\(\int (a+a \sec (e+f x)) (c-c \sec (e+f x))^n \, dx\) [135]

Optimal result

Integrand size = 24, antiderivative size = 99 \[ \int (a+a \sec (e+f x)) (c-c \sec (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+n} c \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2}-n,1,\frac {5}{2},\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac {1}{2}-n} (a+a \sec (e+f x)) (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{3 f} \]

[Out]

1/3*2^(1/2+n)*c*AppellF1(3/2,1,1/2-n,5/2,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(1-sec(f*x+e))^(1/2-n)*(a+a*sec(f*x+
e))*(c-c*sec(f*x+e))^(-1+n)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3997, 142, 141} \[ \int (a+a \sec (e+f x)) (c-c \sec (e+f x))^n \, dx=\frac {c 2^{n+\frac {1}{2}} \tan (e+f x) (a \sec (e+f x)+a) (1-\sec (e+f x))^{\frac {1}{2}-n} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2}-n,1,\frac {5}{2},\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{3 f} \]

[In]

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

[Out]

(2^(1/2 + n)*c*AppellF1[3/2, 1/2 - n, 1, 5/2, (1 + Sec[e + f*x])/2, 1 + Sec[e + f*x]]*(1 - Sec[e + f*x])^(1/2
- n)*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(-1 + n)*Tan[e + f*x])/(3*f)

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 142

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*(b*(c/(b*c -
a*d)) + b*d*(x/(b*c - a*d)))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 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 {(a c \tan (e+f x)) \text {Subst}\left (\int \frac {\sqrt {a+a x} (c-c x)^{-\frac {1}{2}+n}}{x} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = -\frac {\left (2^{-\frac {1}{2}+n} a c (c-c \sec (e+f x))^{-1+n} \left (\frac {c-c \sec (e+f x)}{c}\right )^{\frac {1}{2}-n} \tan (e+f x)\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}-\frac {x}{2}\right )^{-\frac {1}{2}+n} \sqrt {a+a x}}{x} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)}} \\ & = \frac {2^{\frac {1}{2}+n} c \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2}-n,1,\frac {5}{2},\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac {1}{2}-n} (a+a \sec (e+f x)) (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{3 f} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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