Integrand size = 15, antiderivative size = 122 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {5 a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}} \]
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Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {52, 65, 223, 212} \[ \int x^{5/2} \sqrt {a+b x} \, dx=-\frac {5 a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}}+\frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x} \]
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Rule 52
Rule 65
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^{7/2} \sqrt {a+b x}+\frac {1}{8} a \int \frac {x^{5/2}}{\sqrt {a+b x}} \, dx \\ & = \frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^2\right ) \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{48 b} \\ & = -\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}+\frac {\left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{64 b^2} \\ & = \frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{128 b^3} \\ & = \frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{64 b^3} \\ & = \frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^3} \\ & = \frac {5 a^3 \sqrt {x} \sqrt {a+b x}}{64 b^3}-\frac {5 a^2 x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {a x^{5/2} \sqrt {a+b x}}{24 b}+\frac {1}{4} x^{7/2} \sqrt {a+b x}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (15 a^3-10 a^2 b x+8 a b^2 x^2+48 b^3 x^3\right )+30 a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{192 b^{7/2}} \]
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Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\frac {\left (48 b^{3} x^{3}+8 a \,b^{2} x^{2}-10 a^{2} b x +15 a^{3}\right ) \sqrt {x}\, \sqrt {b x +a}}{192 b^{3}}-\frac {5 a^{4} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\) | \(98\) |
default | \(\frac {x^{\frac {5}{2}} \left (b x +a \right )^{\frac {3}{2}}}{4 b}-\frac {5 a \left (\frac {x^{\frac {3}{2}} \left (b x +a \right )^{\frac {3}{2}}}{3 b}-\frac {a \left (\frac {\sqrt {x}\, \left (b x +a \right )^{\frac {3}{2}}}{2 b}-\frac {a \left (\sqrt {x}\, \sqrt {b x +a}+\frac {a \sqrt {x \left (b x +a \right )}\, \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b x +a}\, \sqrt {x}\, \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\) | \(128\) |
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Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.33 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\left [\frac {15 \, a^{4} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (48 \, b^{4} x^{3} + 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{4}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, b^{4} x^{3} + 8 \, a b^{3} x^{2} - 10 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{4}}\right ] \]
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Time = 29.00 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.25 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {1 + \frac {b x}{a}}} + \frac {7 \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} - \frac {5 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} + \frac {b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (88) = 176\).
Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.46 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\frac {5 \, a^{4} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {\frac {15 \, \sqrt {b x + a} a^{4} b^{3}}{\sqrt {x}} + \frac {73 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{192 \, {\left (b^{7} - \frac {4 \, {\left (b x + a\right )} b^{6}}{x} + \frac {6 \, {\left (b x + a\right )}^{2} b^{5}}{x^{2}} - \frac {4 \, {\left (b x + a\right )}^{3} b^{4}}{x^{3}} + \frac {{\left (b x + a\right )}^{4} b^{3}}{x^{4}}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (88) = 176\).
Time = 152.74 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.84 \[ \int x^{5/2} \sqrt {a+b x} \, dx=\frac {\frac {8 \, {\left (\frac {15 \, a^{3} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} - \frac {13 \, a}{b^{2}}\right )} + \frac {33 \, a^{2}}{b^{2}}\right )}\right )} a {\left | b \right |}}{b^{2}} - \frac {{\left (\frac {105 \, a^{4} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {5}{2}}} - {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3}} - \frac {25 \, a}{b^{3}}\right )} + \frac {163 \, a^{2}}{b^{3}}\right )} - \frac {279 \, a^{3}}{b^{3}}\right )} \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} {\left | b \right |}}{b}}{192 \, b} \]
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Timed out. \[ \int x^{5/2} \sqrt {a+b x} \, dx=\int x^{5/2}\,\sqrt {a+b\,x} \,d x \]
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