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

Optimal. Leaf size=84 \[ \frac{\sin ^2(a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac{1-m}{2}} \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m+3}{2},\frac{m+5}{2},\sin ^2(a+b x)\right )}{b (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.0765246, antiderivative size = 84, 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 = {4310, 2577} \[ \frac{\sin ^2(a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac{1-m}{2}} \, _2F_1\left (\frac{1-m}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(a+b x)\right )}{b (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

\begin{align*} \int \sin ^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 ^m(a+b x) \sin ^{2+m}(a+b x) \, dx\\ &=\frac{\cos ^2(a+b x)^{\frac{1-m}{2}} \, _2F_1\left (\frac{1-m}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(a+b x)\right ) \sin ^2(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (3+m)}\\ \end{align*}

Mathematica [C]  time = 3.59562, size = 602, normalized size = 7.17 \[ \frac{16 (m+3) \sin ^3\left (\frac{1}{2} (a+b x)\right ) \cos ^5\left (\frac{1}{2} (a+b x)\right ) \sin ^m(2 (a+b x)) \left (F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-F_1\left (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )}{b (m+1) \left (-2 (m+3) \cos ^2\left (\frac{1}{2} (a+b x)\right ) F_1\left (\frac{m+1}{2};-m,2 m+3;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+2 (\cos (a+b x)-1) \left (m F_1\left (\frac{m+3}{2};1-m,2 (m+1);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-m F_1\left (\frac{m+3}{2};1-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-2 m F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-3 F_1\left (\frac{m+3}{2};-m,2 (m+2);\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+2 m F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+2 F_1\left (\frac{m+3}{2};-m,2 m+3;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )+(m+3) (\cos (a+b x)+1) F_1\left (\frac{m+1}{2};-m,2 (m+1);\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(16*(3 + m)*(AppellF1[(1 + m)/2, -m, 2*(1 + m), (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 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]^5*Sin[(a + b*x)
/2]^3*Sin[2*(a + b*x)]^m)/(b*(1 + m)*(-2*(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*(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, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a +
b*x)/2]^2] - 3*AppellF1[(3 + m)/2, -m, 2*(2 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 2*m*A
ppellF1[(3 + m)/2, -m, 2*(2 + m), (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*AppellF1[(3 + m)/2,
-m, 3 + 2*m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*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]) + (3 + m)*AppellF1[(1 + m)/2, -m, 2*(1 + m)
, (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x])))

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Maple [F]  time = 0.848, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \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(sin(b*x+a)^2*sin(2*b*x+2*a)^m,x)

[Out]

int(sin(b*x+a)^2*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} \sin \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(cos(b*x + a)^2 - 1)*sin(2*b*x + 2*a)^m, 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(sin(b*x+a)**2*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} \sin \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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