\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\) [287]

Optimal result

Integrand size = 37, antiderivative size = 192 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {2 a (B c-7 A d-6 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{105 d f \sqrt {a+a \sin (e+f x)}}+\frac {4 (5 c-d) (B c-7 A d-6 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}+\frac {2 d (B c-7 A d-6 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a+a \sin (e+f x)}} \] Output:

2/105*a*(-7*A*d+B*c-6*B*d)*(15*c^2+10*c*d+7*d^2)*cos(f*x+e)/d/f/(a+a*sin(f 
*x+e))^(1/2)+4/105*(5*c-d)*(-7*A*d+B*c-6*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^ 
(1/2)/f+2/35*d*(-7*A*d+B*c-6*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f-2/ 
7*a*B*cos(f*x+e)*(c+d*sin(f*x+e))^3/d/f/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.92 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (420 A c^2+280 B c^2+560 A c d+532 B c d+266 A d^2+228 B d^2-6 d (14 B c+7 A d+6 B d) \cos (2 (e+f x))+\left (56 A d (5 c+2 d)+B \left (140 c^2+224 c d+141 d^2\right )\right ) \sin (e+f x)-15 B d^2 \sin (3 (e+f x))\right )}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

-1/210*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*( 
420*A*c^2 + 280*B*c^2 + 560*A*c*d + 532*B*c*d + 266*A*d^2 + 228*B*d^2 - 6* 
d*(14*B*c + 7*A*d + 6*B*d)*Cos[2*(e + f*x)] + (56*A*d*(5*c + 2*d) + B*(140 
*c^2 + 224*c*d + 141*d^2))*Sin[e + f*x] - 15*B*d^2*Sin[3*(e + f*x)]))/(f*( 
Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {3042, 3460, 3042, 3240, 27, 3042, 3230, 3042, 3125}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2dx\)

\(\Big \downarrow \) 3460

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2dx}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2dx}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3240

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {2 \int \frac {1}{2} \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {\int \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {\int \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {\frac {1}{3} a \left (15 c^2+10 c d+7 d^2\right ) \int \sqrt {\sin (e+f x) a+a}dx-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {\frac {1}{3} a \left (15 c^2+10 c d+7 d^2\right ) \int \sqrt {\sin (e+f x) a+a}dx-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3125

\(\displaystyle -\frac {(-7 A d+B c-6 B d) \left (\frac {-\frac {2 a^2 \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )}{7 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^3}{7 d f \sqrt {a \sin (e+f x)+a}}\)

Input:

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

Output:

(-2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(7*d*f*Sqrt[a + a*Sin[e + f*x 
]]) - ((B*c - 7*A*d - 6*B*d)*((-2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3 
/2))/(5*a*f) + ((-2*a^2*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(3*f*Sqrt[ 
a + a*Sin[e + f*x]]) - (4*a*(5*c - d)*d*Cos[e + f*x]*Sqrt[a + a*Sin[e + f* 
x]])/(3*f))/(5*a)))/(7*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

rule 3230

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] + Simp[(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 3240

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^2, x_Symbol] :> Simp[(-d^2)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ 
(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^ 
m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x] 
, x], 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, -1]
 

rule 3460

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + 
b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b 
*d*(2*n + 3))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] 
 /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - 
 b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]
 

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (-15 B \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) d^{2}+\left (-21 A \,d^{2}-42 B c d -18 B \,d^{2}\right ) \cos \left (f x +e \right )^{2}+\left (70 A c d +28 A \,d^{2}+35 B \,c^{2}+56 B c d +39 B \,d^{2}\right ) \sin \left (f x +e \right )+105 A \,c^{2}+140 A c d +77 A \,d^{2}+70 B \,c^{2}+154 B c d +66 B \,d^{2}\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(161\)
parts	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) d \left (A d +2 B c \right ) \left (3 \sin \left (f x +e \right )^{2}+4 \sin \left (f x +e \right )+8\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) c \left (2 A d +B c \right ) \left (\sin \left (f x +e \right )+2\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 A \,c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) a}{\cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B \,d^{2} \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (5 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}+8 \sin \left (f x +e \right )+16\right )}{35 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(254\)

Input:

int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x,method=_R 
ETURNVERBOSE)
 

Output:

