Optimal. Leaf size=76 \[ \frac {x^{m+1} \text {Chi}(b x)}{m+1}-\frac {x^m (-b x)^{-m} \Gamma (m+1,-b x)}{2 b (m+1)}+\frac {x^m (b x)^{-m} \Gamma (m+1,b x)}{2 b (m+1)} \]
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Rubi [A] time = 0.08, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6533, 12, 3307, 2181} \[ -\frac {x^m (-b x)^{-m} \text {Gamma}(m+1,-b x)}{2 b (m+1)}+\frac {x^m (b x)^{-m} \text {Gamma}(m+1,b x)}{2 b (m+1)}+\frac {x^{m+1} \text {Chi}(b x)}{m+1} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2181
Rule 3307
Rule 6533
Rubi steps
\begin {align*} \int x^m \text {Chi}(b x) \, dx &=\frac {x^{1+m} \text {Chi}(b x)}{1+m}-\frac {b \int \frac {x^m \cosh (b x)}{b} \, dx}{1+m}\\ &=\frac {x^{1+m} \text {Chi}(b x)}{1+m}-\frac {\int x^m \cosh (b x) \, dx}{1+m}\\ &=\frac {x^{1+m} \text {Chi}(b x)}{1+m}-\frac {\int e^{-b x} x^m \, dx}{2 (1+m)}-\frac {\int e^{b x} x^m \, dx}{2 (1+m)}\\ &=\frac {x^{1+m} \text {Chi}(b x)}{1+m}-\frac {x^m (-b x)^{-m} \Gamma (1+m,-b x)}{2 b (1+m)}+\frac {x^m (b x)^{-m} \Gamma (1+m,b x)}{2 b (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 74, normalized size = 0.97 \[ \frac {x^{m+1} \text {Chi}(b x)}{m+1}-\frac {-x^{m+1} (-b x)^{-m-1} \Gamma (m+1,-b x)-x^{m+1} (b x)^{-m-1} \Gamma (m+1,b x)}{2 (m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \operatorname {Chi}\left (b x\right ), x\right ) \]
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 \[ \int x^{m} {\rm Chi}\left (b x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[ \int x^{m} \Chi \left (b x \right )\, dx \]
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 \[ \int x^{m} {\rm Chi}\left (b x\right )\,{d x} \]
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 \[ \int x^m\,\mathrm {coshint}\left (b\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.56, size = 649, normalized size = 8.54 \[ \frac {4 \cdot 2^{m} b^{- m} m x \sqrt {e^{- 2 m \log {\relax (2 )}} e^{m \log {\left (b^{2} x^{2} \right )}}} \log {\left (b^{2} x^{2} \right )} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {8 \cdot 2^{m} \gamma b^{- m} m x \sqrt {e^{- 2 m \log {\relax (2 )}} e^{m \log {\left (b^{2} x^{2} \right )}}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {4 \cdot 2^{m} b^{- m} x \sqrt {e^{- 2 m \log {\relax (2 )}} e^{m \log {\left (b^{2} x^{2} \right )}}} \log {\left (b^{2} x^{2} \right )} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} - \frac {8 \cdot 2^{m} b^{- m} x \sqrt {e^{- 2 m \log {\relax (2 )}} e^{m \log {\left (b^{2} x^{2} \right )}}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {8 \cdot 2^{m} \gamma b^{- m} x \sqrt {e^{- 2 m \log {\relax (2 )}} e^{m \log {\left (b^{2} x^{2} \right )}}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {b^{2} m^{2} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {m}{2} + \frac {3}{2} \\ \frac {3}{2}, 2, 2, \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {2 b^{2} m x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {m}{2} + \frac {3}{2} \\ \frac {3}{2}, 2, 2, \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {b^{2} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {m}{2} + \frac {3}{2} \\ \frac {3}{2}, 2, 2, \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{8 m^{2} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 16 m \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) + 8 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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