Integrand size = 24, antiderivative size = 88 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=\frac {c x}{a^3}-\frac {2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac {8 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))} \]
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Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3988, 3862, 4007, 4004, 3879, 3881} \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=-\frac {8 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}-\frac {3 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac {2 c \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac {c x}{a^3} \]
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Rule 3862
Rule 3879
Rule 3881
Rule 3988
Rule 4004
Rule 4007
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (\frac {c}{(1+\sec (e+f x))^3}-\frac {c \sec (e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3} \\ & = \frac {c \int \frac {1}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac {c \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3} \\ & = -\frac {2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {c \int \frac {-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac {(2 c) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3} \\ & = -\frac {2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}+\frac {c \int \frac {15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac {(2 c) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3} \\ & = \frac {c x}{a^3}-\frac {2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac {2 c \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}-\frac {(22 c) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3} \\ & = \frac {c x}{a^3}-\frac {2 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac {3 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac {8 c \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.50 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=\frac {c \cot ^5(e+f x) \left (16+3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right )-60 \sec (e+f x)+5 \sec ^2(e+f x)+60 \sec ^3(e+f x)-24 \sec ^5(e+f x)\right )}{15 a^3 f} \]
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Time = 0.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.57
method | result | size |
parallelrisch | \(-\frac {c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-10 f x +20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{10 a^{3} f}\) | \(50\) |
derivativedivides | \(\frac {c \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{2 f \,a^{3}}\) | \(58\) |
default | \(\frac {c \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{2 f \,a^{3}}\) | \(58\) |
norman | \(\frac {\frac {c x}{a}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 a f}-\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{10 a f}}{a^{2}}\) | \(70\) |
risch | \(\frac {c x}{a^{3}}-\frac {2 i c \left (20 \,{\mathrm e}^{4 i \left (f x +e \right )}+55 \,{\mathrm e}^{3 i \left (f x +e \right )}+75 \,{\mathrm e}^{2 i \left (f x +e \right )}+45 \,{\mathrm e}^{i \left (f x +e \right )}+13\right )}{5 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(77\) |
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Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.41 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=\frac {5 \, c f x \cos \left (f x + e\right )^{3} + 15 \, c f x \cos \left (f x + e\right )^{2} + 15 \, c f x \cos \left (f x + e\right ) + 5 \, c f x - {\left (13 \, c \cos \left (f x + e\right )^{2} + 19 \, c \cos \left (f x + e\right ) + 8 \, c\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
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\[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=- \frac {c \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {1}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx\right )}{a^{3}} \]
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Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.81 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=-\frac {c {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3}}\right )} + \frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
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Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {10 \, {\left (f x + e\right )} c}{a^{3}} - \frac {a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 5 \, a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 20 \, a^{12} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15}}}{10 \, f} \]
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Time = 14.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^3} \, dx=\frac {c\,x}{a^3}-\frac {\frac {13\,c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{5}-\frac {7\,c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{10}+\frac {c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{10}}{a^3\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]
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