\(\int \cot (c+d x) (a+a \sec (c+d x))^n \, dx\) [223]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^n}{2 d n}+\frac {\operatorname {Hypergeometric2F1}(1,n,1+n,1+\sec (c+d x)) (a+a \sec (c+d x))^n}{d n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3965, 88, 67, 70} \[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\frac {(a \sec (c+d x)+a)^n \operatorname {Hypergeometric2F1}(1,n,n+1,\sec (c+d x)+1)}{d n}-\frac {(a \sec (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (1,n,n+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{2 d n} \]

[In]

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

[Out]

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

Rule 67

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

Rule 70

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

Rule 88

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 3965

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]

Rubi steps \begin{align*} \text {integral}& = \frac {a^2 \text {Subst}\left (\int \frac {(a+a x)^{-1+n}}{x (-a+a x)} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a \text {Subst}\left (\int \frac {(a+a x)^{-1+n}}{x} \, dx,x,\sec (c+d x)\right )}{d}+\frac {a^2 \text {Subst}\left (\int \frac {(a+a x)^{-1+n}}{-a+a x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {\operatorname {Hypergeometric2F1}\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 {\operatorname {Hypergeometric2F1}(1,n,1+n,1+\sec (c+d x)) (a+a \sec (c+d x))^n}{d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {\left (\operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+\sec (c+d x))\right )-2 \operatorname {Hypergeometric2F1}(1,n,1+n,1+\sec (c+d x))\right ) (a (1+\sec (c+d x)))^n}{2 d n} \]

[In]

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

[Out]

-1/2*((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)/(d*n)

Maple [F]

\[\int \cot \left (d x +c \right ) \left (a +a \sec \left (d x +c \right )\right )^{n}d x\]

[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)

Fricas [F]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right ) \,d x } \]

[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]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \cot {\left (c + d x \right )}\, dx \]

[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)

Maxima [F]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right ) \,d x } \]

[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)

Giac [F]

\[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right ) \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cot (c+d x) (a+a \sec (c+d x))^n \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

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

[Out]

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