Optimal. Leaf size=150 \[ -\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}-\frac {4 a x^{2+m} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m} \]
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Rubi [A]
time = 0.59, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6259, 6874,
371, 864, 822} \begin {gather*} -\frac {3 x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2}-\frac {4 a x^{m+2} \, _2F_1\left (\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 822
Rule 864
Rule 6259
Rule 6874
Rubi steps
\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} x^m \, dx &=\int \frac {x^m (1-a x)^2}{(1+a x) \sqrt {1-a^2 x^2}} \, dx\\ &=\int \left (-\frac {3 x^m}{\sqrt {1-a^2 x^2}}+\frac {a x^{1+m}}{\sqrt {1-a^2 x^2}}+\frac {4 x^m}{(1+a x) \sqrt {1-a^2 x^2}}\right ) \, dx\\ &=-\left (3 \int \frac {x^m}{\sqrt {1-a^2 x^2}} \, dx\right )+4 \int \frac {x^m}{(1+a x) \sqrt {1-a^2 x^2}} \, dx+a \int \frac {x^{1+m}}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}+4 \int \frac {x^m (1-a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}+4 \int \frac {x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx-(4 a) \int \frac {x^{1+m}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{1+m}-\frac {4 a x^{2+m} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{2+m}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 0.04, size = 55, normalized size = 0.37 \begin {gather*} -\frac {x^{1+m} \left (F_1\left (1+m;-\frac {1}{2},\frac {1}{2};2+m;a x,-a x\right )-2 F_1\left (1+m;-\frac {1}{2},\frac {3}{2};2+m;a x,-a x\right )\right )}{1+m} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.76, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (a x +1\right )^{3}}\, dx\]
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} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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