3.2.23 \(\int \sin ^3(a+b x) \sin ^m(2 a+2 b x) \, dx\) [123]

Optimal. Leaf size=84 \[ \frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {4+m}{2};\frac {6+m}{2};\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)} \]

[Out]

(cos(b*x+a)^2)^(1/2-1/2*m)*hypergeom([2+1/2*m, 1/2-1/2*m],[3+1/2*m],sin(b*x+a)^2)*sin(b*x+a)^3*sin(2*b*x+2*a)^
m*tan(b*x+a)/b/(4+m)

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Rubi [A]
time = 0.06, 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 = {4395, 2657} \begin {gather*} \frac {\sin ^3(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+4}{2};\frac {m+6}{2};\sin ^2(a+b x)\right )}{b (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^m,x]

[Out]

((Cos[a + b*x]^2)^((1 - m)/2)*Hypergeometric2F1[(1 - m)/2, (4 + m)/2, (6 + m)/2, Sin[a + b*x]^2]*Sin[a + b*x]^
3*Sin[2*a + 2*b*x]^m*Tan[a + b*x])/(b*(4 + 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 ^3(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 ^{3+m}(a+b x) \, dx\\ &=\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {4+m}{2};\frac {6+m}{2};\sin ^2(a+b x)\right ) \sin ^3(a+b x) \sin ^m(2 a+2 b x) \tan (a+b x)}{b (4+m)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 5.72, size = 602, normalized size = 7.17 \begin {gather*} \frac {32 (4+m) \left (F_1\left (1+\frac {m}{2};-m,3+2 m;2+\frac {m}{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 (1+\frac {m}{2};-m,4+2 m;2+\frac {m}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) \cos ^6\left (\frac {1}{2} (a+b x)\right ) \sin ^4\left (\frac {1}{2} (a+b x)\right ) \sin ^m(2 (a+b x))}{b (2+m) \left (-2 (4+m) F_1\left (1+\frac {m}{2};-m,4+2 m;2+\frac {m}{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 (m F_1\left (2+\frac {m}{2};1-m,3+2 m;3+\frac {m}{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 (2+\frac {m}{2};1-m,4+2 m;3+\frac {m}{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 (2+\frac {m}{2};-m,4+2 m;3+\frac {m}{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 (2+\frac {m}{2};-m,4+2 m;3+\frac {m}{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 (2+\frac {m}{2};-m,5+2 m;3+\frac {m}{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 (2+\frac {m}{2};-m,5+2 m;3+\frac {m}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right ) (-1+\cos (a+b x))+(4+m) F_1\left (1+\frac {m}{2};-m,3+2 m;2+\frac {m}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (1+\cos (a+b x))\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^m,x]

[Out]

(32*(4 + m)*(AppellF1[1 + m/2, -m, 3 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - AppellF1[1 + m
/2, -m, 4 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Cos[(a + b*x)/2]^6*Sin[(a + b*x)/2]^4*Sin[
2*(a + b*x)]^m)/(b*(2 + m)*(-2*(4 + m)*AppellF1[1 + m/2, -m, 4 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b
*x)/2]^2]*Cos[(a + b*x)/2]^2 + 2*(m*AppellF1[2 + m/2, 1 - m, 3 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b
*x)/2]^2] - m*AppellF1[2 + m/2, 1 - m, 4 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 3*AppellF1
[2 + m/2, -m, 4 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 2*m*AppellF1[2 + m/2, -m, 4 + 2*m,
3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[2 + m/2, -m, 5 + 2*m, 3 + m/2, Tan[(a + b*x)/2]
^2, -Tan[(a + b*x)/2]^2] - 2*m*AppellF1[2 + m/2, -m, 5 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2
])*(-1 + Cos[a + b*x]) + (4 + m)*AppellF1[1 + m/2, -m, 3 + 2*m, 2 + 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 = 0.57, size = 0, normalized size = 0.00 \[\int \left (\sin ^{3}\left (x b +a \right )\right ) \left (\sin ^{m}\left (2 x b +2 a \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^m,x)

[Out]

int(sin(b*x+a)^3*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)^3*sin(2*b*x+2*a)^m,x, algorithm="maxima")

[Out]

integrate(sin(2*b*x + 2*a)^m*sin(b*x + a)^3, 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)^3*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*sin(b*x + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{3}{\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)**3*sin(2*b*x+2*a)**m,x)

[Out]

Integral(sin(a + b*x)**3*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)^3*sin(2*b*x+2*a)^m,x, algorithm="giac")

[Out]

integrate(sin(2*b*x + 2*a)^m*sin(b*x + a)^3, 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 )}^3\,{\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)^3*sin(2*a + 2*b*x)^m,x)

[Out]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^m, x)

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