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

Optimal. Leaf size=82 \[ \frac {\sin (a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(a+b x)\right )}{b (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 82, 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} \[ \frac {\sin (a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {m+2}{2};\frac {m+4}{2};\sin ^2(a+b x)\right )}{b (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[a + b*x]^2)^((1 - m)/2)*Hypergeometric2F1[(1 - m)/2, (2 + m)/2, (4 + m)/2, Sin[a + b*x]^2]*Sin[a + b*x]*
Sin[2*a + 2*b*x]^m*Tan[a + b*x])/(b*(2 + 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 \sin (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 {2+m}{2};\frac {4+m}{2};\sin ^2(a+b x)\right ) \sin (a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (2+m)}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 152, normalized size = 1.85 \[ -\frac {i 2^{-m-1} e^{i (a+b x)} \left (-i e^{-2 i (a+b x)} \left (-1+e^{4 i (a+b x)}\right )\right )^{m+1} \left ((1-2 m) \, _2F_1\left (1,\frac {1}{4} (2 m+3);\frac {1}{4} (3-2 m);e^{4 i (a+b x)}\right )+(2 m+1) e^{2 i (a+b x)} \, _2F_1\left (1,\frac {1}{4} (2 m+5);\frac {1}{4} (5-2 m);e^{4 i (a+b x)}\right )\right )}{b \left (4 m^2-1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*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)*Hy
pergeometric2F1[1, (3 + 2*m)/4, (3 - 2*m)/4, E^((4*I)*(a + b*x))] + E^((2*I)*(a + b*x))*(1 + 2*m)*Hypergeometr
ic2F1[1, (5 + 2*m)/4, (5 - 2*m)/4, E^((4*I)*(a + b*x))]))/(b*(-1 + 4*m^2))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 5.04, size = 0, normalized size = 0.00 \[ \int \sin \left (b x +a \right ) \left (\sin ^{m}\left (2 b x +2 a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(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 \[ \int \sin \left (2 \, b x + 2 \, a\right )^{m} \sin \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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