\(\int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx\) [126]

Optimal result

Integrand size = 18, antiderivative size = 72 \[ \int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m}{2},\frac {2+m}{2},\sin ^2(a+b x)\right ) \sec (a+b x) \sin ^m(2 a+2 b x)}{b m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4395, 2657} \[ \int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {\sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m}{2},\frac {m+2}{2},\sin ^2(a+b x)\right )}{b m} \]

[In]

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

[Out]

((Cos[a + b*x]^2)^((1 - m)/2)*Hypergeometric2F1[(1 - m)/2, m/2, (2 + m)/2, Sin[a + b*x]^2]*Sec[a + b*x]*Sin[2*
a + 2*b*x]^m)/(b*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 ^{-1+m}(a+b x) \, dx \\ & = \frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m}{2},\frac {2+m}{2},\sin ^2(a+b x)\right ) \sec (a+b x) \sin ^m(2 a+2 b x)}{b m} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.40 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.53 \[ \int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx=\frac {2 (2+m) \operatorname {AppellF1}\left (\frac {m}{2},-m,2 m,\frac {2+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \cos ^2\left (\frac {1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b m \left ((2+m) \operatorname {AppellF1}\left (\frac {m}{2},-m,2 m,\frac {2+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))-4 m \left (\operatorname {AppellF1}\left (\frac {2+m}{2},1-m,2 m,\frac {4+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+2 \operatorname {AppellF1}\left (\frac {2+m}{2},-m,1+2 m,\frac {4+m}{2},\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (a+b x)\right )\right )} \]

[In]

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

[Out]

(2*(2 + m)*AppellF1[m/2, -m, 2*m, (2 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2*Sin[2
*(a + b*x)]^m)/(b*m*((2 + m)*AppellF1[m/2, -m, 2*m, (2 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + C
os[a + b*x]) - 4*m*(AppellF1[(2 + m)/2, 1 - m, 2*m, (4 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*Ap
pellF1[(2 + m)/2, -m, 1 + 2*m, (4 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sin[(a + b*x)/2]^2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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