\(\int (a (b \csc (c+d x))^p)^n \, dx\) [69]

Optimal result

Integrand size = 14, antiderivative size = 82 \[ \int \left (a (b \csc (c+d x))^p\right )^n \, dx=\frac {\cos (c+d x) \left (a (b \csc (c+d x))^p\right )^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\sin ^2(c+d x)\right ) \sin (c+d x)}{d (1-n p) \sqrt {\cos ^2(c+d x)}} \]

[Out]

cos(d*x+c)*(a*(b*csc(d*x+c))^p)^n*hypergeom([1/2, -1/2*n*p+1/2],[-1/2*n*p+3/2],sin(d*x+c)^2)*sin(d*x+c)/d/(-n*
p+1)/(cos(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4208, 3857, 2722} \[ \int \left (a (b \csc (c+d x))^p\right )^n \, dx=\frac {\sin (c+d x) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\sin ^2(c+d x)\right ) \left (a (b \csc (c+d x))^p\right )^n}{d (1-n p) \sqrt {\cos ^2(c+d x)}} \]

[In]

Int[(a*(b*Csc[c + d*x])^p)^n,x]

[Out]

(Cos[c + d*x]*(a*(b*Csc[c + d*x])^p)^n*Hypergeometric2F1[1/2, (1 - n*p)/2, (3 - n*p)/2, Sin[c + d*x]^2]*Sin[c
+ d*x])/(d*(1 - n*p)*Sqrt[Cos[c + d*x]^2])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 4208

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sec[e + f*x])^n)^
FracPart[p]/(c*Sec[e + f*x])^(n*FracPart[p])), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \left ((b \csc (c+d x))^{-n p} \left (a (b \csc (c+d x))^p\right )^n\right ) \int (b \csc (c+d x))^{n p} \, dx \\ & = \left (\left (a (b \csc (c+d x))^p\right )^n \left (\frac {\sin (c+d x)}{b}\right )^{n p}\right ) \int \left (\frac {\sin (c+d x)}{b}\right )^{-n p} \, dx \\ & = \frac {\cos (c+d x) \left (a (b \csc (c+d x))^p\right )^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\sin ^2(c+d x)\right ) \sin (c+d x)}{d (1-n p) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \left (a (b \csc (c+d x))^p\right )^n \, dx=-\frac {\cos (c+d x) \left (a (b \csc (c+d x))^p\right )^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+n p),\frac {3}{2},\cos ^2(c+d x)\right ) \sin (c+d x) \sin ^2(c+d x)^{\frac {1}{2} (-1+n p)}}{d} \]

[In]

Integrate[(a*(b*Csc[c + d*x])^p)^n,x]

[Out]

-((Cos[c + d*x]*(a*(b*Csc[c + d*x])^p)^n*Hypergeometric2F1[1/2, (1 + n*p)/2, 3/2, Cos[c + d*x]^2]*Sin[c + d*x]
*(Sin[c + d*x]^2)^((-1 + n*p)/2))/d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(((b*csc(d*x + c))^p*a)^n, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a*(b/sin(c + d*x))^p)^n,x)

[Out]

int((a*(b/sin(c + d*x))^p)^n, x)