3.126 \(\int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx\)

Optimal. Leaf size=72 \[ \frac{\sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac{1-m}{2}} \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m}{2},\frac{m+2}{2},\sin ^2(a+b x)\right )}{b m} \]

[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)

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Rubi [A]  time = 0.0702846, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4310, 2577} \[ \frac{\sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac{1-m}{2}} \, _2F_1\left (\frac{1-m}{2},\frac{m}{2};\frac{m+2}{2};\sin ^2(a+b x)\right )}{b m} \]

Antiderivative was successfully verified.

[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 4310

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]

Rule 2577

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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \csc (a+b x) \sin ^m(2 a+2 b x) \, dx &=\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}} \, _2F_1\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]  time = 0.889277, size = 254, normalized size = 3.53 \[ \frac{2 (m+2) \cos ^2\left (\frac{1}{2} (a+b x)\right ) \sin ^m(2 (a+b x)) F_1\left (\frac{m}{2};-m,2 m;\frac{m+2}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )}{b m \left ((m+2) (\cos (a+b x)+1) F_1\left (\frac{m}{2};-m,2 m;\frac{m+2}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 m \sin ^2\left (\frac{1}{2} (a+b x)\right ) \left (F_1\left (\frac{m+2}{2};1-m,2 m;\frac{m+4}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+2 F_1\left (\frac{m+2}{2};-m,2 m+1;\frac{m+4}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[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))

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Maple [F]  time = 0.517, size = 0, normalized size = 0. \begin{align*} \int \csc \left ( bx+a \right ) \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{m}{\left (2 a + 2 b x \right )} \csc{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)