Integrand size = 21, antiderivative size = 105 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
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Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3919, 144, 143} \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}} \]
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Rule 143
Rule 144
Rule 3919
Rubi steps \begin{align*} \text {integral}& = -\frac {\tan (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = -\frac {\left (\sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \\ & = \frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(310\) vs. \(2(105)=210\).
Time = 14.39 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.95 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {15 (a-b)^2 (a+b) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) \cos (c+d x) \cot ^3(c+d x) (1+\sec (c+d x)) (b-b \sec (c+d x)) (a+b \sec (c+d x))^{2/3}}{b^2 (-a+b) d \left (3 (a-b) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {3}{2},\frac {8}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) (b+a \cos (c+d x))+(a+b) \left (10 (a-b) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) \cos (c+d x)+3 \operatorname {AppellF1}\left (\frac {5}{3},\frac {3}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) (b+a \cos (c+d x))\right )\right )} \]
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\[\int \frac {\sec \left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt [3]{a + b \sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]
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