Integrand size = 29, antiderivative size = 102 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]
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Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4085, 3881, 3879} \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]
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Rule 3879
Rule 3881
Rule 4085
Rubi steps \begin{align*} \text {integral}& = \frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2} \, dx}{5 a} \\ & = \frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(2 c+3 d) \int \frac {\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{15 a^2} \\ & = \frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \\ \end{align*}
Time = 0.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \left (5 (8 c+3 d) \sin \left (\frac {f x}{2}\right )-15 (2 c+d) \sin \left (e+\frac {f x}{2}\right )+20 c \sin \left (e+\frac {3 f x}{2}\right )+15 d \sin \left (e+\frac {3 f x}{2}\right )-15 c \sin \left (2 e+\frac {3 f x}{2}\right )+7 c \sin \left (2 e+\frac {5 f x}{2}\right )+3 d \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f (1+\cos (e+f x))^3} \]
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Time = 0.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(\frac {\left (\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c}{3}+5 c +5 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{20 a^{3} f}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 c \,{\mathrm e}^{4 i \left (f x +e \right )}+30 c \,{\mathrm e}^{3 i \left (f x +e \right )}+15 d \,{\mathrm e}^{3 i \left (f x +e \right )}+40 \,{\mathrm e}^{2 i \left (f x +e \right )} c +15 d \,{\mathrm e}^{2 i \left (f x +e \right )}+20 \,{\mathrm e}^{i \left (f x +e \right )} c +15 d \,{\mathrm e}^{i \left (f x +e \right )}+7 c +3 d \right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(114\) |
norman | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{20 a f}-\frac {\left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a f}+\frac {\left (5 c +3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{12 a f}-\frac {\left (13 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{60 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) a^{2}}\) | \(117\) |
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Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left ({\left (7 \, c + 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 3 \, d\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
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\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\int \frac {c \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
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Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{60 \, a^{3} f} \]
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Time = 13.51 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,c+15\,d-10\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-3\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{60\,a^3\,f} \]
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