Optimal. Leaf size=96 \[ -\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \]
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Rubi [A]
time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {49, 52, 65, 223,
212} \begin {gather*} \frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {a+b x}} \end {gather*}
Antiderivative was successfully verified.
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Rule 49
Rule 52
Rule 65
Rule 212
Rule 223
Rubi steps
\begin {align*} \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx &=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{b}\\ &=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}-\frac {(15 a) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{4 b^2}\\ &=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \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 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 76, normalized size = 0.79 \begin {gather*} \frac {\sqrt {x} \left (-15 a^2-5 a b x+2 b^2 x^2\right )}{4 b^3 \sqrt {a+b x}}-\frac {15 a^2 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{4 b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Mathics [A]
time = 7.15, size = 93, normalized size = 0.97 \begin {gather*} \frac {15 a^{\frac {5}{2}} b^6 \text {ArcSinh}\left [\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ] \left (\frac {a+b x}{a}\right )^{\frac {3}{2}}-15 a b^{\frac {13}{2}} \sqrt {x} \left (a+b x\right )-5 b^{\frac {15}{2}} x^{\frac {3}{2}} \left (a+b x\right )+\frac {2 b^{\frac {17}{2}} x^{\frac {5}{2}} \left (a+b x\right )}{a}}{4 \sqrt {a} b^{\frac {19}{2}} \left (\frac {a+b x}{a}\right )^{\frac {3}{2}}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.13, size = 119, normalized size = 1.24
method | result | size |
risch | \(-\frac {\left (-2 b x +7 a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{3}}+\frac {\left (\frac {15 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right )}{8 b^{\frac {7}{2}}}-\frac {2 a^{2} \sqrt {\left (x +\frac {a}{b}\right )^{2} b -a \left (x +\frac {a}{b}\right )}}{b^{4} \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\) | \(119\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.35, size = 131, normalized size = 1.36 \begin {gather*} -\frac {8 \, a^{2} b^{2} - \frac {25 \, {\left (b x + a\right )} a^{2} b}{x} + \frac {15 \, {\left (b x + a\right )}^{2} a^{2}}{x^{2}}}{4 \, {\left (\frac {\sqrt {b x + a} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + a\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} - \frac {15 \, a^{2} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\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 = 0.32, size = 175, normalized size = 1.82 \begin {gather*} \left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 5.74, size = 105, normalized size = 1.09 \begin {gather*} - \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}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.01, size = 132, normalized size = 1.38 \begin {gather*} 2 \left (\frac {2 \left (\left (\frac {\frac {1}{16}\cdot 2 b^{4} \sqrt {x} \sqrt {x}}{b^{5}}-\frac {\frac {1}{16}\cdot 5 b^{3} a}{b^{5}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{16}\cdot 15 b^{2} a^{2}}{b^{5}}\right ) \sqrt {x} \sqrt {a+b x}}{a+b x}-\frac {30 a^{2} \ln \left |\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right |}{16 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 (a+b\,x\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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