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

Optimal. Leaf size=99 \[ \frac {c 2^{n+\frac {1}{2}} \tan (e+f x) (a \sec (e+f x)+a) (1-\sec (e+f x))^{\frac {1}{2}-n} F_1\left (\frac {3}{2};\frac {1}{2}-n,1;\frac {5}{2};\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{3 f} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3912, 137, 136} \[ \frac {c 2^{n+\frac {1}{2}} \tan (e+f x) (a \sec (e+f x)+a) (1-\sec (e+f x))^{\frac {1}{2}-n} F_1\left (\frac {3}{2};\frac {1}{2}-n,1;\frac {5}{2};\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{3 f} \]

Antiderivative was successfully verified.

[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 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 137

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 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 (a+a \sec (e+f x)) (c-c \sec (e+f x))^n \, dx &=-\frac {(a c \tan (e+f x)) \operatorname {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 ) \operatorname {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 F_1\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*}

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Mathematica [F]  time = 1.60, size = 0, normalized size = 0.00 \[ \int (a+a \sec (e+f x)) (c-c \sec (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[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]

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 2.10, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (f x +e \right )\right ) \left (c -c \sec \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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