\(\int \cot (c+d x) (a+b \sec (c+d x))^n \, dx\) [356]

Optimal result

Integrand size = 19, antiderivative size = 162 \[ \int \cot (c+d x) (a+b \sec (c+d x))^n \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a-b) d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a+b) d (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3970, 975, 67, 845, 70} \[ \int \cot (c+d x) (a+b \sec (c+d x))^n \, dx=-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)}+\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

[In]

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

[Out]

-1/2*(Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a - b)]*(a + b*Sec[c + d*x])^(1 + n))/((a - b)*
d*(1 + n)) - (Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c + d*x])^(1 + n))/(
2*(a + b)*d*(1 + n)) + (Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n
))/(a*d*(1 + 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 845

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 975

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 3970

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

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.81 \[ \int \cot (c+d x) (a+b \sec (c+d x))^n \, dx=-\frac {\left (a (a+b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right )+(a-b) \left (a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right )-2 (a+b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right )\right )\right ) (a+b \sec (c+d x))^{1+n}}{2 a (a-b) (a+b) d (1+n)} \]

[In]

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

[Out]

-1/2*((a*(a + b)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a - b)] + (a - b)*(a*Hypergeometric2
F1[1, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a + b)] - 2*(a + b)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*Sec[c
+ d*x])/a]))*(a + b*Sec[c + d*x])^(1 + n))/(a*(a - b)*(a + b)*d*(1 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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