Optimal. Leaf size=86 \[ -\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]
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Rubi [A]
time = 0.01, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 52, 56, 221}
\begin {gather*} \frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {15 \sqrt {x} \sqrt {b x+2}}{2 b^3}+\frac {5 x^{3/2} \sqrt {b x+2}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {b x+2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 49
Rule 52
Rule 56
Rule 221
Rubi steps
\begin {align*} \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{b}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}-\frac {15 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{2 b^2}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{2 b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.09, size = 65, normalized size = 0.76 \begin {gather*} \frac {\sqrt {x} \left (-30-5 b x+b^2 x^2\right )}{2 b^3 \sqrt {2+b x}}-\frac {15 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Mathics [A]
time = 6.71, size = 75, normalized size = 0.87 \begin {gather*} \frac {30 b^6 \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (2+b x\right )^{\frac {3}{2}}-30 b^{\frac {13}{2}} \sqrt {x} \left (2+b x\right )-5 b^{\frac {15}{2}} x^{\frac {3}{2}} \left (2+b x\right )+b^{\frac {17}{2}} x^{\frac {5}{2}} \left (2+b x\right )}{2 b^{\frac {19}{2}} \left (2+b x\right )^{\frac {3}{2}}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.13, size = 63, normalized size = 0.73
method | result | size |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-\frac {7}{2} x^{2} b^{2}+\frac {35}{2} b x +105\right )}{14 \sqrt {\frac {b x}{2}+1}}+15 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\) | \(63\) |
risch | \(\frac {\left (b x -7\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b^{3}}+\frac {\left (\frac {15 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{2 b^{\frac {7}{2}}}-\frac {8 \sqrt {\left (x +\frac {2}{b}\right )^{2} b -2 x -\frac {4}{b}}}{b^{4} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\) | \(106\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.35, size = 119, normalized size = 1.38 \begin {gather*} -\frac {8 \, b^{2} - \frac {25 \, {\left (b x + 2\right )} b}{x} + \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} - \frac {15 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, 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 = 152, normalized size = 1.77 \begin {gather*} \left [\frac {15 \, {\left (b x + 2\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}, -\frac {30 \, {\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 5.27, size = 80, normalized size = 0.93 \begin {gather*} \frac {x^{\frac {5}{2}}}{2 b \sqrt {b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x + 2}} - \frac {15 \sqrt {x}}{b^{3} \sqrt {b x + 2}} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.01, size = 123, normalized size = 1.43 \begin {gather*} 2 \left (\frac {2 \left (\left (\frac {\frac {1}{8} b^{4} \sqrt {x} \sqrt {x}}{b^{5}}-\frac {\frac {1}{8}\cdot 5 b^{3}}{b^{5}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{8}\cdot 30 b^{2}}{b^{5}}\right ) \sqrt {x} \sqrt {b x+2}}{b x+2}-\frac {15 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{2 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 \frac {x^{5/2}}{{\left (b\,x+2\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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