Integrand size = 23, antiderivative size = 260 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {2 \sqrt {2} \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 b d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
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Time = 0.41 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3920, 4092, 3919, 144, 143} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=-\frac {2 \sqrt {2} \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{5 b d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac {2 \sqrt {2} a \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{5 b d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 d} \]
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Rule 143
Rule 144
Rule 3919
Rule 3920
Rule 4092
Rubi steps \begin{align*} \text {integral}& = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2}{5} \int \frac {\sec (c+d x) (b+a \sec (c+d x))}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {(2 a) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{5 b}-\frac {\left (2 \left (a^2-b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{5 b} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {(2 a \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {\left (2 \left (a^2-b^2\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}-\frac {\left (2 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac {\left (2 \left (a^2-b^2\right ) \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 )}{5 b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 d}+\frac {2 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {2 \sqrt {2} \left (a^2-b^2\right ) \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)}{5 b 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. \(10909\) vs. \(2(260)=520\).
Time = 42.67 (sec) , antiderivative size = 10909, normalized size of antiderivative = 41.96 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\text {Result too large to show} \]
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\[\int \sec \left (d x +c \right )^{2} \left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]
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\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{2} \,d x } \]
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\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}} \sec ^{2}{\left (c + d x \right )}\, dx \]
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\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{2} \,d x } \]
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\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sec \left (d x + c\right )^{2} \,d x } \]
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Timed out. \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{2/3} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}}{{\cos \left (c+d\,x\right )}^2} \,d x \]
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