Integrand size = 12, antiderivative size = 163 \[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x) \left (8+\sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\arctan \left (\sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^4} \]
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Time = 0.38 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6472, 1818, 1828, 653, 209} \[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\frac {\arctan \left (\sqrt {\frac {1-a x}{a x+1}}\right )}{4 a^4}-\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)^4}{4 a^4}+\frac {\left (9 \sqrt {\frac {1-a x}{a x+1}}+4\right ) (a x+1)^3}{12 a^4}-\frac {\left (5 \sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)^2}{8 a^4}+\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)}{8 a^4} \]
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Rule 209
Rule 653
Rule 1818
Rule 1828
Rule 6472
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3}{\frac {1}{a x}+\sqrt {\frac {1-a x}{1+a x}}+\frac {\sqrt {\frac {1-a x}{1+a x}}}{a x}} \, dx \\ & = -\frac {4 \text {Subst}\left (\int \frac {(-1+x)^4 x (1+x)^2}{\left (1+x^2\right )^5} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a^4} \\ & = -\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {\text {Subst}\left (\int \frac {8-8 x-48 x^2+16 x^3+16 x^4-8 x^5}{\left (1+x^2\right )^4} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{2 a^4} \\ & = -\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}-\frac {\text {Subst}\left (\int \frac {24-144 x-96 x^2+48 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4} \\ & = -\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\text {Subst}\left (\int \frac {24-192 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{48 a^4} \\ & = -\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x) \left (8+\sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^4} \\ & = -\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x) \left (8+\sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\arctan \left (\sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^4} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.60 \[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\frac {8 a^3 x^3+3 a \sqrt {\frac {1-a x}{1+a x}} \left (x+a x^2-2 a^2 x^3-2 a^3 x^4\right )-3 i \log \left (-2 i a x+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)\right )}{24 a^4} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.92 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74
method | result | size |
default | \(a \left (\frac {x^{3}}{3 a^{2}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (2 \,\operatorname {csgn}\left (a \right ) a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}-\sqrt {-a^{2} x^{2}+1}\, x \,\operatorname {csgn}\left (a \right ) a +\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a x}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) \operatorname {csgn}\left (a \right )}{8 a^{4} \sqrt {-a^{2} x^{2}+1}}\right )\) | \(120\) |
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Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.58 \[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\frac {8 \, a^{3} x^{3} - 3 \, {\left (2 \, a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 3 \, \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{24 \, a^{4}} \]
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\[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=a \int \frac {x^{4}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]
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\[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\int { \frac {x^{3}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}} \,d x } \]
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\[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\int { \frac {x^{3}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}} \,d x } \]
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Time = 25.41 (sec) , antiderivative size = 795, normalized size of antiderivative = 4.88 \[ \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx=\frac {\ln \left (\frac {a\,\sqrt {\frac {1}{a\,x}+1}-\frac {1}{x}+a\,\sqrt {\frac {1}{a\,x}-1}\,1{}\mathrm {i}}{2\,a-2\,a\,\sqrt {\frac {1}{a\,x}+1}+\frac {1}{x}}\right )\,3{}\mathrm {i}}{8\,a^4}+\frac {\frac {1{}\mathrm {i}}{1024\,a^4}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,3{}\mathrm {i}}{128\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,53{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6\,87{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8\,657{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}\,121{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{12}}}+\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{8\,a^4}+\frac {\frac {1{}\mathrm {i}}{32\,a^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}}-\frac {\ln \left (\frac {2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}-\frac {2}{x}+a\,\sqrt {-\frac {a-\frac {1}{x}}{a}}\,2{}\mathrm {i}}{2\,a+\frac {1}{x}-2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}}\right )\,1{}\mathrm {i}}{2\,a^4}+\frac {x^3}{3\,a}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,1{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4} \]
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