Optimal. Leaf size=99 \[ \frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},4;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},3;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]
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Rubi [A] time = 0.12, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5266, 12, 4471, 2251} \[ \frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},4;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},3;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2251
Rule 4471
Rule 5266
Rubi steps
\begin {align*} \int e^{\sec ^{-1}(a x)} x^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^x \sec ^3(x) \tan (x)}{a^2} \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int e^x \sec ^3(x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {16 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4}-\frac {8 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=-\frac {(8 i) \operatorname {Subst}\left (\int \frac {e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3} \, dx,x,\sec ^{-1}(a x)\right )}{a^3}+\frac {(16 i) \operatorname {Subst}\left (\int \frac {e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4} \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},3;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}+\frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac {3}{2}-\frac {i}{2},4;\frac {5}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 95, normalized size = 0.96 \[ \frac {e^{\sec ^{-1}(a x)} \left (a^4 x^4 \left (\cos \left (2 \sec ^{-1}(a x)\right )-\sin \left (2 \sec ^{-1}(a x)\right )+5\right )-(4+4 i) \left (a x \sqrt {1-\frac {1}{a^2 x^2}}-i\right ) \, _2F_1\left (\frac {1}{2}-\frac {i}{2},1;\frac {3}{2}-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )\right )}{12 a^4 x} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.34, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )}, x\right ) \]
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 \[ \int x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.05, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\mathrm {arcsec}\left (a x \right )} x^{2}\, dx \]
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 \[ \int x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )}\,{d x} \]
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 \[ \int x^2\,{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )} \,d x \]
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 \[ \int x^{2} e^{\operatorname {asec}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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