Optimal. Leaf size=169 \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^{m+2} (c-c \sec (e+f x))^{-m-3}}{a^2 f (2 m+1) \left (4 m^2+16 m+15\right )}+\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-3}}{a f \left (4 m^2+8 m+3\right )}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.373941, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3951, 3950} \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^{m+2} (c-c \sec (e+f x))^{-m-3}}{a^2 f (2 m+1) \left (4 m^2+16 m+15\right )}+\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-3}}{a f \left (4 m^2+8 m+3\right )}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3951
Rule 3950
Rubi steps
\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \, dx &=-\frac{(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}-\frac{2 \int \sec (e+f x) (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \, dx}{a (1+2 m)}\\ &=-\frac{(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}+\frac{2 (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a f \left (3+8 m+4 m^2\right )}+\frac{2 \int \sec (e+f x) (a+a \sec (e+f x))^{2+m} (c-c \sec (e+f x))^{-3-m} \, dx}{a^2 \left (3+8 m+4 m^2\right )}\\ &=-\frac{(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}+\frac{2 (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a f \left (3+8 m+4 m^2\right )}-\frac{2 (a+a \sec (e+f x))^{2+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a^2 f (5+2 m) \left (3+8 m+4 m^2\right )}\\ \end{align*}
Mathematica [C] time = 9.02579, size = 321, normalized size = 1.9 \[ -\frac{i 2^{m+3} \left (1+e^{i (e+f x)}\right ) \left (-i e^{-\frac{1}{2} i (e+f x)} \left (-1+e^{i (e+f x)}\right )\right )^{-2 m} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-m} \left (\frac{\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}\right )^m \left (\left (4 m^2+12 m+7\right ) e^{4 i (e+f x)}+\left (8 m^2+24 m+22\right ) e^{2 i (e+f x)}-4 (2 m+3) e^{i (e+f x)}-4 (2 m+3) e^{3 i (e+f x)}+4 m^2+12 m+7\right ) \sin ^{-2 (-m-3)}\left (\frac{e}{2}+\frac{f x}{2}\right ) \sec ^{m+3}(e+f x) (\sec (e+f x)+1)^{-m} (a (\sec (e+f x)+1))^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1) (2 m+3) (2 m+5) \left (-1+e^{i (e+f x)}\right )^5} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.658, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{-3-m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.53933, size = 211, normalized size = 1.25 \begin{align*} \frac{{\left ({\left (4 \, m^{2} + 8 \, m + 3\right )} \left (-a\right )^{m} - \frac{2 \,{\left (4 \, m^{2} + 12 \, m + 5\right )} \left (-a\right )^{m} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (4 \, m^{2} + 16 \, m + 15\right )} \left (-a\right )^{m} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} c^{-m - 3}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{4 \,{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )^{2 \, m} \sin \left (f x + e\right )^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 0.512713, size = 296, normalized size = 1.75 \begin{align*} -\frac{{\left ({\left (4 \, m^{2} + 12 \, m + 7\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (2 \, m + 3\right )} \cos \left (f x + e\right ) + 2\right )} \left (\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \left (\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}\right )^{-m - 3} \sin \left (f x + e\right )}{{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right )^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )}^{m}{\left (-c \sec \left (f x + e\right ) + c\right )}^{-m - 3} \sec \left (f x + e\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]