3.329 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=217 \[ -\frac{2 \sqrt{2} a (A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};-\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} a B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};-\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt{\sin (e+f x)+1}} \]

[Out]

(-4*Sqrt[2]*a*B*AppellF1[1/2, -3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x
]*(c + d*Sin[e + f*x])^n)/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n) - (2*Sqrt[2]*a*(A - B)*A
ppellF1[1/2, -1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e +
f*x])^n)/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

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Rubi [A]  time = 0.300444, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2968, 3017, 2755, 139, 138, 2784} \[ -\frac{2 \sqrt{2} a (A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};-\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} a B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};-\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt{\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]

[Out]

(-4*Sqrt[2]*a*B*AppellF1[1/2, -3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x
]*(c + d*Sin[e + f*x])^n)/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n) - (2*Sqrt[2]*a*(A - B)*A
ppellF1[1/2, -1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e +
f*x])^n)/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3017

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Dist[A - C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Dist[C, Int[(a
+ b*Sin[e + f*x])^m*(1 + Sin[e + f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A - B + C, 0] &&
!IntegerQ[2*m]

Rule 2755

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*C
os[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]]), Subst[Int[((a + b*x)^m*Sqrt[1 + (d*x)/c])/Sqrt
[1 - (d*x)/c], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b
^2, 0] &&  !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 2784

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[(a^m*Cos[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]]), Subst[Int[((1 + (b*x)/a)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[1 - (b*x)/a], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0]
&& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx &=\int (c+d \sin (e+f x))^n \left (a A+(a A+a B) \sin (e+f x)+a B \sin ^2(e+f x)\right ) \, dx\\ &=(a (A-B)) \int (1+\sin (e+f x)) (c+d \sin (e+f x))^n \, dx+(a B) \int (1+\sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx\\ &=\frac{(a (A-B) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} (c+d x)^n}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{(a B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} (c+d x)^n}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=\frac{\left (a (A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} \left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (a B \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} \left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=-\frac{4 \sqrt{2} a B F_1\left (\frac{1}{2};-\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{f \sqrt{1+\sin (e+f x)}}-\frac{2 \sqrt{2} a (A-B) F_1\left (\frac{1}{2};-\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{f \sqrt{1+\sin (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 9.05416, size = 0, normalized size = 0. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]

[Out]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n, x]

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Maple [F]  time = 0.48, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (B a \cos \left (f x + e\right )^{2} -{\left (A + B\right )} a \sin \left (f x + e\right ) -{\left (A + B\right )} a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

integral(-(B*a*cos(f*x + e)^2 - (A + B)*a*sin(f*x + e) - (A + B)*a)*(d*sin(f*x + e) + c)^n, 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((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)