\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx\) [599]

Optimal result

Integrand size = 29, antiderivative size = 79 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} \sqrt {c-d} f} \] Output:

-2^(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)/(c-d)^(1/2)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(283\) vs. \(2(79)=158\).

Time = 2.20 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.58 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \sqrt {a (1+\sin (e+f x))} \sqrt {c+d \sin (e+f x)} \left (\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}\right )} \] Input:

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]),x]
 

Output:

(Log[1 + Tan[(e + f*x)/2]] - 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]])/(f*Sqrt[ 
a*(1 + Sin[e + f*x])]*Sqrt[c + d*Sin[e + f*x]]*(Sec[(e + f*x)/2]^2/(2 + 2* 
Tan[(e + f*x)/2]) - (-1/2*((c - d)*Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 
+ Cos[e + f*x])^(-1))^(3/2)*(d + d*Cos[e + f*x] + c*Sin[e + f*x]))/Sqrt[c 
+ d*Sin[e + f*x]])/(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sq 
rt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2])))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]),x]
 

Output:

-((Sqrt[2]*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]*Sqrt[c - d]*f))
 

Defintions of rubi rules used

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 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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(312\) vs. \(2(64)=128\).

Time = 6.04 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.96

method	result	size
default	\(\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sqrt {\frac {c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}\, \ln \left (-\frac {4 \left (\sqrt {\frac {c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}\, \sqrt {2 c -2 d}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {2 c -2 d}\, \sqrt {\frac {c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}-\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) c +c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) d -d \sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \sqrt {2 c -2 d}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \sqrt {c +2 d \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}}\)	\(313\)

Input:

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

Output:

2/f/(2*c-2*d)^(1/2)*(cos(1/2*f*x+1/2*e)+1)*(cos(1/2*f*x+1/2*e)+sin(1/2*f*x 
+1/2*e))*((c+2*d*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)^2)^(1/2)*ln(-4*(((c+2*d*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)^2)^(1/2)*(2*c-2*d)^(1/2)*cos(1/2*f*x+1/2*e)+(2*c-2*d)^(1/ 
2)*((c+2*d*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)^2 
)^(1/2)-cos(1/2*f*x+1/2*e)*c+c*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*d-d*s 
in(1/2*f*x+1/2*e))/(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)*a)^(1/2)/(c+2*d*sin(1/2*f*x+1/2*e)*cos(1/2* 
f*x+1/2*e))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 475, normalized size of antiderivative = 6.01 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx=\left [\frac {\sqrt {2} \log \left (\frac {{\left (c^{2} - 14 \, c d + 17 \, d^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (13 \, c^{2} - 22 \, c d - 3 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - \frac {4 \, \sqrt {2} {\left ({\left (c^{2} - 4 \, c d + 3 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, c^{2} + 8 \, c d - 4 \, d^{2} - {\left (3 \, c^{2} - 4 \, c d + d^{2}\right )} \cos \left (f x + e\right ) + {\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} + {\left (c^{2} - 4 \, c d + 3 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{\sqrt {a c - a d}} - 4 \, c^{2} - 8 \, c d - 4 \, d^{2} - 2 \, {\left (9 \, c^{2} - 14 \, c d + 9 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (c^{2} - 14 \, c d + 17 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, c^{2} - 8 \, c d - 4 \, d^{2} + 2 \, {\left (7 \, c^{2} - 18 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) - 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 4}\right )}{4 \, \sqrt {a c - a d} f}, \frac {\sqrt {2} \sqrt {-\frac {1}{a c - a d}} \arctan \left (-\frac {\sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left ({\left (c - 3 \, d\right )} \sin \left (f x + e\right ) - 3 \, c + d\right )} \sqrt {d \sin \left (f x + e\right ) + c} \sqrt {-\frac {1}{a c - a d}}}{4 \, {\left (d \cos \left (f x + e\right ) \sin \left (f x + e\right ) + c \cos \left (f x + e\right )\right )}}\right )}{2 \, f}\right ] \] Input:

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

Output:

[1/4*sqrt(2)*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 - (13*c^2 - 22*c* 
d - 3*d^2)*cos(f*x + e)^2 - 4*sqrt(2)*((c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^ 
2 - 4*c^2 + 8*c*d - 4*d^2 - (3*c^2 - 4*c*d + d^2)*cos(f*x + e) + (4*c^2 - 
8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*s 
in(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/sqrt(a*c - a*d) - 4*c^2 - 8*c*d 
- 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^ 
2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 + 2*(7*c^2 - 18*c*d + 7*d^2)*cos 
(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e 
)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4))/(sqrt(a*c - 
a*d)*f), 1/2*sqrt(2)*sqrt(-1/(a*c - a*d))*arctan(-1/4*sqrt(2)*sqrt(a*sin(f 
*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)*s 
qrt(-1/(a*c - a*d))/(d*cos(f*x + e)*sin(f*x + e) + c*cos(f*x + e)))/f]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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