Optimal. Leaf size=169 \[ \frac {2 x^{1+m} (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {(1+2 m) x^{1+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{(1+m) \sqrt {c-a^2 c x^2}}-\frac {2 a (1+m) x^{2+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{(2+m) \sqrt {c-a^2 c x^2}} \]
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Rubi [A]
time = 0.20, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6286, 1820,
822, 372, 371} \begin {gather*} -\frac {(2 m+1) \sqrt {1-a^2 x^2} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{(m+1) \sqrt {c-a^2 c x^2}}-\frac {2 a (m+1) \sqrt {1-a^2 x^2} x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{(m+2) \sqrt {c-a^2 c x^2}}+\frac {2 (a x+1) x^{m+1}}{\sqrt {c-a^2 c x^2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 372
Rule 822
Rule 1820
Rule 6286
Rubi steps
\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)} x^m}{\sqrt {c-a^2 c x^2}} \, dx &=c \int \frac {x^m (1+a x)^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {2 x^{1+m} (1+a x)}{\sqrt {c-a^2 c x^2}}-\int \frac {x^m (1+2 m+2 a (1+m) x)}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {2 x^{1+m} (1+a x)}{\sqrt {c-a^2 c x^2}}-(2 a (1+m)) \int \frac {x^{1+m}}{\sqrt {c-a^2 c x^2}} \, dx-(1+2 m) \int \frac {x^m}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {2 x^{1+m} (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\left (2 a (1+m) \sqrt {1-a^2 x^2}\right ) \int \frac {x^{1+m}}{\sqrt {1-a^2 x^2}} \, dx}{\sqrt {c-a^2 c x^2}}-\frac {\left ((1+2 m) \sqrt {1-a^2 x^2}\right ) \int \frac {x^m}{\sqrt {1-a^2 x^2}} \, dx}{\sqrt {c-a^2 c x^2}}\\ &=\frac {2 x^{1+m} (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {(1+2 m) x^{1+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{(1+m) \sqrt {c-a^2 c x^2}}-\frac {2 a (1+m) x^{2+m} \sqrt {1-a^2 x^2} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{(2+m) \sqrt {c-a^2 c x^2}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 0.08, size = 66, normalized size = 0.39 \begin {gather*} \frac {x^{1+m} \sqrt {1-a^2 x^2} F_1\left (1+m;\frac {3}{2},-\frac {1}{2};2+m;a x,-a x\right )}{(1+m) \sqrt {-1+a x} \sqrt {-c (1+a x)}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +1\right )^{2} x^{m}}{\left (-a^{2} x^{2}+1\right ) \sqrt {-a^{2} c \,x^{2}+c}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x^{m}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x x^{m}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx \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 = -0.01 \begin {gather*} \int -\frac {x^m\,{\left (a\,x+1\right )}^2}{\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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