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

Optimal. Leaf size=84 \[ \frac {\sin ^2(a+b x) \tan (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {m+3}{2};\frac {m+5}{2};\sin ^2(a+b x)\right )}{b (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 84, 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 = {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 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 ^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*}

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Mathematica [C]  time = 3.51, 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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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{m}, x\right ) \]

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

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)

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maple [F]  time = 6.23, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{2}\left (b x +a \right )\right ) \left (\sin ^{m}\left (2 b x +2 a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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