\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [85]

Optimal result

Integrand size = 36, antiderivative size = 122 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \sqrt {2} a (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f} \] Output:

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

Mathematica [A] (verified)

Time = 4.68 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.36 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 \sqrt {2} (A+B) \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \sqrt {-c (1+\sin (e+f x))}+\sqrt {c} (6 A+9 B-B \cos (2 (e+f x))+2 (3 A+5 B) \sin (e+f x))\right )}{3 \sqrt {c} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

-1/3*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*Sqrt[2]*(A + B)*ArcTan[Sq 
rt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sqrt[c])]*Sqrt[-(c*(1 + Sin[e + f*x]) 
)] + Sqrt[c]*(6*A + 9*B - B*Cos[2*(e + f*x)] + 2*(3*A + 5*B)*Sin[e + f*x]) 
))/(Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x 
]])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3446, 3042, 3337, 27, 3042, 3230, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3337

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

\(\displaystyle a c \left (\frac {-\frac {12 c (A+B) \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}-\frac {2 c (3 A+5 B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}}{3 c^2}+\frac {2 B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c^2 f}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle a c \left (\frac {\frac {6 \sqrt {2} \sqrt {c} (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{f}-\frac {2 c (3 A+5 B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}}{3 c^2}+\frac {2 B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c^2 f}\right )\)

Input:

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

Output:

a*c*((2*B*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*c^2*f) + ((6*Sqrt[2]*( 
A + B)*Sqrt[c]*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + 
f*x]])])/f - (2*(3*A + 5*B)*c*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]]))/ 
(3*c^2))
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[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 3337

Int[cos[(e_.) + (f_.)*(x_)]^2*((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 + 2)/(b^2*f*(m + 3))), x] - Simp[1/(b^2*(m + 3))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(b*d*(m + 2) - a*c*(m + 3) + (b*c*(m + 3) - 
 a*d*(m + 4))*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[ 
a^2 - b^2, 0] && GeQ[m, -3/2] && LtQ[m, 0]
 

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[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] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a \left (-3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) A -3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) B +B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}+3 A c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}+3 B c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(158\)
parts	\(-\frac {A a \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {B a \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (-3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(268\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (105) = 210\).

Time = 0.09 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.08 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left ({\left (A + B\right )} a c \cos \left (f x + e\right ) - {\left (A + B\right )} a c \sin \left (f x + e\right ) + {\left (A + B\right )} a c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + 2 \, {\left (B a \cos \left (f x + e\right )^{2} - {\left (3 \, A + 4 \, B\right )} a \cos \left (f x + e\right ) - {\left (3 \, A + 5 \, B\right )} a - {\left (B a \cos \left (f x + e\right ) + {\left (3 \, A + 5 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algo 
rithm="fricas")
 

Output:

1/3*(3*sqrt(2)*((A + B)*a*c*cos(f*x + e) - (A + B)*a*c*sin(f*x + e) + (A + 
 B)*a*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2 
)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3* 
cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos( 
f*x + e) - 2))/sqrt(c) + 2*(B*a*cos(f*x + e)^2 - (3*A + 4*B)*a*cos(f*x + e 
) - (3*A + 5*B)*a - (B*a*cos(f*x + e) + (3*A + 5*B)*a)*sin(f*x + e))*sqrt( 
-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algo 
rithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(c)*a*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) - 1),x)*a + int 
((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) - 1),x)*b + int 
((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) - 1),x)*a + int((s 
qrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) - 1),x)*b))/c