Integrand size = 14, antiderivative size = 179 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {3 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \]
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Time = 0.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6306, 101, 156, 12, 95, 304, 209, 212} \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {3 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {11 x \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{24 a^2}+\frac {1}{3} x^3 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}-\frac {5 x^2 \sqrt [4]{1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/4}}{12 a} \]
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Rule 12
Rule 95
Rule 101
Rule 156
Rule 209
Rule 212
Rule 304
Rule 6306
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\sqrt [4]{1-\frac {x}{a}}}{x^4 \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {1}{3} \text {Subst}\left (\int \frac {-\frac {5}{2 a}+\frac {2 x}{a^2}}{x^3 \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {1}{6} \text {Subst}\left (\int \frac {-\frac {11}{4 a^2}+\frac {5 x}{2 a^3}}{x^2 \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {1}{6} \text {Subst}\left (\int -\frac {9}{8 a^3 x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {3 \text {Subst}\left (\int \frac {1}{x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{16 a^3} \\ & = \frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {3 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^3} \\ & = \frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3-\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \\ & = \frac {11 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x}{24 a^2}-\frac {5 \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^2}{12 a}+\frac {1}{3} \sqrt [4]{1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/4} x^3+\frac {3 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.31 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.17 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {e^{-\frac {5}{2} \coth ^{-1}(a x)} \left (-22034705-26688365 e^{2 \coth ^{-1}(a x)}-3731255 e^{4 \coth ^{-1}(a x)}+3122405 e^{6 \coth ^{-1}(a x)}+22034705 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+17244920 e^{2 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-9077530 e^{4 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )-7043960 e^{6 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+446985 e^{8 \coth ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 \coth ^{-1}(a x)}\right )+256 e^{6 \coth ^{-1}(a x)} \left (685+1090 e^{2 \coth ^{-1}(a x)}+437 e^{4 \coth ^{-1}(a x)}\right ) \, _4F_3\left (\frac {7}{4},2,2,2;1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+2048 e^{6 \coth ^{-1}(a x)} \left (21+38 e^{2 \coth ^{-1}(a x)}+17 e^{4 \coth ^{-1}(a x)}\right ) \, _5F_4\left (\frac {7}{4},2,2,2,2;1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{6 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+8192 e^{8 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{10 \coth ^{-1}(a x)} \, _6F_5\left (\frac {7}{4},2,2,2,2,2;1,1,1,1,\frac {19}{4};e^{2 \coth ^{-1}(a x)}\right )\right )}{221760 a^3} \]
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\[\int x^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}}d x\]
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Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.57 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 11\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \]
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\[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\int x^{2} \sqrt [4]{\frac {a x - 1}{a x + 1}}\, dx \]
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Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {4 \, {\left (29 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 6 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{48} \, a {\left (\frac {18 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {9 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {9 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} - \frac {4 \, {\left (\frac {6 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {29 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]
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Time = 4.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int e^{-\frac {1}{2} \coth ^{-1}(a x)} x^2 \, dx=\frac {\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{4}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}+\frac {29\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{12}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}-\frac {3\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {3\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3} \]
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