3.189 \(\int \cos (a+b x) \sin ^m(2 a+2 b x) \, dx\)

Optimal. Leaf size=83 \[ -\frac{\cos (a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m+2}{2},\frac{m+4}{2},\cos ^2(a+b x)\right )}{b (m+2)} \]

[Out]

-((Cos[a + b*x]*Cot[a + b*x]*Hypergeometric2F1[(1 - m)/2, (2 + m)/2, (4 + m)/2, Cos[a + b*x]^2]*(Sin[a + b*x]^
2)^((1 - m)/2)*Sin[2*a + 2*b*x]^m)/(b*(2 + m)))

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Rubi [A]  time = 0.0608929, antiderivative size = 83, 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 = {4309, 2576} \[ -\frac{\cos (a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \, _2F_1\left (\frac{1-m}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(a+b x)\right )}{b (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[a + b*x]*Cot[a + b*x]*Hypergeometric2F1[(1 - m)/2, (2 + m)/2, (4 + m)/2, Cos[a + b*x]^2]*(Sin[a + b*x]^
2)^((1 - m)/2)*Sin[2*a + 2*b*x]^m)/(b*(2 + m)))

Rule 4309

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

Rule 2576

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

Rubi steps

\begin{align*} \int \cos (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 ^{1+m}(a+b x) \sin ^m(a+b x) \, dx\\ &=-\frac{\cos (a+b x) \cot (a+b x) \, _2F_1\left (\frac{1-m}{2},\frac{2+m}{2};\frac{4+m}{2};\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x)}{b (2+m)}\\ \end{align*}

Mathematica [C]  time = 0.244453, size = 149, normalized size = 1.8 \[ \frac{2^{-m-1} e^{i (a+b x)} \left (-i e^{-2 i (a+b x)} \left (-1+e^{4 i (a+b x)}\right )\right )^{m+1} \left ((2 m-1) \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (3-2 m),e^{4 i (a+b x)}\right )+(2 m+1) e^{2 i (a+b x)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+5),\frac{1}{4} (5-2 m),e^{4 i (a+b x)}\right )\right )}{b \left (4 m^2-1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 - m)*E^(I*(a + b*x))*(((-I)*(-1 + E^((4*I)*(a + b*x))))/E^((2*I)*(a + b*x)))^(1 + m)*((-1 + 2*m)*Hyperg
eometric2F1[1, (3 + 2*m)/4, (3 - 2*m)/4, E^((4*I)*(a + b*x))] + E^((2*I)*(a + b*x))*(1 + 2*m)*Hypergeometric2F
1[1, (5 + 2*m)/4, (5 - 2*m)/4, E^((4*I)*(a + b*x))]))/(b*(-1 + 4*m^2))

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Maple [F]  time = 0.914, size = 0, normalized size = 0. \begin{align*} \int \cos \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(cos(b*x+a)*sin(2*b*x+2*a)^m,x)

[Out]

int(cos(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} \cos \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(2*b*x + 2*a)^m*cos(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} \cos \left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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} \cos \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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