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

Optimal. Leaf size=199 \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) \left (A (m+2) (c m+c+d m)+B \left (c m (m+2)+d \left (m^2+m+1\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{\cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

((B*d - (B*c + A*d)*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (2^(1/2 + m)*(A*(2 + m
)*(c + c*m + d*m) + B*(c*m*(2 + m) + d*(1 + m + m^2)))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 -
Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (B*d*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m))

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Rubi [A]  time = 0.361758, antiderivative size = 198, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2968, 3023, 2751, 2652, 2651} \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac{\cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*d - (B*c + A*d)*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (2^(1/2 + m)*(B*c*m*(2
+ m) + A*(2 + m)*(c + c*m + d*m) + B*d*(1 + m + m^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 -
Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)*(2 + m)) - (B*d*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m))

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 3023

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

Rule 2751

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

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x))^m \left (A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\int (a+a \sin (e+f x))^m (a (B d (1+m)+A c (2+m))-a (B d-(B c+A d) (2+m)) \sin (e+f x)) \, dx}{a (2+m)}\\ &=\frac{(B d-(B c+A d) (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (B c m (2+m)+A (2+m) (c+c m+d m)+B d \left (1+m+m^2\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{(B d-(B c+A d) (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}+\frac{\left (\left (B c m (2+m)+A (2+m) (c+c m+d m)+B d \left (1+m+m^2\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=\frac{(B d-(B c+A d) (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{2^{\frac{1}{2}+m} \left (B c m (2+m)+A (2+m) (c+c m+d m)+B d \left (1+m+m^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)}\\ \end{align*}

Mathematica [A]  time = 3.40543, size = 212, normalized size = 1.07 \[ -\frac{\csc ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^m (a (\sin (e+f x)+1))^m \left (-\frac{2}{5} (A-B) (c-d) \tan ^5\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{5}{2},m+3;\frac{7}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-\frac{4}{3} (A c-B d) \tan ^3\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{3}{2},m+3;\frac{5}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-2 (A+B) (c+d) \tan \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac{1}{2},m+3;\frac{3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((Csc[(2*e + Pi + 2*f*x)/4]^2)^m*(a*(1 + Sin[e + f*x]))^m*(-2*(A + B)*(c + d)*Hypergeometric2F1[1/2, 3 + m,
3/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4] - (4*(A*c - B*d)*Hypergeometric2F1[3/2, 3 + m, 5/
2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4]^3)/3 - (2*(A - B)*(c - d)*Hypergeometric2F1[5/2, 3
+ m, 7/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4]^5)/5))/f)

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Maple [F]  time = 2.088, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),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 (d \sin \left (f x + e\right ) + c\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin{\left (e + f x \right )}\right ) \left (c + d \sin{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))*(c + d*sin(e + f*x)), x)

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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 (d \sin \left (f x + e\right ) + c\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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