Integrand size = 24, antiderivative size = 198 \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\sqrt {3} b \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]
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Time = 0.07 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1024, 402, 455, 57, 631, 210, 31} \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {\sqrt {3} b \arctan \left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]
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Rule 31
Rule 57
Rule 210
Rule 402
Rule 455
Rule 631
Rule 1024
Rubi steps \begin{align*} \text {integral}& = a \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx+b \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx \\ & = \frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}+\frac {1}{2} b \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right ) \\ & = \frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {1}{4} (3 b) \text {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {(3 b) \text {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \\ & = \frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{2\ 2^{2/3}} \\ & = \frac {a \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\sqrt {3} b \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {a \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {a \text {arctanh}(x)}{6\ 2^{2/3}}+\frac {a \text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}}-\frac {b \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 b \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 10.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.73 \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{6} b x^2 \operatorname {AppellF1}\left (1,\frac {1}{3},1,2,x^2,-\frac {x^2}{3}\right )-\frac {9 a x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{1-x^2} \left (3+x^2\right ) \left (-9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )+2 x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},x^2,-\frac {x^2}{3}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},x^2,-\frac {x^2}{3}\right )\right )\right )} \]
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\[\int \frac {b x +a}{\left (-x^{2}+1\right )^{\frac {1}{3}} \left (x^{2}+3\right )}d x\]
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Exception generated. \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int \frac {a + b x}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \]
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\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int { \frac {b x + a}{{\left (x^{2} + 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {a+b x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int \frac {a+b\,x}{{\left (1-x^2\right )}^{1/3}\,\left (x^2+3\right )} \,d x \]
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