Optimal. Leaf size=150 \[ -\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}} \]
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Rubi [A]
time = 0.03, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222}
\begin {gather*} \frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{8} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{24 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 52
Rule 56
Rule 222
Rubi steps
\begin {align*} \int x^{5/2} (2-b x)^{5/2} \, dx &=\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5}{6} \int x^{5/2} (2-b x)^{3/2} \, dx\\ &=\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{2} \int x^{5/2} \sqrt {2-b x} \, dx\\ &=\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{8} \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx\\ &=-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{24 b}\\ &=-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{16 b^2}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{16 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 98, normalized size = 0.65 \begin {gather*} \frac {\sqrt {x} \sqrt {2-b x} \left (-15-5 b x-2 b^2 x^2+54 b^3 x^3-40 b^4 x^4+8 b^5 x^5\right )}{48 b^3}+\frac {5 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{8 (-b)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in
optimal.
time = 170.18, size = 253, normalized size = 1.69 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I \left (-30 b^6 \text {ArcCosh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (-2+b x\right )^2+30 b^{\frac {13}{2}} \sqrt {x} \left (-2+b x\right )^{\frac {3}{2}}-5 b^{\frac {15}{2}} x^{\frac {3}{2}} \left (-2+b x\right )^{\frac {3}{2}}-b^{\frac {17}{2}} x^{\frac {5}{2}} \left (-2+b x\right )^{\frac {3}{2}}+2 b^{\frac {19}{2}} x^{\frac {7}{2}} \left (-55+67 b x-28 b^2 x^2+4 b^3 x^3\right ) \left (-2+b x\right )^{\frac {3}{2}}\right )}{48 b^{\frac {19}{2}} \left (-2+b x\right )^2},\text {Abs}\left [b x\right ]>2\right \}\right \},\frac {5 \text {ArcSin}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ]}{8 b^{\frac {7}{2}}}-\frac {5 \sqrt {x}}{8 b^3 \sqrt {2-b x}}+\frac {5 x^{\frac {3}{2}}}{48 b^2 \sqrt {2-b x}}+\frac {x^{\frac {5}{2}}}{48 b \sqrt {2-b x}}+\frac {55 x^{\frac {7}{2}}}{24 \sqrt {2-b x}}-\frac {67 b x^{\frac {9}{2}}}{24 \sqrt {2-b x}}+\frac {7 b^2 x^{\frac {11}{2}}}{6 \sqrt {2-b x}}-\frac {b^3 x^{\frac {13}{2}}}{6 \sqrt {2-b x}}\right ] \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.13, size = 157, normalized size = 1.05
method | result | size |
meijerg | \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (-56 b^{5} x^{5}+280 b^{4} x^{4}-378 b^{3} x^{3}+14 x^{2} b^{2}+35 b x +105\right ) \sqrt {-\frac {b x}{2}+1}}{336 b^{3}}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{8 b^{\frac {7}{2}}}}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\) | \(105\) |
risch | \(-\frac {\left (8 b^{5} x^{5}-40 b^{4} x^{4}+54 b^{3} x^{3}-2 x^{2} b^{2}-5 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{48 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +2}}\) | \(131\) |
default | \(-\frac {x^{\frac {5}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{6 b}+\frac {-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{6 b}+\frac {5 \left (-\frac {3 \sqrt {x}\, \left (-b x +2\right )^{\frac {7}{2}}}{20 b}+\frac {3 \left (\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\right )}{20 b}\right )}{6 b}}{b}\) | \(157\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.37, size = 209, normalized size = 1.39 \begin {gather*} \frac {\frac {15 \, \sqrt {-b x + 2} b^{5}}{\sqrt {x}} + \frac {85 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}} - \frac {198 \, {\left (-b x + 2\right )}^{\frac {7}{2}} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (-b x + 2\right )}^{\frac {9}{2}} b}{x^{\frac {9}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {11}{2}}}{x^{\frac {11}{2}}}}{24 \, {\left (b^{9} - \frac {6 \, {\left (b x - 2\right )} b^{8}}{x} + \frac {15 \, {\left (b x - 2\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x - 2\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x - 2\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x - 2\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x - 2\right )}^{6} b^{3}}{x^{6}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{8 \, b^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.33, size = 173, normalized size = 1.15 \begin {gather*} \left [\frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{48 \, b^{4}}, \frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{48 \, b^{4}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 242 vs.
\(2 (105) = 210\).
time = 0.02, size = 522, normalized size = 3.48 \begin {gather*} 2 b^{2} \left (2 \left (\left (\left (\left (\left (\frac {\frac {1}{5443200}\cdot 226800 b^{10} \sqrt {x} \sqrt {x}}{b^{10}}-\frac {\frac {1}{5443200}\cdot 45360 b^{9}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 102060 b^{8}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 238140 b^{7}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 595350 b^{6}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 1786050 b^{5}}{b^{10}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {21 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{16 b^{5} \sqrt {-b}}\right )-8 b \left (2 \left (\left (\left (\left (\frac {\frac {1}{100800}\cdot 5040 b^{8} \sqrt {x} \sqrt {x}}{b^{8}}-\frac {\frac {1}{100800}\cdot 1260 b^{7}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 2940 b^{6}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 7350 b^{5}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 22050 b^{4}}{b^{8}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {7 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{8 b^{4} \sqrt {-b}}\right )+8 \left (2 \left (\left (\left (\frac {\frac {1}{2880}\cdot 180 b^{6} \sqrt {x} \sqrt {x}}{b^{6}}-\frac {\frac {1}{2880}\cdot 60 b^{5}}{b^{6}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{2880}\cdot 150 b^{4}}{b^{6}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{2880}\cdot 450 b^{3}}{b^{6}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {5 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{8 b^{3} \sqrt {-b}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,{\left (2-b\,x\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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