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

Optimal. Leaf size=85 \[ -\frac{\cos ^2(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m+3}{2},\frac{m+5}{2},\cos ^2(a+b x)\right )}{b (m+3)} \]

[Out]

-((Cos[a + b*x]^2*Cot[a + b*x]*Hypergeometric2F1[(1 - m)/2, (3 + m)/2, (5 + m)/2, Cos[a + b*x]^2]*(Sin[a + b*x
]^2)^((1 - m)/2)*Sin[2*a + 2*b*x]^m)/(b*(3 + m)))

________________________________________________________________________________________

Rubi [A]  time = 0.0707474, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4309, 2576} \[ -\frac{\cos ^2(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac{1-m}{2}} \sin ^m(2 a+2 b x) \, _2F_1\left (\frac{1-m}{2},\frac{m+3}{2};\frac{m+5}{2};\cos ^2(a+b x)\right )}{b (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[a + b*x]^2*Cot[a + b*x]*Hypergeometric2F1[(1 - m)/2, (3 + m)/2, (5 + m)/2, Cos[a + b*x]^2]*(Sin[a + b*x
]^2)^((1 - m)/2)*Sin[2*a + 2*b*x]^m)/(b*(3 + 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 ^2(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 ^{2+m}(a+b x) \sin ^m(a+b x) \, dx\\ &=-\frac{\cos ^2(a+b x) \cot (a+b x) \, _2F_1\left (\frac{1-m}{2},\frac{3+m}{2};\frac{5+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 (3+m)}\\ \end{align*}

Mathematica [C]  time = 7.96872, size = 890, normalized size = 10.47 \[ \frac{4 (m+3) \left (4 F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-F_1\left (\frac{m+1}{2};-m,2 m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \cos ^3\left (\frac{1}{2} (a+b x)\right ) \cos ^2(a+b x) \sin \left (\frac{1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (m+1) \left (8 (m+3) F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{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 )-2 (m+3) F_1\left (\frac{m+1}{2};-m,2 m+1;\frac{m+3}{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 )-8 (m+3) F_1\left (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{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 )+2 \left (4 m F_1\left (\frac{m+3}{2};1-m,2 (m+1);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-m F_1\left (\frac{m+3}{2};1-m,2 m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 m F_1\left (\frac{m+3}{2};1-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-2 m F_1\left (\frac{m+3}{2};-m,2 (m+1);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-F_1\left (\frac{m+3}{2};-m,2 (m+1);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-8 m F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-12 F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+8 m F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+8 F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) (\cos (a+b x)-1)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(3 + m)*(4*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - AppellF
1[(1 + m)/2, -m, 1 + 2*m, (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(1 + m)/2, -m, 3 +
2*m, (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Cos[(a + b*x)/2]^3*Cos[a + b*x]^2*Sin[(a + b*x)/2]*S
in[2*(a + b*x)]^m)/(b*(1 + m)*(8*(3 + m)*AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(a + b*x)/2]^2, -Ta
n[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2 - 2*(3 + m)*AppellF1[(1 + m)/2, -m, 1 + 2*m, (3 + m)/2, Tan[(a + b*x)/2]^
2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2 - 8*(3 + m)*AppellF1[(1 + m)/2, -m, 3 + 2*m, (3 + m)/2, Tan[(a + b*
x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2 + 2*(4*m*AppellF1[(3 + m)/2, 1 - m, 2*(1 + m), (5 + m)/2, Tan
[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - m*AppellF1[(3 + m)/2, 1 - m, 1 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -
Tan[(a + b*x)/2]^2] - 4*m*AppellF1[(3 + m)/2, 1 - m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]
^2] - AppellF1[(3 + m)/2, -m, 2*(1 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 2*m*AppellF1[(3
+ m)/2, -m, 2*(1 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 12*AppellF1[(3 + m)/2, -m, 2*(2
+ m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 8*m*AppellF1[(3 + m)/2, -m, 2*(2 + m), (5 + m)/2,
Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 8*AppellF1[(3 + m)/2, -m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -
Tan[(a + b*x)/2]^2] + 8*m*AppellF1[(3 + m)/2, -m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]
)*(-1 + Cos[a + b*x])))

________________________________________________________________________________________

Maple [F]  time = 0.875, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{2} \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)^2*sin(2*b*x+2*a)^m,x)

[Out]

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

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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 )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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