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 \[ \frac{(A n+B n+B) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac{\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac{1}{2}-n} (a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (\frac{1}{2}-n,-n,1-n,-\frac{2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]

[Out]

(A*(a + a*Sec[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]))

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Rubi [A]  time = 0.255129, antiderivative size = 164, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 1.07748, size = 111, normalized size = 0.68 \[ \frac{\sin (c+d x) \sec ^{-n}(c+d x) (a (\sec (c+d x)+1))^n \left (\frac{(A n+B n+B) \left (-\cot ^2\left (\frac{1}{2} (c+d x)\right )\right )^{\frac{1}{2}-n} \text{Hypergeometric2F1}\left (\frac{1}{2}-n,-n,1-n,\csc ^2\left (\frac{1}{2} (c+d x)\right )\right )}{n (\cos (c+d x)+1)}+A\right )}{d (n+1)} \]

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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Maple [F]  time = 1.14, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{-1-n} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+B\sec \left ( dx+c \right ) \right ) \, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)