^{2} B \,x^{2}-45 A \,b^{2} x +20 B a b x +18 a b A -8 a^{2} B \right )}{315 b^{3}}\) | \(47\) |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{3}}\) | \(52\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{3}}\) | \(52\) |
trager | \(-\frac {2 \left (-35 B \,x^{4} b^{4}-45 A \,x^{3} b^{4}-50 B \,x^{3} a \,b^{3}-72 A \,x^{2} a \,b^{3}-3 B \,x^{2} a^{2} b^{2}-9 A x \,a^{2} b^{2}+4 B x \,a^{3} b +18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{3}}\) | \(95\) |
risch | \(-\frac {2 \left (-35 B \,x^{4} b^{4}-45 A \,x^{3} b^{4}-50 B \,x^{3} a \,b^{3}-72 A \,x^{2} a \,b^{3}-3 B \,x^{2} a^{2} b^{2}-9 A x \,a^{2} b^{2}+4 B x \,a^{3} b +18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{3}}\) | \(95\) |
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Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (35 \, B b^{4} x^{4} + 8 \, B a^{4} - 18 \, A a^{3} b + 5 \, {\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{3} + 3 \, {\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{2} - {\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x + a}}{315 \, b^{3}} \]
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Time = 0.65 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (A b - 2 B a\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- A a b + B a^{2}\right )}{5 b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} B - 45 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 63 \, {\left (B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{315 \, b^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (56) = 112\).
Time = 0.28 (sec) , antiderivative size = 275, normalized size of antiderivative = 4.10 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (\frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A a^{2}}{b} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B a^{2}}{b^{2}} + \frac {42 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A a}{b} + \frac {18 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B a}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A}{b} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B}{b^{2}}\right )}}{315 \, b} \]
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Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2\,{\left (a+b\,x\right )}^{5/2}\,\left (63\,B\,a^2+35\,B\,{\left (a+b\,x\right )}^2-63\,A\,a\,b+45\,A\,b\,\left (a+b\,x\right )-90\,B\,a\,\left (a+b\,x\right )\right )}{315\,b^3} \]
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