\(\int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx\) [598]

Optimal result

Integrand size = 29, antiderivative size = 141 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} \sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f} \] Output:

-2*d^(1/2)*arctan(a^(1/2)*d^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c+d*s 
in(f*x+e))^(1/2))/a^(1/2)/f-2^(1/2)*(c-d)^(1/2)*arctanh(1/2*a^(1/2)*(c-d)^ 
(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/a^ 
(1/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 14.38 (sec) , antiderivative size = 1251, normalized size of antiderivative = 8.87 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx =\text {Too large to display} \] Input:

Integrate[Sqrt[c + d*Sin[e + f*x]]/Sqrt[a + a*Sin[e + f*x]],x]
 

Output:

((Sqrt[2]*Sqrt[c - d]*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*Sqrt[c - d]*Log[ 
c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x 
]] + (-c + d)*Tan[(e + f*x)/2]] - I*Sqrt[d]*(Log[(2*(c - I*d + (1 - I)*Sqr 
t[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + ((-I 
)*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(I + Tan[(e + f*x)/2]))] - Log[(2*(c 
+ I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*S 
in[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(-I + Tan[(e + f*x)/2 
]))]))*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a*(1 + Sin[e + f*x])]*((Sqrt[c - 
d]*Sec[(e + f*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sqrt[2]*Sqrt[c 
- d]*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*d*Cos[e + f*x]*Sqrt[( 
1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqrt[c - d]*((1 + Cos[ 
e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]))/(c - d + 2*S 
qrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + 
d)*Tan[(e + f*x)/2]) - I*Sqrt[d]*((d^(3/2)*(I + Tan[(e + f*x)/2])*((2*(((( 
-I)*c + d)*Sec[(e + f*x)/2]^2)/2 + ((1 - I)*d^(3/2)*Cos[e + f*x]*Sqrt[(1 + 
 Cos[e + f*x])^(-1)])/(Sqrt[2]*Sqrt[c + d*Sin[e + f*x]]) + ((1 - I)*Sqrt[d 
]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/S 
qrt[2]))/(d^(3/2)*(I + Tan[(e + f*x)/2])) - (Sec[(e + f*x)/2]^2*(c - I*d + 
 (1 - I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + 
f*x]] + ((-I)*c + d)*Tan[(e + f*x)/2]))/(d^(3/2)*(I + Tan[(e + f*x)/2])...
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3256, 3042, 3254, 218, 3261, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 d \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle (c-d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}\)

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} \sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}\)

Input:

Int[Sqrt[c + d*Sin[e + f*x]]/Sqrt[a + a*Sin[e + f*x]],x]
 

Output:

(-2*Sqrt[d]*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x] 
]*Sqrt[c + d*Sin[e + f*x]])])/(Sqrt[a]*f) - (Sqrt[2]*Sqrt[c - d]*ArcTanh[( 
Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c 
 + d*Sin[e + f*x]])])/(Sqrt[a]*f)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221

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

rule 3042

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

rule 3254

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

rule 3256

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

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 
 EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(3453\) vs. \(2(114)=228\).

Time = 10.02 (sec) , antiderivative size = 3454, normalized size of antiderivative = 24.50

\[\text {output too large to display}\]

Input:

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

Output:

-1/f/(d^2/c^2)^(1/2)/(-(d^2/c^2)^(1/2)*c)^(1/2)/(2*c-2*d)^(1/2)/(c-d)^2/c* 
((2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+1)*c*(2*c-2*d)^(1/2)*arctan(((d^ 
2/c^2)^(1/2)*c^2-d^2)*c*((d^2/c^2)^(1/2)-1)/(((d^2/c^2)^(1/2)*c^4+6*(d^2/c 
^2)^(1/2)*d^2*c^2+d^4*(d^2/c^2)^(1/2)-4*d^2*c^2-4*d^4)*c)^(1/2)*((d^2/c^2) 
^(1/2)*c*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e)*d)/((d^2/c^2)^(1/2)*c*sin(1 
/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*d)/(d*(c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f 
*x+1/2*e))/(2*(d^2/c^2)^(1/2)*c*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+d))^ 
(1/2))*(((d^2/c^2)^(1/2)*c^4+6*(d^2/c^2)^(1/2)*d^2*c^2+d^4*(d^2/c^2)^(1/2) 
-4*d^2*c^2-4*d^4)*c)^(1/2)*(d*(c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e) 
)/(2*(d^2/c^2)^(1/2)*c*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+d))^(1/2)*(-( 
d^2/c^2)^(1/2)*c)^(1/2)*(d^2/c^2)^(1/2)+(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+ 
1/2*e)+1)*(2*c-2*d)^(1/2)*d*arctan(((d^2/c^2)^(1/2)*c^2-d^2)*c*((d^2/c^2)^ 
(1/2)-1)/(((d^2/c^2)^(1/2)*c^4+6*(d^2/c^2)^(1/2)*d^2*c^2+d^4*(d^2/c^2)^(1/ 
2)-4*d^2*c^2-4*d^4)*c)^(1/2)*((d^2/c^2)^(1/2)*c*sin(1/2*f*x+1/2*e)-cos(1/2 
*f*x+1/2*e)*d)/((d^2/c^2)^(1/2)*c*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*d) 
/(d*(c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))/(2*(d^2/c^2)^(1/2)*c*sin 
(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+d))^(1/2))*(((d^2/c^2)^(1/2)*c^4+6*(d^2 
/c^2)^(1/2)*d^2*c^2+d^4*(d^2/c^2)^(1/2)-4*d^2*c^2-4*d^4)*c)^(1/2)*(d*(c+2* 
d*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e))/(2*(d^2/c^2)^(1/2)*c*sin(1/2*f*x+ 
1/2*e)*cos(1/2*f*x+1/2*e)+d))^(1/2)*(-(d^2/c^2)^(1/2)*c)^(1/2)+(-2*sin(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (114) = 228\).

Time = 0.39 (sec) , antiderivative size = 2040, normalized size of antiderivative = 14.47 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fric 
as")
 

Output:

[1/4*(2*sqrt(2)*sqrt((c - d)/a)*log((2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sq 
rt(d*sin(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - 
 (c - 3*d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + 
e) - 2*c - 2*d)*sin(f*x + e) - 2*c - 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) 
+ 2)*sin(f*x + e) - cos(f*x + e) - 2)) + sqrt(-d/a)*log((128*d^4*cos(f*x + 
 e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4 
*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3 
*d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 - 8*(16*d^3*cos(f*x + e)^ 
4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 
 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^ 
2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c* 
d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 
+ 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin( 
f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d^3 + 289 
*d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 
 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 - 6*c*d^ 
3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - 9*d^4)*cos( 
f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)))/f, 1/2*(sqrt(2 
)*sqrt((c - d)/a)*log((2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + 
 e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - (c - 3*d)*...
 

Sympy [F]

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

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

Output:

Integral(sqrt(c + d*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)
 

Maxima [F]

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

integrate((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

integrate(sqrt(d*sin(f*x + e) + c)/sqrt(a*sin(f*x + e) + a), x)
                                                                                    
                                                                                    
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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