\(\int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx\) [127]

Optimal result

Integrand size = 20, antiderivative size = 85 \[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=-\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \csc (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sin ^2(a+b x)\right ) \sec (a+b x) \sin ^m(2 a+2 b x)}{b (1-m)} \]

[Out]

-(cos(b*x+a)^2)^(1/2-1/2*m)*csc(b*x+a)*hypergeom([-1/2+1/2*m, 1/2-1/2*m],[1/2+1/2*m],sin(b*x+a)^2)*sec(b*x+a)*
sin(2*b*x+2*a)^m/b/(1-m)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4395, 2657} \[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=-\frac {\csc (a+b x) \sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m-1}{2},\frac {m+1}{2},\sin ^2(a+b x)\right )}{b (1-m)} \]

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^m,x]

[Out]

-(((Cos[a + b*x]^2)^((1 - m)/2)*Csc[a + b*x]*Hypergeometric2F1[(1 - m)/2, (-1 + m)/2, (1 + m)/2, Sin[a + b*x]^
2]*Sec[a + b*x]*Sin[2*a + 2*b*x]^m)/(b*(1 - m)))

Rule 2657

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b^(2*IntPart[
(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 4395

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[(g*Sin[c + d
*x])^p/(Cos[a + b*x]^p*(f*Sin[a + b*x])^p), Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b
, c, d, f, g, n, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72 \[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),m,\frac {1+m}{2},-\tan ^2(a+b x)\right ) \sec ^2(a+b x)^m \sin ^m(2 (a+b x))}{b (-1+m)} \]

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^m,x]

[Out]

(Cot[a + b*x]*Hypergeometric2F1[(-1 + m)/2, m, (1 + m)/2, -Tan[a + b*x]^2]*(Sec[a + b*x]^2)^m*Sin[2*(a + b*x)]
^m)/(b*(-1 + m))

Maple [F]

\[\int \csc \left (x b +a \right )^{2} \sin \left (2 x b +2 a \right )^{m}d x\]

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^m,x)

[Out]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^m,x)

Fricas [F]

\[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )^{2} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^m,x, algorithm="fricas")

[Out]

integral(sin(2*b*x + 2*a)^m*csc(b*x + a)^2, x)

Sympy [F]

\[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\int \sin ^{m}{\left (2 a + 2 b x \right )} \csc ^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**m,x)

[Out]

Integral(sin(2*a + 2*b*x)**m*csc(a + b*x)**2, x)

Maxima [F]

\[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )^{2} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^m,x, algorithm="maxima")

[Out]

integrate(sin(2*b*x + 2*a)^m*csc(b*x + a)^2, x)

Giac [F]

\[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )^{2} \,d x } \]

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^m,x, algorithm="giac")

[Out]

integrate(sin(2*b*x + 2*a)^m*csc(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(a+b x) \sin ^m(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^m}{{\sin \left (a+b\,x\right )}^2} \,d x \]

[In]

int(sin(2*a + 2*b*x)^m/sin(a + b*x)^2,x)

[Out]

int(sin(2*a + 2*b*x)^m/sin(a + b*x)^2, x)