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

Optimal. Leaf size=72 \[ \frac {\sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {m}{2};\frac {m+2}{2};\sin ^2(a+b x)\right )}{b m} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.87, 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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right ), x\right ) \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F]  time = 4.34, size = 0, normalized size = 0.00 \[ \int \csc \left (b x +a \right ) \left (\sin ^{m}\left (2 b x +2 a \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (2 \, b x + 2 \, a\right )^{m} \csc \left (b x + a\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^m}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin ^{m}{\left (2 a + 2 b x \right )} \csc {\left (a + b x \right )}\, dx \]

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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