Integrand size = 23, antiderivative size = 371 \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {37\ 3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
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Time = 0.72 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3909, 4095, 4086, 3913, 3912, 65, 231} \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {37\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac {3 \tan (c+d x) \sec ^2(c+d x)}{8 d \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{40 a d}+\frac {99 \tan (c+d x)}{80 d \sqrt [3]{a \sec (c+d x)+a}} \]
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Rule 65
Rule 231
Rule 3909
Rule 3912
Rule 3913
Rule 4086
Rule 4095
Rubi steps \begin{align*} \text {integral}& = \frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \int \frac {\sec ^2(c+d x) \left (2 a-\frac {1}{3} a \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{8 a} \\ & = \frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {9 \int \frac {\sec (c+d x) \left (-\frac {2 a^2}{9}+\frac {11}{3} a^2 \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{40 a^2} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}-\frac {37}{80} \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}-\frac {\left (37 \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{80 \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {(37 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{80 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {(111 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{40 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {37\ 3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.40 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.42 \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {\left (-4 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )+16 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )-7 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )+3 \sec ^2(c+d x) \sqrt [6]{1+\sec (c+d x)}\right ) \tan (c+d x)}{8 d \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a (1+\sec (c+d x))}} \]
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\[\int \frac {\sec \left (d x +c \right )^{4}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]
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