Optimal. Leaf size=148 \[ \frac{c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac{c^4 \tan (e+f x) \sec ^2(e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}-\frac{23 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}+\frac{14 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac{3 c^4 \tan (e+f x)}{a^3 f (\sec (e+f x)+1)^3}+\frac{c^4 x}{a^3} \]
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Rubi [A] time = 0.606057, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3903, 3777, 3922, 3919, 3794, 3796, 3797, 3799, 4000, 3816, 4008, 3998, 3770} \[ \frac{c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac{c^4 \tan (e+f x) \sec ^2(e+f x)}{5 a^3 f (\sec (e+f x)+1)^3}-\frac{23 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)}+\frac{14 c^4 \tan (e+f x)}{5 a^3 f (\sec (e+f x)+1)^2}-\frac{3 c^4 \tan (e+f x)}{a^3 f (\sec (e+f x)+1)^3}+\frac{c^4 x}{a^3} \]
Antiderivative was successfully verified.
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Rule 3903
Rule 3777
Rule 3922
Rule 3919
Rule 3794
Rule 3796
Rule 3797
Rule 3799
Rule 4000
Rule 3816
Rule 4008
Rule 3998
Rule 3770
Rubi steps
\begin{align*} \int \frac{(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx &=\frac{\int \left (\frac{c^4}{(1+\sec (e+f x))^3}-\frac{4 c^4 \sec (e+f x)}{(1+\sec (e+f x))^3}+\frac{6 c^4 \sec ^2(e+f x)}{(1+\sec (e+f x))^3}-\frac{4 c^4 \sec ^3(e+f x)}{(1+\sec (e+f x))^3}+\frac{c^4 \sec ^4(e+f x)}{(1+\sec (e+f x))^3}\right ) \, dx}{a^3}\\ &=\frac{c^4 \int \frac{1}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac{c^4 \int \frac{\sec ^4(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac{\left (4 c^4\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}-\frac{\left (4 c^4\right ) \int \frac{\sec ^3(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}+\frac{\left (6 c^4\right ) \int \frac{\sec ^2(e+f x)}{(1+\sec (e+f x))^3} \, dx}{a^3}\\ &=-\frac{3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac{c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}-\frac{c^4 \int \frac{(2-5 \sec (e+f x)) \sec ^2(e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac{c^4 \int \frac{-5+2 \sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac{\left (4 c^4\right ) \int \frac{\sec (e+f x) (-3+5 \sec (e+f x))}{(1+\sec (e+f x))^2} \, dx}{5 a^3}-\frac{\left (8 c^4\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}+\frac{\left (18 c^4\right ) \int \frac{\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{5 a^3}\\ &=-\frac{3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac{c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac{14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}+\frac{c^4 \int \frac{15-7 \sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac{c^4 \int \frac{\sec (e+f x) (-14+15 \sec (e+f x))}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac{\left (8 c^4\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}+\frac{\left (6 c^4\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{5 a^3}-\frac{\left (28 c^4\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c^4 x}{a^3}-\frac{3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac{c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac{14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac{6 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))}+\frac{c^4 \int \sec (e+f x) \, dx}{a^3}-\frac{\left (22 c^4\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}-\frac{\left (29 c^4\right ) \int \frac{\sec (e+f x)}{1+\sec (e+f x)} \, dx}{15 a^3}\\ &=\frac{c^4 x}{a^3}+\frac{c^4 \tanh ^{-1}(\sin (e+f x))}{a^3 f}-\frac{3 c^4 \tan (e+f x)}{a^3 f (1+\sec (e+f x))^3}-\frac{c^4 \sec ^2(e+f x) \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^3}+\frac{14 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))^2}-\frac{23 c^4 \tan (e+f x)}{5 a^3 f (1+\sec (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.17546, size = 231, normalized size = 1.56 \[ \frac{c^4 (\cos (e+f x)-1)^4 \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right ) \left (8 \tan \left (\frac{e}{2}\right ) \cot ^3\left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )-4 \tan \left (\frac{e}{2}\right ) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right )+5 \cot ^5\left (\frac{1}{2} (e+f x)\right ) \left (-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+f x\right )-\sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (8 \cos (e+f x)+3 \cos (2 (e+f x))+9) \csc ^5\left (\frac{1}{2} (e+f x)\right )\right )}{10 a^3 f (\cos (e+f x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.103, size = 110, normalized size = 0.7 \begin{align*} -{\frac{4\,{c}^{4}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-4\,{\frac{{c}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}+2\,{\frac{{c}^{4}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}}+{\frac{{c}^{4}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{c}^{4}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.59596, size = 535, normalized size = 3.61 \begin{align*} -\frac{c^{4}{\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{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + c^{4}{\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{4 \, c^{4}{\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{4 \, c^{4}{\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{18 \, c^{4}{\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 = 1.1315, size = 601, normalized size = 4.06 \begin{align*} \frac{10 \, c^{4} f x \cos \left (f x + e\right )^{3} + 30 \, c^{4} f x \cos \left (f x + e\right )^{2} + 30 \, c^{4} f x \cos \left (f x + e\right ) + 10 \, c^{4} f x + 5 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 5 \,{\left (c^{4} \cos \left (f x + e\right )^{3} + 3 \, c^{4} \cos \left (f x + e\right )^{2} + 3 \, c^{4} \cos \left (f x + e\right ) + c^{4}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 16 \,{\left (3 \, c^{4} \cos \left (f x + e\right )^{2} + 4 \, c^{4} \cos \left (f x + e\right ) + 3 \, c^{4}\right )} \sin \left (f x + e\right )}{10 \,{\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^{4} \left (\int - \frac{4 \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{6 \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{4 \sec ^{3}{\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 ^{4}{\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 = 1.45611, size = 144, normalized size = 0.97 \begin{align*} \frac{\frac{5 \,{\left (f x + e\right )} c^{4}}{a^{3}} + \frac{5 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{5 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac{4 \,{\left (a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 5 \, a^{12} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{15}}}{5 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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