Optimal. Leaf size=91 \[ \frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},3;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}-\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},2;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2} \]
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Rubi [A] time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5266, 12, 4471, 2251} \[ \frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},3;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}-\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},2;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2} \]
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 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^x \sec ^2(x) \tan (x)}{a} \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int e^x \sec ^2(x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {8 i e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^3}-\frac {4 i e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^2}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a^2}\\ &=-\frac {(4 i) \operatorname {Subst}\left (\int \frac {e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^2} \, dx,x,\sec ^{-1}(a x)\right )}{a^2}+\frac {(8 i) \operatorname {Subst}\left (\int \frac {e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^3} \, dx,x,\sec ^{-1}(a x)\right )}{a^2}\\ &=-\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},2;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}+\frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1-\frac {i}{2},3;2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 107, normalized size = 1.18 \[ \frac {\left (\frac {1}{5}+\frac {i}{10}\right ) e^{\sec ^{-1}(a x)} \left ((-2+i) a x \left (\sqrt {1-\frac {1}{a^2 x^2}}-a x\right )+(1+2 i) \, _2F_1\left (-\frac {i}{2},1;1-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )-e^{2 i \sec ^{-1}(a x)} \, _2F_1\left (1,1-\frac {i}{2};2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )\right )}{a^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x e^{\left (\operatorname {arcsec}\left (a x\right )\right )}, x\right ) \]
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 \[ \int x e^{\left (\operatorname {arcsec}\left (a x\right )\right )}\,{d x} \]
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 \[ \int {\mathrm e}^{\mathrm {arcsec}\left (a x \right )} x\, dx \]
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 \[ \int x e^{\left (\operatorname {arcsec}\left (a x\right )\right )}\,{d x} \]
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 \[ \int x\,{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )} \,d x \]
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 \[ \int x e^{\operatorname {asec}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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