\(\int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx\) [131]

Optimal result

Integrand size = 24, antiderivative size = 92 \[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2}+n,\frac {1}{2}-m,1,\frac {3}{2}+n,\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}} \]

[Out]

2^(1/2+m)*AppellF1(1/2+n,1,1/2-m,3/2+n,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n
)/(1+sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3997, 141} \[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\frac {2^{m+\frac {1}{2}} \tan (e+f x) (c-c \sec (e+f x))^n \operatorname {AppellF1}\left (n+\frac {1}{2},\frac {1}{2}-m,1,n+\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {\sec (e+f x)+1}} \]

[In]

Int[(1 + Sec[e + f*x])^m*(c - c*Sec[e + f*x])^n,x]

[Out]

(2^(1/2 + m)*AppellF1[1/2 + n, 1/2 - m, 1, 3/2 + n, (1 - Sec[e + f*x])/2, 1 - Sec[e + f*x]]*(c - c*Sec[e + f*x
])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[1 + Sec[e + f*x]])

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])), Subst[Int[(a + b*x)^(m - 1/2)*((c
+ d*x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(c \tan (e+f x)) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m} (c-c x)^{-\frac {1}{2}+n}}{x} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1+\sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = \frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2}+n,\frac {1}{2}-m,1,\frac {3}{2}+n,\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}} \\ \end{align*}

Mathematica [F]

\[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx \]

[In]

Integrate[(1 + Sec[e + f*x])^m*(c - c*Sec[e + f*x])^n,x]

[Out]

Integrate[(1 + Sec[e + f*x])^m*(c - c*Sec[e + f*x])^n, x]

Maple [F]

\[\int \left (\sec \left (f x +e \right )+1\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{n}d x\]

[In]

int((sec(f*x+e)+1)^m*(c-c*sec(f*x+e))^n,x)

[Out]

int((sec(f*x+e)+1)^m*(c-c*sec(f*x+e))^n,x)

Fricas [F]

\[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int { {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sec(f*x+e))^m*(c-c*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((-c*sec(f*x + e) + c)^n*(sec(f*x + e) + 1)^m, x)

Sympy [F]

\[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{m}\, dx \]

[In]

integrate((1+sec(f*x+e))**m*(c-c*sec(f*x+e))**n,x)

[Out]

Integral((-c*(sec(e + f*x) - 1))**n*(sec(e + f*x) + 1)**m, x)

Maxima [F]

\[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int { {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sec(f*x+e))^m*(c-c*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((-c*sec(f*x + e) + c)^n*(sec(f*x + e) + 1)^m, x)

Giac [F]

\[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int { {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sec(f*x+e))^m*(c-c*sec(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((-c*sec(f*x + e) + c)^n*(sec(f*x + e) + 1)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx=\int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^m\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int((1/cos(e + f*x) + 1)^m*(c - c/cos(e + f*x))^n,x)

[Out]

int((1/cos(e + f*x) + 1)^m*(c - c/cos(e + f*x))^n, x)