3.304 \(\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=253 \[ -\frac{(-A n+C n+C) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac{\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac{1}{2}-n} (a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (\frac{1}{2}-n,-n,1-n,-\frac{2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)}+\frac{C 2^{n+\frac{3}{2}} \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac{1}{2};n+1,-n-\frac{1}{2};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d} \]

[Out]

(A*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n) - ((C - A*n + C*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])) + (2^(3/2 + n)*C*AppellF1[1
/2, 1 + n, -1/2 - n, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*(a + a*Sec[c +
d*x])^n*Tan[c + d*x])/d

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Rubi [A]  time = 0.521556, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4087, 4023, 3828, 3825, 132, 133} \[ -\frac{(-A n+C n+C) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac{\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac{1}{2}-n} (a \sec (c+d x)+a)^n \, _2F_1\left (\frac{1}{2}-n,-n;1-n;-\frac{2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)}+\frac{C 2^{n+\frac{3}{2}} \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac{1}{2};n+1,-n-\frac{1}{2};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

(A*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n) - ((C - A*n + C*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])) + (2^(3/2 + n)*C*AppellF1[1
/2, 1 + n, -1/2 - n, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(1 + Sec[c + d*x])^(-1/2 - n)*(a + a*Sec[c +
d*x])^n*Tan[c + d*x])/d

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4023

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

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac{\int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (a A n+a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}\\ &=\frac{A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac{C \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a}+\left (-C+\frac{A n}{1+n}\right ) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx\\ &=\frac{A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\left (C (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))^{1+n} \, dx+\left (\left (-C+\frac{A n}{1+n}\right ) (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\\ &=\frac{A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac{\left (C (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 \sqrt{1-\sec (c+d x)}}+\frac{\left (\left (-C+\frac{A n}{1+n}\right ) (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 \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{\left (C-\frac{A n}{1+n}\right ) \, _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+\sec (c+d x))}+\frac{2^{\frac{3}{2}+n} C F_1\left (\frac{1}{2};1+n,-\frac{1}{2}-n;\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d}\\ \end{align*}

Mathematica [F]  time = 24.6505, size = 0, normalized size = 0. \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

Integrate[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2), x]

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Maple [F]  time = 0.376, 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+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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+C*sec(d*x+c)^2),x)

[Out]

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + 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 (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + 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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + 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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1), x)