3.84 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=73 \[ \frac {2 a c^2 (5 A+B) \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

2/15*a*(5*A+B)*c^2*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-2/5*a*B*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2967, 2856, 2673} \[ \frac {2 a c^2 (5 A+B) \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(5*A + B)*c^2*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^(3/2)) - (2*a*B*c*Cos[e + f*x]^3)/(5*f*Sqrt[c -
c*Sin[e + f*x]])

Rule 2673

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

Rule 2856

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] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2967

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*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx &=(a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{5} (a (5 A+B) c) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=\frac {2 a (5 A+B) c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.42, size = 87, normalized size = 1.19 \[ \frac {2 a \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 (5 A+3 B \sin (e+f x)-2 B)}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(5*A - 2*B + 3*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*f*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2]))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 130, normalized size = 1.78 \[ -\frac {2 \, {\left (3 \, B a \cos \left (f x + e\right )^{3} + {\left (5 \, A + 4 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (5 \, A + B\right )} a \cos \left (f x + e\right ) - 2 \, {\left (5 \, A + B\right )} a + {\left (3 \, B a \cos \left (f x + e\right )^{2} - {\left (5 \, A + B\right )} a \cos \left (f x + e\right ) - 2 \, {\left (5 \, A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

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-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*B*a*cos(f*x + e)^3 + (5*A + 4*B)*a*cos(f*x + e)^2 - (5*A + B)*a*cos(f*x + e) - 2*(5*A + B)*a + (3*B*a
*cos(f*x + e)^2 - (5*A + B)*a*cos(f*x + e) - 2*(5*A + B)*a)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x
+ e) - f*sin(f*x + e) + f)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*c)*(2*f*(2*
A*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(2*f*x+2*exp(1)+pi))/
(2*f)^2-2*f*(4*A*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi))+2*B*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x
-pi)+1/2*exp(1))/(2*f)^2+6*f*(-2*A*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-2*B*a*sign(sin(1/2*(f*x+exp(1))-1/4*pi
)))*sin(1/4*(6*f*x+6*exp(1)-pi))/(6*f)^2-24*B*a*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(6*f*x+6*exp(1)+p
i))/(12*f)^2+40*B*a*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(10*f*x+10*exp(1)-pi))/(20*f)^2)

________________________________________________________________________________________

maple [A]  time = 1.31, size = 63, normalized size = 0.86 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) c \left (1+\sin \left (f x +e \right )\right )^{2} a \left (3 B \sin \left (f x +e \right )+5 A -2 B \right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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-c*sin(f*x+e))^(1/2),x)

[Out]

-2/15*(sin(f*x+e)-1)*c*(1+sin(f*x+e))^2*a*(3*B*sin(f*x+e)+5*A-2*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\,{d x} \]

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-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2),x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int A \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \]

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-c*sin(f*x+e))**(1/2),x)

[Out]

a*(Integral(A*sqrt(-c*sin(e + f*x) + c), x) + Integral(A*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral
(B*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral(B*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x))

________________________________________________________________________________________