\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) [337]

Optimal result

Integrand size = 33, antiderivative size = 199 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {(B d-(B c+A d) (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (A (2+m) (c+c m+d m)+B \left (c m (2+m)+d \left (1+m+m^2\right )\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {B d \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)} \]

[Out]

(B*d-(A*d+B*c)*(2+m))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m)-2^(1/2+m)*(A*(2+m)*(c*m+d*m+c)+B*(c*m*(2+m)+
d*(m^2+m+1)))*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x
+e))^m/f/(m^2+3*m+2)-B*d*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3047, 3102, 2830, 2731, 2730} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+2) (c m+c+d m)+B c m (m+2)+B d \left (m^2+m+1\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2)}+\frac {\cos (e+f x) (B d-(m+2) (A d+B c)) (a \sin (e+f x)+a)^m}{f (m+1) (m+2)}-\frac {B d \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2)} \]

[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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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*S
in[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 3047

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 3102

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]

Rubi steps \begin{align*} \text {integral}& = \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) \operatorname {Hypergeometric2F1}\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 [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 22.75 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.59 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {(a (1+\sin (e+f x)))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {2 (2 A c+B d) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {2 i (B c+A d) \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {2 i (B c+A d) \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {B d \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}+\frac {B d \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}\right )}{4 f} \]

[In]

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

[Out]

-1/4*((a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))*((-2*(2*A*c + B*d)*Hypergeometric2F1[1, 1
+ m, 1 - m, I*Cos[e + f*x] - Sin[e + f*x]])/m - ((2*I)*(B*c + A*d)*Hypergeometric2F1[1, m, -m, I*Cos[e + f*x]
- Sin[e + f*x]]*(Cos[e + f*x] - I*Sin[e + f*x]))/(1 + m) + ((2*I)*(B*c + A*d)*Hypergeometric2F1[1, 2 + m, 2 -
m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[e + f*x] + I*Sin[e + f*x]))/(-1 + m) + (B*d*Hypergeometric2F1[1, -1 +
m, -1 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]))/(2 + m) + (B*d*Hypergeometr
ic2F1[1, 3 + m, 3 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]))/(-2 + m)))/f

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )d x\]

[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)

Fricas [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[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)

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )\, dx \]

[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)

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[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)

Giac [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]

[In]

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

[Out]

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