Integrand size = 23, antiderivative size = 192 \[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\frac {a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1+n),1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},\frac {n}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right ) f} \]
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Time = 0.38 (sec) , antiderivative size = 192, 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} \[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\frac {a \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac {n-1}{2}} (d \sec (e+f x))^n \operatorname {AppellF1}\left (\frac {1}{2},\frac {n-1}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {b \sin (e+f x) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n \operatorname {AppellF1}\left (\frac {1}{2},\frac {n}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \]
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Rule 440
Rule 2902
Rule 3268
Rule 3954
Rubi steps \begin{align*} \text {integral}& = \left (\cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{1-n}(e+f x)}{b+a \cos (e+f x)} \, dx \\ & = -\left (\left (a \cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{2-n}(e+f x)}{b^2-a^2 \cos ^2(e+f x)} \, dx\right )+\left (b \cos ^n(e+f x) (d \sec (e+f x))^n\right ) \int \frac {\cos ^{1-n}(e+f x)}{b^2-a^2 \cos ^2(e+f x)} \, dx \\ & = -\frac {\left (a \cos ^{2 \left (\frac {1}{2}-\frac {n}{2}\right )+n}(e+f x) \cos ^2(e+f x)^{-\frac {1}{2}+\frac {n}{2}} (d \sec (e+f x))^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1-n}{2}}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (b \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{-n/2}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (-1+n),1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (-1+n)} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},\frac {n}{2},1,\frac {3}{2},\sin ^2(e+f x),\frac {a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos ^2(e+f x)^{n/2} (d \sec (e+f x))^n \sin (e+f x)}{\left (a^2-b^2\right ) f} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(5280\) vs. \(2(192)=384\).
Time = 35.15 (sec) , antiderivative size = 5280, normalized size of antiderivative = 27.50 \[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\text {Result too large to show} \]
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\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{n}}{a +b \sec \left (f x +e \right )}d x\]
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\[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{n}}{a + b \sec {\left (e + f x \right )}}\, dx \]
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\[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(d \sec (e+f x))^n}{a+b \sec (e+f x)} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n}{a+\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]
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