3.223 \(\int \cot (c+d x) (a+a \sec (c+d x))^n \, dx\)

Optimal. Leaf size=74 \[ \frac{(a \sec (c+d x)+a)^n \text{Hypergeometric2F1}(1,n,n+1,\sec (c+d x)+1)}{d n}-\frac{(a \sec (c+d x)+a)^n \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (\sec (c+d x)+1)\right )}{2 d n} \]

[Out]

-(Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(2*d*n) + (Hypergeometric2F1[1,
n, 1 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^n)/(d*n)

________________________________________________________________________________________

Rubi [A]  time = 0.0608846, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3880, 86, 65, 68} \[ \frac{(a \sec (c+d x)+a)^n \, _2F_1(1,n;n+1;\sec (c+d x)+1)}{d n}-\frac{(a \sec (c+d x)+a)^n \, _2F_1\left (1,n;n+1;\frac{1}{2} (\sec (c+d x)+1)\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*(a + a*Sec[c + d*x])^n,x]

[Out]

-(Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(2*d*n) + (Hypergeometric2F1[1,
n, 1 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^n)/(d*n)

Rule 3880

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(d*b^(m - 1)
)^(-1), Subst[Int[((-a + b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x, x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx &=\frac{a^2 \operatorname{Subst}\left (\int \frac{(a+a x)^{-1+n}}{x (-a+a x)} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{(a+a x)^{-1+n}}{x} \, dx,x,\sec (c+d x)\right )}{d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{(a+a x)^{-1+n}}{-a+a x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{\, _2F_1\left (1,n;1+n;\frac{1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^n}{2 d n}+\frac{\, _2F_1(1,n;1+n;1+\sec (c+d x)) (a+a \sec (c+d x))^n}{d n}\\ \end{align*}

Mathematica [A]  time = 0.0413445, size = 57, normalized size = 0.77 \[ -\frac{(a (\sec (c+d x)+1))^n \left (\text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (\sec (c+d x)+1)\right )-2 \text{Hypergeometric2F1}(1,n,n+1,\sec (c+d x)+1)\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*(a + a*Sec[c + d*x])^n,x]

[Out]

-((Hypergeometric2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2] - 2*Hypergeometric2F1[1, n, 1 + n, 1 + Sec[c + d*x]])*
(a*(1 + Sec[c + d*x]))^n)/(2*d*n)

________________________________________________________________________________________

Maple [F]  time = 0.33, size = 0, normalized size = 0. \begin{align*} \int \cot \left ( dx+c \right ) \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+a*sec(d*x+c))^n,x)

[Out]

int(cot(d*x+c)*(a+a*sec(d*x+c))^n,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*cot(d*x + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*cot(d*x + c), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{n} \cot{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))**n,x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*cot(c + d*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*cot(d*x + c), x)