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

Optimal. Leaf size=38 \[ \frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]

[Out]

(A*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n)

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Rubi [A]  time = 0.981481, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 6, integrand size = 102, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {4023, 3828, 3825, 132, 133, 4086} \[ \frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

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])

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 \left (\frac{\sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a (B+A n+B n)-a C (1+n) \sec (c+d x))}{a (1+n)}+\sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )\right ) \, dx &=\frac{\int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a (B+A n+B n)-a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}+\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n \left (A+B \sec (c+d x)+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{C \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a}+\frac{\int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (a (B+A n+B n)+a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}-\left (B-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)}+\frac{C \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^{1+n} \, dx}{a}+\frac{(B+A n+B n-C (1+n)) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx}{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 (B-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)}+\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+\frac{\left ((B+A n+B n-C (1+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{\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 (B-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 (B-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}+\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 ((B+A n+B n-C (1+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{\left (B-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{(B+A n+B n-C (1+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 = 0.154944, size = 38, normalized size = 1. \[ \frac{A \sin (c+d x) \sec ^{-n}(c+d x) (a (\sec (c+d x)+1))^n}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(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 = 1.303, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( -a \left ( An+Bn+B \right ) -aC \left ( 1+n \right ) \sec \left ( dx+c \right ) \right ) }{ \left ( 1+n \right ) a \left ( \sec \left ( dx+c \right ) \right ) ^{n}}}+ \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 ) +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((a+a*sec(d*x+c))^n*(-a*(A*n+B*n+B)-a*C*(1+n)*sec(d*x+c))/a/(1+n)/(sec(d*x+c)^n)+sec(d*x+c)^(-1-n)*(a+a*sec
(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [B]  time = 19.4958, size = 419, normalized size = 11.03 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (-{\left (d n + d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) - c\right ) \sin \left (c n\right ) -{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (-{\left (d n - d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) + c\right ) \sin \left (c n\right ) -{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (c n\right ) \sin \left (-{\left (d n + d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) - c\right ) +{\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )}^{n} A a^{n} \cos \left (c n\right ) \sin \left (-{\left (d n - d\right )} x + 2 \, n \arctan \left (\sin \left (d x + c\right ), \cos \left (d x + c\right ) + 1\right ) + c\right )}{2 \,{\left ({\left (d n + d\right )} 2^{n} \cos \left (c n\right )^{2} +{\left (d n + d\right )} 2^{n} \sin \left (c n\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(-a*(A*n+B*n+B)-a*C*(1+n)*sec(d*x+c))/a/(1+n)/(sec(d*x+c)^n)+sec(d*x+c)^(-1-n)*(a
+a*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2*((cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)^n*A*a^n*cos(-(d*n + d)*x + 2*n*arctan2(sin(d*x + c
), cos(d*x + c) + 1) - c)*sin(c*n) - (cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)^n*A*a^n*cos(-(d*n
- d)*x + 2*n*arctan2(sin(d*x + c), cos(d*x + c) + 1) + c)*sin(c*n) - (cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(
d*x + c) + 1)^n*A*a^n*cos(c*n)*sin(-(d*n + d)*x + 2*n*arctan2(sin(d*x + c), cos(d*x + c) + 1) - c) + (cos(d*x
+ c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)^n*A*a^n*cos(c*n)*sin(-(d*n - d)*x + 2*n*arctan2(sin(d*x + c), co
s(d*x + c) + 1) + c))/((d*n + d)*2^n*cos(c*n)^2 + (d*n + d)*2^n*sin(c*n)^2)

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Fricas [A]  time = 0.595195, size = 142, normalized size = 3.74 \begin{align*} \frac{A \left (\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}\right )^{n} \frac{1}{\cos \left (d x + c\right )}^{-n - 1} \sin \left (d x + c\right )}{{\left (d n + d\right )} \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(-a*(A*n+B*n+B)-a*C*(1+n)*sec(d*x+c))/a/(1+n)/(sec(d*x+c)^n)+sec(d*x+c)^(-1-n)*(a
+a*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

A*((a*cos(d*x + c) + a)/cos(d*x + c))^n*(1/cos(d*x + c))^(-n - 1)*sin(d*x + c)/((d*n + d)*cos(d*x + c))

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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((a+a*sec(d*x+c))**n*(-a*(A*n+B*n+B)-a*C*(1+n)*sec(d*x+c))/a/(1+n)/(sec(d*x+c)**n)+sec(d*x+c)**(-1-n)
*(a+a*sec(d*x+c))**n*(A+B*sec(d*x+c)+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} + 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} - \frac{{\left (C a{\left (n + 1\right )} \sec \left (d x + c\right ) +{\left (A n + B n + B\right )} a\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{a{\left (n + 1\right )} \sec \left (d x + c\right )^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(-a*(A*n+B*n+B)-a*C*(1+n)*sec(d*x+c))/a/(1+n)/(sec(d*x+c)^n)+sec(d*x+c)^(-1-n)*(a
+a*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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