\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\) [197]

Optimal result

Integrand size = 36, antiderivative size = 145 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {2^{\frac {1}{2}+m} a^3 c^2 (B (2-m)-A (3+m)) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-3+m}}{5 f (3+m)}-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)} \]

[Out]

1/5*2^(1/2+m)*a^3*c^2*(B*(2-m)-A*(3+m))*cos(f*x+e)^5*hypergeom([5/2, 1/2-m],[7/2],1/2-1/2*sin(f*x+e))*(1+sin(f
*x+e))^(1/2-m)*(a+a*sin(f*x+e))^(-3+m)/f/(3+m)-a^2*B*c^2*cos(f*x+e)^5*(a+a*sin(f*x+e))^(-2+m)/f/(3+m)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2939, 2768, 72, 71} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {a^3 c^2 2^{m+\frac {1}{2}} (B (2-m)-A (m+3)) \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f (m+3)}-\frac {a^2 B c^2 \cos ^5(e+f x) (a \sin (e+f x)+a)^{m-2}}{f (m+3)} \]

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^2,x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 2939

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

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} (A+B \sin (e+f x)) \, dx \\ & = -\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\left (a^2 c^2 \left (A-\frac {B (2-m)}{3+m}\right )\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} \, dx \\ & = -\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac {\left (a^4 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x)\right ) \text {Subst}\left (\int (a-a x)^{3/2} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}} \\ & = -\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac {\left (2^{-\frac {1}{2}+m} a^4 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}} \\ & = -\frac {2^{\frac {1}{2}+m} a^3 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{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))^{-3+m}}{5 f}-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 13.09 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.06 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {i c^2 (a (1+\sin (e+f x)))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {4 i (3 A-2 B) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {(8 A-7 B) \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {(8 A-7 B) \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {2 i (A-2 B) \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}+\frac {2 (A-2 B) \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (2 (e+f x))+\sin (2 (e+f x)))}{2+m}-\frac {B \operatorname {Hypergeometric2F1}(1,-2+m,-2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))-i \sin (3 (e+f x)))}{3+m}+\frac {B \operatorname {Hypergeometric2F1}(1,4+m,4-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))}{-3+m}\right )}{8 f} \]

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^2,x]

[Out]

((I/8)*c^2*(a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))*(((-4*I)*(3*A - 2*B)*Hypergeometric2
F1[1, 1 + m, 1 - m, I*Cos[e + f*x] - Sin[e + f*x]])/m - ((8*A - 7*B)*Hypergeometric2F1[1, m, -m, I*Cos[e + f*x
] - Sin[e + f*x]]*(Cos[e + f*x] - I*Sin[e + f*x]))/(1 + m) + ((8*A - 7*B)*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) + ((2*I)*(A - 2*B)*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) + (2*(A - 2*B
)*Hypergeometric2F1[1, -1 + m, -1 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]))
/(2 + m) - (B*Hypergeometric2F1[1, -2 + m, -2 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[3*(e + f*x)] - I*Sin[3*
(e + f*x)]))/(3 + m) + (B*Hypergeometric2F1[1, 4 + m, 4 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[3*(e + f*x)]
+ I*Sin[3*(e + f*x)]))/(-3 + m)))/f

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=c^{2} \left (\int A \left (a \sin {\left (e + f x \right )} + a\right )^{m}\, dx + \int \left (- 2 A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\right )\, dx + \int A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- 2 B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \]

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**2,x)

[Out]

c**2*(Integral(A*(a*sin(e + f*x) + a)**m, x) + Integral(-2*A*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integr
al(A*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2, x) + Integral(B*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integ
ral(-2*B*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2, x) + Integral(B*(a*sin(e + f*x) + a)**m*sin(e + f*x)**3, x))

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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