\(\int \csc (c+d x) (a+b \sec (c+d x))^n \, dx\) [271]

Optimal result

Integrand size = 19, antiderivative size = 115 \[ \int \csc (c+d x) (a+b \sec (c+d x))^n \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a-b) d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a+b) d (1+n)} \]

[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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3959, 88, 70} \[ \int \csc (c+d x) (a+b \sec (c+d x))^n \, dx=\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)} \]

[In]

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

[Out]

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

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 3959

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[-f^(-1), Subs
t[Int[(-1 + x)^((p - 1)/2)*(1 + x)^((p - 1)/2)*((a + b*x)^m/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[{a,
b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.15 \[ \int \csc (c+d x) (a+b \sec (c+d x))^n \, dx=\frac {\left (\operatorname {Hypergeometric2F1}\left (1,-n,1-n,\frac {(a+b) \cos (c+d x)}{b+a \cos (c+d x)}\right )-2^n \operatorname {Hypergeometric2F1}\left (-n,-n,1-n,\frac {(-a+b) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{2 b}\right ) \left (\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{b}\right )^{-n}\right ) (a+b \sec (c+d x))^n}{2 d n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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