3.2.76 \(\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\) [176]

3.2.76.1 Optimal result
3.2.76.2 Mathematica [C] (verified)
3.2.76.3 Rubi [A] (verified)
3.2.76.4 Maple [B] (verified)
3.2.76.5 Fricas [F]
3.2.76.6 Sympy [F]
3.2.76.7 Maxima [A] (verification not implemented)
3.2.76.8 Giac [A] (verification not implemented)
3.2.76.9 Mupad [F(-1)]

3.2.76.1 Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {(A-B) c \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}} \]

output
(A-B)*c*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)-B*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(1/2)
 
3.2.76.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.14 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((A-B) \left (-i f x+2 \log \left (i+e^{i (e+f x)}\right )\right )+B \sin (e+f x)\right ) \sqrt {c-c \sin (e+f x)}}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))}} \]

input
Integrate[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/Sqrt[a + a*Sin[e 
 + f*x]],x]
 
output
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((A - B)*((-I)*f*x + 2*Log[I + E^(I 
*(e + f*x))]) + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(f*(Cos[(e + f*x 
)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])])
 
3.2.76.3 Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3217

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

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

3.2.76.3.1 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]
 
3.2.76.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(305\) vs. \(2(88)=176\).

Time = 3.39 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.19

method result size
parts \(-\frac {A \left (2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right )}{f \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}+\frac {B \left (2 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \cos \left (f x +e \right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-\cos \left (f x +e \right ) \sin \left (f x +e \right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{f \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) \(306\)
default \(\frac {\left (A \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 A \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+A \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 A \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-B \cos \left (f x +e \right ) \sin \left (f x +e \right )-B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 B \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+B \left (\cos ^{2}\left (f x +e \right )\right )-B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 B \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 A \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-B \sin \left (f x +e \right )-B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 B \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-B \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{f \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) \(347\)

input
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x,metho 
d=_RETURNVERBOSE)
 
output
-A/f*(2*ln(-cot(f*x+e)+csc(f*x+e)+1)-ln(2/(1+cos(f*x+e))))*(-c*(sin(f*x+e) 
-1))^(1/2)*(cos(f*x+e)+sin(f*x+e)+1)/(-cos(f*x+e)+sin(f*x+e)-1)/(a*(1+sin( 
f*x+e)))^(1/2)+B/f*(2*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)+2*ln(-cot(f* 
x+e)+csc(f*x+e)+1)*sin(f*x+e)-ln(2/(1+cos(f*x+e)))*cos(f*x+e)-ln(2/(1+cos( 
f*x+e)))*sin(f*x+e)+cos(f*x+e)^2-cos(f*x+e)*sin(f*x+e)+2*ln(-cot(f*x+e)+cs 
c(f*x+e)+1)-ln(2/(1+cos(f*x+e)))-sin(f*x+e)-1)*(-c*(sin(f*x+e)-1))^(1/2)/( 
-cos(f*x+e)+sin(f*x+e)-1)/(a*(1+sin(f*x+e)))^(1/2)
 
3.2.76.5 Fricas [F]

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

input
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x 
, algorithm="fricas")
 
output
integral((B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)/sqrt(a*sin(f*x + e 
) + a), x)
 
3.2.76.6 Sympy [F]

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

input
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**(1/2) 
,x)
 
output
Integral(sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin(e + f*x))/sqrt(a*(sin(e + 
f*x) + 1)), x)
 
3.2.76.7 Maxima [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.83 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {B {\left (\frac {2 \, \sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {a}} - \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {a}} - \frac {2 \, \sqrt {a} \sqrt {c} \sin \left (f x + e\right )}{{\left (a + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - A {\left (\frac {2 \, \sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {a}} - \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {a}}\right )}}{f} \]

input
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x 
, algorithm="maxima")
 
output
(B*(2*sqrt(c)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/sqrt(a) - sqrt(c)*l 
og(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/sqrt(a) - 2*sqrt(a)*sqrt(c)*si 
n(f*x + e)/((a + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1) 
)) - A*(2*sqrt(c)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/sqrt(a) - sqrt( 
c)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/sqrt(a)))/f
 
3.2.76.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} B \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (A \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {a} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {c}}{2 \, f} \]

input
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x 
, algorithm="giac")
 
output
1/2*sqrt(2)*(2*sqrt(2)*B*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 
 1/2*f*x + 1/2*e)^2/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2 
)*(A*sqrt(a)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*sqrt(a)*sgn(sin(-1/4* 
pi + 1/2*f*x + 1/2*e)))*log(-4*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 4)/(a*sg 
n(cos(-1/4*pi + 1/2*f*x + 1/2*e))))*sqrt(c)/f
 
3.2.76.9 Mupad [F(-1)]

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

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