Integrand size = 20, antiderivative size = 134 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}-\frac {3 a (4 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \]
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Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {3 a (4 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}+\frac {3 \sqrt {x} \sqrt {a+b x} (4 A b-5 a B)}{4 b^3}-\frac {x^{3/2} \sqrt {a+b x} (4 A b-5 a B)}{2 a b^2}+\frac {2 x^{5/2} (A b-a B)}{a b \sqrt {a+b x}} \]
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Rule 52
Rule 65
Rule 79
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}-\frac {\left (2 \left (2 A b-\frac {5 a B}{2}\right )\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{a b} \\ & = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}+\frac {(3 (4 A b-5 a B)) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{4 b^2} \\ & = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}-\frac {(3 a (4 A b-5 a B)) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^3} \\ & = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}-\frac {(3 a (4 A b-5 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}-\frac {(3 a (4 A b-5 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^3} \\ & = \frac {2 (A b-a B) x^{5/2}}{a b \sqrt {a+b x}}+\frac {3 (4 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{4 b^3}-\frac {(4 A b-5 a B) x^{3/2} \sqrt {a+b x}}{2 a b^2}-\frac {3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \\ \end{align*}
Time = 0.58 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-15 a^2 B+a b (12 A-5 B x)+2 b^2 x (2 A+B x)\right )}{4 b^3 \sqrt {a+b x}}+\frac {3 a (-4 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{7/2}} \]
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Time = 1.47 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {\left (2 b B x +4 A b -7 B a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{3}}-\frac {a \left (12 A \sqrt {b}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {15 B a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-\frac {16 \left (A b -B a \right ) \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{3} \sqrt {x}\, \sqrt {b x +a}}\) | \(164\) |
default | \(-\frac {\left (-4 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+12 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{2} x -8 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b x +10 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a x +12 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b -24 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a -15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+30 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{2}\right ) \sqrt {x}}{8 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b x +a}}\) | \(244\) |
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Time = 0.23 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.90 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{3} - 4 \, A a^{2} b + {\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} - {\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]
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Time = 15.11 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.36 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=A \left (\frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) + B \left (- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (108) = 216\).
Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.84 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}} - \frac {3 \, \sqrt {b x^{2} + a x} B a^{2}}{b^{4} x + a b^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, {\left (b^{3} x + a b^{2}\right )}} + \frac {3 \, \sqrt {b x^{2} + a x} A a}{b^{3} x + a b^{2}} + \frac {15 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} B a}{4 \, b^{3}} \]
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Time = 15.50 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28 \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B {\left | b \right |}}{b^{5}} - \frac {9 \, B a b^{9} {\left | b \right |} - 4 \, A b^{10} {\left | b \right |}}{b^{14}}\right )} - \frac {3 \, {\left (5 \, B a^{2} {\left | b \right |} - 4 \, A a b {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{8 \, b^{\frac {9}{2}}} - \frac {4 \, {\left (B a^{3} {\left | b \right |} - A a^{2} b {\left | b \right |}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {7}{2}}} \]
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Timed out. \[ \int \frac {x^{3/2} (A+B x)}{(a+b x)^{3/2}} \, dx=\int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^{3/2}} \,d x \]
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