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

Optimal. Leaf size=92 \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (c-c \sec (e+f x))^n F_1\left (n+\frac {1}{2};\frac {1}{2}-m,1;n+\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {\sec (e+f x)+1}} \]

[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]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3912, 136} \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (c-c \sec (e+f x))^n F_1\left (n+\frac {1}{2};\frac {1}{2}-m,1;n+\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[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 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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 3912

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*} \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx &=-\frac {(c \tan (e+f x)) \operatorname {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} F_1\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]  time = 1.10, size = 0, normalized size = 0.00 \[ \int (1+\sec (e+f x))^m (c-c \sec (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[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]

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 2.14, size = 0, normalized size = 0.00 \[ \int \left (1+\sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^m\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________