3.206 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=166 \[ -\frac{2 B c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+2}}{a^2 f (2 m+5) \sqrt{c-c \sin (e+f x)}}+\frac{4 c^2 (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt{c-c \sin (e+f x)}}-\frac{2 c^2 (A-3 B) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.354095, antiderivative size = 192, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2973, 2740, 2738} \[ -\frac{8 c^2 (B (3-2 m)-A (2 m+5)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt{c-c \sin (e+f x)}}-\frac{2 c (B (3-2 m)-A (2 m+5)) \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f (2 m+3) (2 m+5)}-\frac{2 B \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^2*(B*(3 - 2*m) - A*(5 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c
- c*Sin[e + f*x]]) - (2*c*(B*(3 - 2*m) - A*(5 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e +
f*x]])/(f*(3 + 2*m)*(5 + 2*m)) - (2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(f*(5 +
2*m))

Rule 2973

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

Rule 2740

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

Rule 2738

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx &=-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}+\frac{\left (B c \left (-\frac{3}{2}+m\right )+A c \left (\frac{5}{2}+m\right )\right ) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx}{c \left (\frac{5}{2}+m\right )}\\ &=-\frac{2 c (B (3-2 m)-A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)}}{f (3+2 m) (5+2 m)}-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}-\frac{(4 c (B (3-2 m)-A (5+2 m))) \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx}{(3+2 m) (5+2 m)}\\ &=-\frac{8 c^2 (B (3-2 m)-A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (3+2 m) (5+2 m) \sqrt{c-c \sin (e+f x)}}-\frac{2 c (B (3-2 m)-A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)}}{f (3+2 m) (5+2 m)}-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}\\ \end{align*}

Mathematica [A]  time = 1.69844, size = 174, normalized size = 1.05 \[ \frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m \left (-2 (2 m+1) (2 A m+5 A-2 B m-9 B) \sin (e+f x)+8 A m^2+40 A m+50 A+B \left (4 m^2+8 m+3\right ) \cos (2 (e+f x))-4 B m^2-16 B m-39 B\right )}{f (2 m+1) (2 m+3) (2 m+5) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]]*(50*A - 39*B + 40*A
*m - 16*B*m + 8*A*m^2 - 4*B*m^2 + B*(3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] - 2*(1 + 2*m)*(5*A - 9*B + 2*A*m - 2*B*
m)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

________________________________________________________________________________________

Maple [F]  time = 0.325, 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-c\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, 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-c*sin(f*x+e))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [B]  time = 1.69904, size = 672, normalized size = 4.05 \begin{align*} \text{result too large to display} \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-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-2*((a^m*c^(3/2)*(2*m + 5) - a^m*c^(3/2)*(2*m - 3)*sin(f*x + e)/(cos(f*x + e) + 1) - a^m*c^(3/2)*(2*m - 3)*sin
(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^m*c^(3/2)*(2*m + 5)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*A*e^(2*m*log(sin
(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((4*m^2 + 8*m + 3)*(sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)) - 2*(a^m*c^(3/2)*(2*m + 9) - 2*(2*m^2 + 9*m)*a^m*c^(3/2)*sin(f*x + e
)/(cos(f*x + e) + 1) + (4*m^2 + 15)*a^m*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (4*m^2 + 15)*a^m*c^(3/2)
*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 2*(2*m^2 + 9*m)*a^m*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^m*c
^(3/2)*(2*m + 9)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*B*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*lo
g(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 36*m^2 + 46*m + (8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 15)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)))/f

________________________________________________________________________________________

Fricas [A]  time = 2.11756, size = 772, normalized size = 4.65 \begin{align*} \frac{2 \,{\left ({\left (4 \, B c m^{2} + 8 \, B c m + 3 \, B c\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (A + B\right )} c m +{\left (4 \, A c m^{2} + 12 \,{\left (A - B\right )} c m +{\left (5 \, A - 6 \, B\right )} c\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left (5 \, A - 3 \, B\right )} c +{\left (4 \,{\left (A - B\right )} c m^{2} + 4 \,{\left (5 \, A - 3 \, B\right )} c m +{\left (25 \, A - 21 \, B\right )} c\right )} \cos \left (f x + e\right ) +{\left (8 \,{\left (A + B\right )} c m +{\left (4 \, B c m^{2} + 8 \, B c m + 3 \, B c\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left (5 \, A - 3 \, B\right )} c -{\left (4 \,{\left (A - B\right )} c m^{2} + 4 \,{\left (3 \, A - 5 \, B\right )} c m +{\left (5 \, A - 9 \, B\right )} c\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m +{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) -{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \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-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2*((4*B*c*m^2 + 8*B*c*m + 3*B*c)*cos(f*x + e)^3 + 8*(A + B)*c*m + (4*A*c*m^2 + 12*(A - B)*c*m + (5*A - 6*B)*c)
*cos(f*x + e)^2 + 4*(5*A - 3*B)*c + (4*(A - B)*c*m^2 + 4*(5*A - 3*B)*c*m + (25*A - 21*B)*c)*cos(f*x + e) + (8*
(A + B)*c*m + (4*B*c*m^2 + 8*B*c*m + 3*B*c)*cos(f*x + e)^2 + 4*(5*A - 3*B)*c - (4*(A - B)*c*m^2 + 4*(3*A - 5*B
)*c*m + (5*A - 9*B)*c)*cos(f*x + e))*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*m^3 +
36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*
sin(f*x + e) + 15*f)

________________________________________________________________________________________

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))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

sage2