Optimal. Leaf size=159 \[ -\frac {3 x^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};-a^2 x^2\right )}{1+m}+\frac {4 i a x^{2+m} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};-a^2 x^2\right )}{2+m} \]
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Rubi [A]
time = 0.66, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5168, 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 {i 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 i 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 5168
Rule 6874
Rubi steps
\begin {align*} \int e^{3 i \tan ^{-1}(a x)} x^m \, dx &=\int \frac {x^m (1+i a x)^2}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx\\ &=\int \left (-\frac {3 x^m}{\sqrt {1+a^2 x^2}}-\frac {i a x^{1+m}}{\sqrt {1+a^2 x^2}}+\frac {4 x^m}{(1-i 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-i a x) \sqrt {1+a^2 x^2}} \, dx-(i 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 {i 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+i 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 {i 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 i 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 {i 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 i 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 = 0.06, size = 113, normalized size = 0.71 \begin {gather*} -\frac {i x^{1+m} \sqrt {1-i a x} \sqrt {-i+a x} \left (F_1\left (1+m;-\frac {1}{2},\frac {1}{2};2+m;-i a x,i a x\right )-2 F_1\left (1+m;-\frac {1}{2},\frac {3}{2};2+m;-i a x,i a x\right )\right )}{(1+m) \sqrt {1+i a x} \sqrt {i+a x}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.08, size = 146, normalized size = 0.92
method | result | size |
meijerg | \(\frac {x^{1+m} \hypergeom \left (\left [\frac {3}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {3 i a \,x^{2+m} \hypergeom \left (\left [\frac {3}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m}-\frac {3 a^{2} x^{m +3} \hypergeom \left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {m}{2}+\frac {5}{2}\right ], -a^{2} x^{2}\right )}{m +3}-\frac {i a^{3} x^{4+m} \hypergeom \left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [\frac {m}{2}+3\right ], -a^{2} x^{2}\right )}{4+m}\) | \(146\) |
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*} - i \left (\int \frac {i x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \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\,x\,1{}\mathrm {i}\right )}^3}{{\left (a^2\,x^2+1\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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