Integrand size = 23, antiderivative size = 174 \[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}+\frac {a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{3},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d} \]
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Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3954, 2902, 3268, 440} \[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\frac {a \sin (c+d x) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{3},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac {b \sin (c+d x) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}} \]
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Rule 440
Rule 2902
Rule 3268
Rule 3954
Rubi steps \begin{align*} \text {integral}& = \left (\cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\cos ^{\frac {4}{3}}(c+d x)}{b+a \cos (c+d x)} \, dx \\ & = -\left (\left (a \cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\cos ^{\frac {7}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\right )+\left (b \cos ^{\frac {2}{3}}(c+d x) \sec ^{\frac {2}{3}}(c+d x)\right ) \int \frac {\cos ^{\frac {4}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx \\ & = \frac {b \text {Subst}\left (\int \frac {\sqrt [6]{1-x^2}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}-\frac {\left (a \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x)\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{2/3}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d} \\ & = -\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}+\frac {a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {2}{3},1,\frac {3}{2},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac {2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(7542\) vs. \(2(174)=348\).
Time = 84.74 (sec) , antiderivative size = 7542, normalized size of antiderivative = 43.34 \[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\text {Result too large to show} \]
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\[\int \frac {1}{\sec \left (d x +c \right )^{\frac {1}{3}} \left (a +b \sec \left (d x +c \right )\right )}d x\]
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Timed out. \[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right ) \sqrt [3]{\sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx=\int \frac {1}{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]
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