Optimal. Leaf size=15 \[ \frac {x^m}{\sqrt {a+b x^2}} \]
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Rubi [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.05, antiderivative size = 123, normalized size of antiderivative = 8.20, number of steps
used = 5, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {372, 371}
\begin {gather*} \frac {x^m \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};-\frac {b x^2}{a}\right )}{\sqrt {a+b x^2}}-\frac {b x^{m+2} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};-\frac {b x^2}{a}\right )}{a (m+2) \sqrt {a+b x^2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 372
Rubi steps
\begin {align*} \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx &=-\left (b \int \frac {x^{1+m}}{\left (a+b x^2\right )^{3/2}} \, dx\right )+m \int \frac {x^{-1+m}}{\sqrt {a+b x^2}} \, dx\\ &=-\frac {\left (b \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {x^{1+m}}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \, dx}{a \sqrt {a+b x^2}}+\frac {\left (m \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {x^{-1+m}}{\sqrt {1+\frac {b x^2}{a}}} \, dx}{\sqrt {a+b x^2}}\\ &=\frac {x^m \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {2+m}{2};-\frac {b x^2}{a}\right )}{\sqrt {a+b x^2}}-\frac {b x^{2+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};-\frac {b x^2}{a}\right )}{a (2+m) \sqrt {a+b x^2}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.04, size = 131, normalized size = 8.73 \begin {gather*} \frac {x^m \sqrt {a+b x^2} \left (a (2+m) \, _2F_1\left (-\frac {1}{2},\frac {m}{2};1+\frac {m}{2};-\frac {b x^2}{a}\right )-b x^2 \left (m \, _2F_1\left (\frac {1}{2},1+\frac {m}{2};2+\frac {m}{2};-\frac {b x^2}{a}\right )+\, _2F_1\left (\frac {3}{2},1+\frac {m}{2};2+\frac {m}{2};-\frac {b x^2}{a}\right )\right )\right )}{a^2 (2+m) \sqrt {1+\frac {b x^2}{a}}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int -\frac {b \,x^{1+m}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {m \,x^{-1+m}}{\sqrt {b \,x^{2}+a}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.33, size = 13, normalized size = 0.87 \begin {gather*} \frac {x^{m}}{\sqrt {b x^{2} + a}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.54, size = 26, normalized size = 1.73 \begin {gather*} \frac {\sqrt {b x^{2} + a} x^{m + 1}}{b x^{3} + a x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] Result contains complex when optimal does not.
time = 2.88, size = 94, normalized size = 6.27 \begin {gather*} \frac {m x^{m} \Gamma \left (\frac {m}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} \\ \frac {m}{2} + 1 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + 1\right )} - \frac {b x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 = -0.07 \begin {gather*} -\int \frac {b\,x^{m+1}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {m\,x^{m-1}}{\sqrt {b\,x^2+a}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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