Optimal. Leaf size=137 \[ \frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}-\frac{x^2 \sqrt{a^2 x^2+1}}{a^2}-\frac{9 i (3 a x+2 i) \sqrt{a^2 x^2+1}}{8 a^4}+\frac{27 \sqrt{a^2 x^2+1}}{4 a^4}+\frac{(1-i a x)^3}{a^4 \sqrt{a^2 x^2+1}}+\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]
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Rubi [A] time = 0.62436, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5060, 1633, 1593, 12, 852, 1635, 1815, 27, 743, 641, 215} \[ \frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}-\frac{x^2 \sqrt{a^2 x^2+1}}{a^2}-\frac{9 i (3 a x+2 i) \sqrt{a^2 x^2+1}}{8 a^4}+\frac{27 \sqrt{a^2 x^2+1}}{4 a^4}+\frac{(1-i a x)^3}{a^4 \sqrt{a^2 x^2+1}}+\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]
Antiderivative was successfully verified.
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Rule 5060
Rule 1633
Rule 1593
Rule 12
Rule 852
Rule 1635
Rule 1815
Rule 27
Rule 743
Rule 641
Rule 215
Rubi steps
\begin{align*} \int e^{-3 i \tan ^{-1}(a x)} x^3 \, dx &=\int \frac{x^3 (1-i a x)^2}{(1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=(i a) \int \frac{\sqrt{1+a^2 x^2} \left (-\frac{i x^3}{a}-x^4\right )}{(1+i a x)^2} \, dx\\ &=(i a) \int \frac{\left (-\frac{i}{a}-x\right ) x^3 \sqrt{1+a^2 x^2}}{(1+i a x)^2} \, dx\\ &=a^2 \int \frac{x^3 \left (1+a^2 x^2\right )^{3/2}}{a^2 (1+i a x)^3} \, dx\\ &=\int \frac{x^3 \left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^3} \, dx\\ &=\int \frac{x^3 (1-i a x)^3}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\int \frac{(1-i a x)^2 \left (-\frac{3 i}{a^3}-\frac{x}{a^2}+\frac{i x^2}{a}\right )}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int \frac{-\frac{12 i}{a}-28 x+27 i a x^2+12 a^2 x^3}{\sqrt{1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int \frac{-36 i a-108 a^2 x+81 i a^3 x^2}{\sqrt{1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int \frac{9 i a (2 i+3 a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{(3 i) \int \frac{(2 i+3 a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{4 a^3}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i+3 a x) \sqrt{1+a^2 x^2}}{8 a^4}-\frac{(3 i) \int \frac{-17 a^2+18 i a^3 x}{\sqrt{1+a^2 x^2}} \, dx}{8 a^5}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}+\frac{27 \sqrt{1+a^2 x^2}}{4 a^4}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i+3 a x) \sqrt{1+a^2 x^2}}{8 a^4}+\frac{(51 i) \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{8 a^3}\\ &=\frac{(1-i a x)^3}{a^4 \sqrt{1+a^2 x^2}}+\frac{27 \sqrt{1+a^2 x^2}}{4 a^4}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}+\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i+3 a x) \sqrt{1+a^2 x^2}}{8 a^4}+\frac{51 i \sinh ^{-1}(a x)}{8 a^4}\\ \end{align*}
Mathematica [A] time = 0.0563022, size = 80, normalized size = 0.58 \[ \sqrt{a^2 x^2+1} \left (-\frac{x^2}{a^2}-\frac{19 i x}{8 a^3}-\frac{4 i}{a^4 (a x-i)}+\frac{6}{a^4}+\frac{i x^3}{4 a}\right )+\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.087, size = 296, normalized size = 2.2 \begin{align*}{\frac{{\frac{i}{4}}x}{{a}^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{{\frac{3\,i}{8}}x}{{a}^{3}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{{\frac{3\,i}{8}}}{{a}^{3}}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-5\,{\frac{1}{{a}^{6}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{5/2} \left ( x-{\frac{i}{a}} \right ) ^{-2}}+4\,{\frac{1}{{a}^{4}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{3/2}}+{\frac{6\,ix}{{a}^{3}}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}+{\frac{6\,i}{{a}^{3}}\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{i}{{a}^{7}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.55073, size = 292, normalized size = 2.13 \begin{align*} \frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{6} x^{2} - 2 i \, a^{5} x - a^{4}} + \frac{3 \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{2 i \, a^{5} x + 2 \, a^{4}} + \frac{6 \, \sqrt{a^{2} x^{2} + 1}}{i \, a^{5} x + a^{4}} + \frac{i \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{4 \, a^{3}} + \frac{3 i \, \sqrt{a^{2} x^{2} + 1} x}{8 \, a^{3}} - \frac{3 i \, \sqrt{-a^{2} x^{2} + 4 i \, a x + 3} x}{2 \, a^{3}} - \frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4}} + \frac{3 i \, \arcsin \left (i \, a x + 2\right )}{2 \, a^{4}} + \frac{63 i \, \operatorname{arsinh}\left (a x\right )}{8 \, a^{4}} + \frac{9 \, \sqrt{a^{2} x^{2} + 1}}{2 \, a^{4}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 4 i \, a x + 3}}{a^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.74957, size = 219, normalized size = 1.6 \begin{align*} \frac{-32 i \, a x - 51 \,{\left (i \, a x + 1\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) +{\left (2 i \, a^{4} x^{4} - 6 \, a^{3} x^{3} - 11 i \, a^{2} x^{2} + 29 \, a x - 80 i\right )} \sqrt{a^{2} x^{2} + 1} - 32}{8 \,{\left (a^{5} x - i \, a^{4}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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