3.4.5 \(\int (\frac {\sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a A n-a C (1+n) \sec (c+d x))}{a (1+n)}+\sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+C \sec ^2(c+d x))) \, dx\) [305]

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

[Out]

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

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Rubi [A]
time = 0.62, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 6, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {4108, 3913, 3910, 134, 138, 4172} \begin {gather*} \frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^n*(-(a*A*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 + 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 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 \left (\frac {\sec ^{-n}(c+d x) (a+a \sec (c+d x))^n (-a A 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+C \sec ^2(c+d x)\right )\right ) \, dx &=\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)}+\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 {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 A n+a C (1+n) \sec (c+d x)) \, dx}{a (1+n)}+\frac {(C-A n+C 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 {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-\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 ((C-A n+C 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)}+\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 {\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 ((C-A n+C n) (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 (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 {(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}+\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 {(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 {\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))}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 38, normalized size = 1.00 \begin {gather*} \frac {A \sec ^{-n}(c+d x) (a (1+\sec (c+d x)))^n \sin (c+d x)}{d (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^n*(-(a*A*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 + 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 = 2.21, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n} \left (-a A n -a C \left (1+n \right ) \sec \left (d x +c \right )\right ) \left (\sec ^{-n}\left (d x +c \right )\right )}{a \left (1+n \right )}+\left (\sec ^{-1-n}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (38) = 76\).
time = 2.45, size = 310, normalized size = 8.16 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*(-a*A*n-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+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 = 3.65, size = 58, normalized size = 1.53 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \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}}-\frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A\,a\,n+\frac {C\,a\,\left (n+1\right )}{\cos \left (c+d\,x\right )}\right )}{a\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^n\,\left (n+1\right )} \,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) - ((a + a/cos(c + d*x))^n*(A*a*n
+ (C*a*(n + 1))/cos(c + d*x)))/(a*(1/cos(c + d*x))^n*(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) - ((a + a/cos(c + d*x))^n*(A*a*n
+ (C*a*(n + 1))/cos(c + d*x)))/(a*(1/cos(c + d*x))^n*(n + 1)), x)

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