2/105*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*(-15*B*cos(f*x+e)^2*sin(f*x+e)*d^2+( 
-21*A*d^2-42*B*c*d-18*B*d^2)*cos(f*x+e)^2+(70*A*c*d+28*A*d^2+35*B*c^2+56*B 
*c*d+39*B*d^2)*sin(f*x+e)+105*A*c^2+140*A*c*d+77*A*d^2+70*B*c^2+154*B*c*d+ 
66*B*d^2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.59 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {2 \, {\left (15 \, B d^{2} \cos \left (f x + e\right )^{4} + 3 \, {\left (14 \, B c d + {\left (7 \, A + 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 35 \, {\left (3 \, A + B\right )} c^{2} - 14 \, {\left (5 \, A + 7 \, B\right )} c d - {\left (49 \, A + 27 \, B\right )} d^{2} - {\left (35 \, B c^{2} + 14 \, {\left (5 \, A + B\right )} c d + {\left (7 \, A + 36 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, {\left (3 \, A + 2 \, B\right )} c^{2} + 14 \, {\left (10 \, A + 11 \, B\right )} c d + 11 \, {\left (7 \, A + 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) + {\left (15 \, B d^{2} \cos \left (f x + e\right )^{3} + 35 \, {\left (3 \, A + B\right )} c^{2} + 14 \, {\left (5 \, A + 7 \, B\right )} c d + {\left (49 \, A + 27 \, B\right )} d^{2} - 3 \, {\left (14 \, B c d + {\left (7 \, A + B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (35 \, B c^{2} + 14 \, {\left (5 \, A + 4 \, B\right )} c d + {\left (28 \, A + 39 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, al 
gorithm="fricas")
 

Output:

2/105*(15*B*d^2*cos(f*x + e)^4 + 3*(14*B*c*d + (7*A + 6*B)*d^2)*cos(f*x + 
e)^3 - 35*(3*A + B)*c^2 - 14*(5*A + 7*B)*c*d - (49*A + 27*B)*d^2 - (35*B*c 
^2 + 14*(5*A + B)*c*d + (7*A + 36*B)*d^2)*cos(f*x + e)^2 - (35*(3*A + 2*B) 
*c^2 + 14*(10*A + 11*B)*c*d + 11*(7*A + 6*B)*d^2)*cos(f*x + e) + (15*B*d^2 
*cos(f*x + e)^3 + 35*(3*A + B)*c^2 + 14*(5*A + 7*B)*c*d + (49*A + 27*B)*d^ 
2 - 3*(14*B*c*d + (7*A + B)*d^2)*cos(f*x + e)^2 - (35*B*c^2 + 14*(5*A + 4* 
B)*c*d + (28*A + 39*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e 
) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)
 

Sympy [F]

\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \] Input:

integrate((a+a*sin(f*x+e))**(1/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2,x)
 

Output:

Integral(sqrt(a*(sin(e + f*x) + 1))*(A + B*sin(e + f*x))*(c + d*sin(e + f* 
x))**2, x)
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, al 
gorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.81 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, al 
gorithm="giac")
 

Output:

1/420*sqrt(2)*(15*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 
7/2*f*x + 7/2*e) + 105*(8*A*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B* 
c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 8*A*c*d*sgn(cos(-1/4*pi + 1/2*f* 
x + 1/2*e)) + 8*B*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*A*d^2*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 35*(4*B*c^2*sgn(cos(-1/4*pi + 1/2*f*x 
+ 1/2*e)) + 8*A*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B*c*d*sgn(cos( 
-1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) 
+ 3*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2 
*e) + 21*(4*B*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*d^2*sgn(cos(-1 
/4*pi + 1/2*f*x + 1/2*e)) + B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin 
(-5/4*pi + 5/2*f*x + 5/2*e))*sqrt(a)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}d x \right ) a \,c^{2}+\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b \,d^{2}+\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a \,d^{2}+2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b c d +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a c d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b \,c^{2}\right ) \] Input:

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

Output:

sqrt(a)*(int(sqrt(sin(e + f*x) + 1),x)*a*c**2 + int(sqrt(sin(e + f*x) + 1) 
*sin(e + f*x)**3,x)*b*d**2 + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x) 
*a*d**2 + 2*int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*b*c*d + 2*int(sq 
rt(sin(e + f*x) + 1)*sin(e + f*x),x)*a*c*d + int(sqrt(sin(e + f*x) + 1)*si 
n(e + f*x),x)*b*c**2)