3.4.4 \(\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [304]

Optimal. Leaf size=253 \[ \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {(C-A n+C n) \, _2F_1\left (\frac {1}{2}-n,-n;1-n;-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+n) (1+\sec (c+d x))}+\frac {2^{\frac {3}{2}+n} C F_1\left (\frac {1}{2};1+n,-\frac {1}{2}-n;\frac {3}{2};1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)}{d} \]

[Out]

A*(a+a*sec(d*x+c))^n*sin(d*x+c)/d/(1+n)/(sec(d*x+c)^n)-(-A*n+C*n+C)*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))+2^(3/2+n)*C*AppellF1(1/2,1+n,-1/2-n,3/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(1+sec(d*x+c))^(-1/
2-n)*(a+a*sec(d*x+c))^n*tan(d*x+c)/d

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Rubi [A]
time = 0.35, antiderivative size = 253, normalized size of antiderivative = 1.00, 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 = {4172, 4108, 3913, 3910, 134, 138} \begin {gather*} -\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} \end {gather*}

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 134

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)/((b*e - a*f)*(m + 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)*((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 138

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

Rule 3910

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 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
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 4108

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 4172

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])

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 ) \text {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 ) \text {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*}

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Mathematica [F]
time = 25.53, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx \end {gather*}

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 = 1.43, size = 0, normalized size = 0.00 \[\int \left (\sec ^{-1-n}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

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+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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