3.2.25 \(\int \sin (a+b x) \sin ^m(2 a+2 b x) \, dx\) [125]

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

[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 = 0.05, 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 = {4395, 2657} \begin {gather*} \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)} \end {gather*}

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 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 4395

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] Result contains complex when optimal does not.
time = 0.57, size = 152, normalized size = 1.85 \begin {gather*} -\frac {i 2^{-1-m} e^{i (a+b x)} \left (-i e^{-2 i (a+b x)} \left (-1+e^{4 i (a+b x)}\right )\right )^{1+m} \left ((1-2 m) \, _2F_1\left (1,\frac {1}{4} (3+2 m);\frac {1}{4} (3-2 m);e^{4 i (a+b x)}\right )+e^{2 i (a+b x)} (1+2 m) \, _2F_1\left (1,\frac {1}{4} (5+2 m);\frac {1}{4} (5-2 m);e^{4 i (a+b x)}\right )\right )}{b \left (-1+4 m^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \sin \left (x b +a \right ) \left (\sin ^{m}\left (2 x b +2 a \right )\right )\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (a + b x \right )} \sin ^{m}{\left (2 a + 2 b x \right )}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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