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

Optimal. Leaf size=85 \[ -\frac {\cos ^2(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac {1-m}{2}} \sin ^m(2 a+2 b x) \, _2F_1\left (\frac {1-m}{2},\frac {m+3}{2};\frac {m+5}{2};\cos ^2(a+b x)\right )}{b (m+3)} \]

[Out]

-cos(b*x+a)^2*cot(b*x+a)*hypergeom([3/2+1/2*m, 1/2-1/2*m],[5/2+1/2*m],cos(b*x+a)^2)*(sin(b*x+a)^2)^(1/2-1/2*m)
*sin(2*b*x+2*a)^m/b/(3+m)

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Rubi [A]  time = 0.07, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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]

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]

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*}

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Mathematica [C]  time = 7.81, 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])))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sin \left (2 \, b x + 2 \, a\right )^{m} \cos \left (b x + a\right )^{2}, x\right ) \]

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)

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

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)

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maple [F]  time = 5.90, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (b x +a \right )\right ) \left (\sin ^{m}\left (2 b x +2 a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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)

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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