Optimal. Leaf size=96 \[ -\frac{23 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)}-\frac{8 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)^2}-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac{c^2 x}{a^3} \]
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Rubi [A] time = 0.303866, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797} \[ -\frac{23 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)}-\frac{8 c^2 \tan (e+f x)}{15 a^3 f (\sec (e+f x)+1)^2}-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}+\frac{c^2 x}{a^3} \]
Antiderivative was successfully verified.
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Rule 3903
Rule 3777
Rule 3922
Rule 3919
Rule 3794
Rule 3796
Rule 3797
Rubi steps
\begin{align*} \int \frac{(c-c \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx &=\frac{\int \left (\frac{c^2}{(1+\sec (e+f x))^3}-\frac{2 c^2 \sec (e+f x)}{(1+\sec (e+f x))^3}+\frac{c^2 \sec ^2(e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3}\\ &=\frac{c^2 \int \frac{1}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac{c^2 \int \frac{\sec ^2(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac{\left (2 c^2\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}\\ &=-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{c^2 \int \frac{-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}+\frac{\left (3 c^2\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac{\left (4 c^2\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}\\ &=-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}+\frac{c^2 \int \frac{15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac{c^2 \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{5 a^3}-\frac{\left (4 c^2\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c^2 x}{a^3}-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}-\frac{c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}-\frac{\left (22 c^2\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c^2 x}{a^3}-\frac{4 c^2 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{8 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^2}-\frac{23 c^2 \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.46073, size = 171, normalized size = 1.78 \[ \frac{c^2 \sec \left (\frac{e}{2}\right ) \sec ^5\left (\frac{1}{2} (e+f x)\right ) \left (360 \sin \left (e+\frac{f x}{2}\right )-280 \sin \left (e+\frac{3 f x}{2}\right )+150 \sin \left (2 e+\frac{3 f x}{2}\right )-86 \sin \left (2 e+\frac{5 f x}{2}\right )+150 f x \cos \left (e+\frac{f x}{2}\right )+75 f x \cos \left (e+\frac{3 f x}{2}\right )+75 f x \cos \left (2 e+\frac{3 f x}{2}\right )+15 f x \cos \left (2 e+\frac{5 f x}{2}\right )+15 f x \cos \left (3 e+\frac{5 f x}{2}\right )-500 \sin \left (\frac{f x}{2}\right )+150 f x \cos \left (\frac{f x}{2}\right )\right )}{480 a^3 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.091, size = 87, normalized size = 0.9 \begin{align*} -{\frac{{c}^{2}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{2\,{c}^{2}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-2\,{\frac{{c}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}+2\,{\frac{{c}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.57184, size = 285, normalized size = 2.97 \begin{align*} -\frac{c^{2}{\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{2 \, c^{2}{\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 \, c^{2}{\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.00222, size = 338, normalized size = 3.52 \begin{align*} \frac{15 \, c^{2} f x \cos \left (f x + e\right )^{3} + 45 \, c^{2} f x \cos \left (f x + e\right )^{2} + 45 \, c^{2} f x \cos \left (f x + e\right ) + 15 \, c^{2} f x -{\left (43 \, c^{2} \cos \left (f x + e\right )^{2} + 54 \, c^{2} \cos \left (f x + e\right ) + 23 \, c^{2}\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{2} \left (\int - \frac{2 \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{\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 + \int \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}\, dx\right )}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.3757, size = 113, normalized size = 1.18 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} c^{2}}{a^{3}} - \frac{3 \, a^{12} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 10 \, a^{12} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, a^{12} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{15}}}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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