\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [135]

Optimal result

Integrand size = 40, antiderivative size = 100 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.47 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left ((A+B) \left (-i f x+2 \log \left (i-e^{i (e+f x)}\right )\right )+B \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

-(((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*((A + 
B)*((-I)*f*x + 2*Log[I - E^(I*(e + f*x))]) + B*Sin[e + f*x]))/(f*(Cos[(e + 
 f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3450, 3042, 3216, 3042, 3146, 16, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

\(\displaystyle -\frac {a (A+B) \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {B \int \sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}dx}{c}\)

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 3042

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

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3216

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

rule 3217

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)]
 

rule 3450

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(314\) vs. \(2(92)=184\).

Time = 4.67 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.15

method	result	size
default	\(-\frac {A \sqrt {4}\, \left (\ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )-\ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}+\frac {2 B \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\ln \left (\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-\ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}}{f \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}}\)	\(315\)
parts	\(-\frac {A \sqrt {4}\, \left (\ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )-\ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}+\frac {2 B \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\ln \left (\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-\ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}}{f \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}}\)	\(315\)

Input:

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

Output:

-A/f*4^(1/2)*(ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)+l 
n(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)-1)-ln(2/(cos(1/4*Pi 
+1/2*f*x+1/2*e)+1)))*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*cos(1/4*Pi+1 
/2*f*x+1/2*e)^2)^(1/2)*cot(1/4*Pi+1/2*f*x+1/2*e)+2*B/f*(sin(1/2*f*x+1/2*e) 
-cos(1/2*f*x+1/2*e))*(sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+ln(2*(sin(1/2* 
f*x+1/2*e)-cos(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1))-ln(2/(cos(1/2*f*x+1 
/2*e)+1)))*((2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+1)*a)^(1/2)/(cos(1/2* 
f*x+1/2*e)+sin(1/2*f*x+1/2*e))/(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)- 
1)*c)^(1/2)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {B {\left (\frac {2 \, \sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {c}} - \frac {\sqrt {a} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {c}} + \frac {2 \, \sqrt {a} \sqrt {c} \sin \left (f x + e\right )}{{\left (c + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + A {\left (\frac {2 \, \sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {c}} - \frac {\sqrt {a} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {c}}\right )}}{f} \] Input:

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

Output:

(B*(2*sqrt(a)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/sqrt(c) - sqrt(a)*l 
og(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/sqrt(c) + 2*sqrt(a)*sqrt(c)*si 
n(f*x + e)/((c + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1) 
)) + A*(2*sqrt(a)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/sqrt(c) - sqrt( 
a)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/sqrt(c)))/f
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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