Optimal. Leaf size=86 \[ \frac {15}{4} b^2 \sqrt {b x-a}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \]
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Rubi [A] time = 0.02, 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 = {47, 50, 63, 205} \begin {gather*} \frac {15}{4} b^2 \sqrt {b x-a}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )-\frac {(b x-a)^{5/2}}{2 x^2}-\frac {5 b (b x-a)^{3/2}}{4 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 47
Rule 50
Rule 63
Rule 205
Rubi steps
\begin {align*} \int \frac {(-a+b x)^{5/2}}{x^3} \, dx &=-\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{4} (5 b) \int \frac {(-a+b x)^{3/2}}{x^2} \, dx\\ &=-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}+\frac {1}{8} \left (15 b^2\right ) \int \frac {\sqrt {-a+b x}}{x} \, dx\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{8} \left (15 a b^2\right ) \int \frac {1}{x \sqrt {-a+b x}} \, dx\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {1}{4} (15 a b) \operatorname {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )\\ &=\frac {15}{4} b^2 \sqrt {-a+b x}-\frac {5 b (-a+b x)^{3/2}}{4 x}-\frac {(-a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )\\ \end {align*}
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Mathematica [C] time = 0.01, size = 38, normalized size = 0.44 \begin {gather*} \frac {2 b^2 (b x-a)^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};1-\frac {b x}{a}\right )}{7 a^3} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.09, size = 76, normalized size = 0.88 \begin {gather*} \frac {\sqrt {b x-a} \left (15 a^2+25 a (b x-a)+8 (b x-a)^2\right )}{4 x^2}-\frac {15}{4} \sqrt {a} b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.90, size = 139, normalized size = 1.62 \begin {gather*} \left [\frac {15 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, x^{2}}, -\frac {15 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - {\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, x^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.04, size = 83, normalized size = 0.97 \begin {gather*} -\frac {15 \, \sqrt {a} b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) - 8 \, \sqrt {b x - a} b^{3} - \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{3} + 7 \, \sqrt {b x - a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 70, normalized size = 0.81 \begin {gather*} -\frac {15 \sqrt {a}\, b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4}+2 \sqrt {b x -a}\, b^{2}+\frac {7 \sqrt {b x -a}\, a^{2}}{4 x^{2}}+\frac {9 \left (b x -a \right )^{\frac {3}{2}} a}{4 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.92, size = 97, normalized size = 1.13 \begin {gather*} -\frac {15}{4} \, \sqrt {a} b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + 2 \, \sqrt {b x - a} b^{2} + \frac {9 \, {\left (b x - a\right )}^{\frac {3}{2}} a b^{2} + 7 \, \sqrt {b x - a} a^{2} b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} + 2 \, {\left (b x - a\right )} a + a^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.09, size = 69, normalized size = 0.80 \begin {gather*} 2\,b^2\,\sqrt {b\,x-a}-\frac {15\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4}+\frac {9\,a\,{\left (b\,x-a\right )}^{3/2}}{4\,x^2}+\frac {7\,a^2\,\sqrt {b\,x-a}}{4\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.93, size = 267, normalized size = 3.10 \begin {gather*} \begin {cases} - \frac {15 i \sqrt {a} b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {i a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {11 i a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} - 1}} - \frac {2 i b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {15 \sqrt {a} b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} + \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {- \frac {a}{b x} + 1}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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