3.276 \(\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=164 \[ \frac {(A n+B n+B) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac {\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a \sec (c+d x)+a)^n \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]

[Out]

A*(a+a*sec(d*x+c))^n*sin(d*x+c)/d/(1+n)/(sec(d*x+c)^n)+(A*n+B*n+B)*hypergeom([-n, 1/2-n],[1-n],-2*sec(d*x+c)/(
1-sec(d*x+c)))*sec(d*x+c)^(1-n)*((1+sec(d*x+c))/(1-sec(d*x+c)))^(1/2-n)*(a+a*sec(d*x+c))^n*sin(d*x+c)/d/n/(1+n
)/(1+sec(d*x+c))

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Rubi [A]  time = 0.26, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4013, 3828, 3825, 132} \[ \frac {(A n+B n+B) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac {\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a \sec (c+d x)+a)^n \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

(A*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n) + ((B + A*n + B*n)*Hypergeometric2F1[1/2 -
n, -n, 1 - n, (-2*Sec[c + d*x])/(1 - Sec[c + d*x])]*Sec[c + d*x]^(1 - n)*((1 + Sec[c + d*x])/(1 - Sec[c + d*x]
))^(1/2 - n)*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*n*(1 + n)*(1 + Sec[c + d*x]))

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 3825

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

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(B+A n+B n) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx}{1+n}\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {\left ((B+A n+B n) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^n \, dx}{1+n}\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {\left ((B+A n+B n) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{-\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d (1+n) \sqrt {1-\sec (c+d x)}}\\ &=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(B+A n+B n) \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+n) (1+\sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 111, normalized size = 0.68 \[ \frac {\sin (c+d x) \sec ^{-n}(c+d x) (a (\sec (c+d x)+1))^n \left (\frac {(A n+B n+B) \left (-\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^{\frac {1}{2}-n} \, _2F_1\left (\frac {1}{2}-n,-n;1-n;\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n (\cos (c+d x)+1)}+A\right )}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]

[Out]

((A + ((B + A*n + B*n)*(-Cot[(c + d*x)/2]^2)^(1/2 - n)*Hypergeometric2F1[1/2 - n, -n, 1 - n, Csc[(c + d*x)/2]^
2])/(n*(1 + Cos[c + d*x])))*(a*(1 + Sec[c + d*x]))^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n)

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{-n - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1), x)

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maple [F]  time = 4.56, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{-1-n}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)

[Out]

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^n)/(1/cos(c + d*x))^(n + 1),x)

[Out]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^n)/(1/cos(c + d*x))^(n + 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(-1-n)*(a+a*sec(d*x+c))**n*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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